Function to cache its argument's return value












13














I want to write a function once that accepts a callback as input and returns a function. When the returned function is called the first time, it should call the callback and return that output. If it is called any additional times, instead of calling the callback again it will simply return the output value from the first time it was called.



Below is what I have tried to do. But I am not getting the results as I expected. I need to understand this concept.



function once(func) {
let num;
function retFunc(x){
num = func(x);
return num;
}
return retFunc;
}

function addByTwo(input){
return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6









share|improve this question




















  • 1




    The num is already a good first step, but how do you remember that it was called a first time already? You always overwrite num = func(x)
    – Bergi
    9 hours ago
















13














I want to write a function once that accepts a callback as input and returns a function. When the returned function is called the first time, it should call the callback and return that output. If it is called any additional times, instead of calling the callback again it will simply return the output value from the first time it was called.



Below is what I have tried to do. But I am not getting the results as I expected. I need to understand this concept.



function once(func) {
let num;
function retFunc(x){
num = func(x);
return num;
}
return retFunc;
}

function addByTwo(input){
return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6









share|improve this question




















  • 1




    The num is already a good first step, but how do you remember that it was called a first time already? You always overwrite num = func(x)
    – Bergi
    9 hours ago














13












13








13


1





I want to write a function once that accepts a callback as input and returns a function. When the returned function is called the first time, it should call the callback and return that output. If it is called any additional times, instead of calling the callback again it will simply return the output value from the first time it was called.



Below is what I have tried to do. But I am not getting the results as I expected. I need to understand this concept.



function once(func) {
let num;
function retFunc(x){
num = func(x);
return num;
}
return retFunc;
}

function addByTwo(input){
return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6









share|improve this question















I want to write a function once that accepts a callback as input and returns a function. When the returned function is called the first time, it should call the callback and return that output. If it is called any additional times, instead of calling the callback again it will simply return the output value from the first time it was called.



Below is what I have tried to do. But I am not getting the results as I expected. I need to understand this concept.



function once(func) {
let num;
function retFunc(x){
num = func(x);
return num;
}
return retFunc;
}

function addByTwo(input){
return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6






javascript






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









Solomon Ucko

625719




625719










asked 9 hours ago









Amit KumarAmit Kumar

1058




1058








  • 1




    The num is already a good first step, but how do you remember that it was called a first time already? You always overwrite num = func(x)
    – Bergi
    9 hours ago














  • 1




    The num is already a good first step, but how do you remember that it was called a first time already? You always overwrite num = func(x)
    – Bergi
    9 hours ago








1




1




The num is already a good first step, but how do you remember that it was called a first time already? You always overwrite num = func(x)
– Bergi
9 hours ago




The num is already a good first step, but how do you remember that it was called a first time already? You always overwrite num = func(x)
– Bergi
9 hours ago












5 Answers
5






active

oldest

votes


















12














You should only assign num = func(x) when num is undefined - that is, on the very first call of retFunc:






function once(func) {
let num;
function retFunc(x){
if (num === undefined) {
num = func(x);
}
return num;
}
return retFunc;
}

function addByTwo(input){
return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6





But this isn't a guaranteed general solution - what if the passed function (addByTwo in your example) results in undefined when called? Then, the === undefined check won't work. So, it might be better to set a flag or something similar, and reassign that flag the first time the callback is called:






function once(func) {
let num;
let done = false;
function retFunc(x){
if (!done) {
done = true;
num = func(x);
}
return num;
}
return retFunc;
}

function returnsUndefinedOn1(input){
return input === 1 ? undefined : input;
}

var onceFunc = once(returnsUndefinedOn1);

console.log(onceFunc(1));
console.log(onceFunc(10));
console.log(onceFunc(9001));








share|improve this answer

















  • 2




    That's the desired behavior - on an input of 1, the function returnsUndefinedOn1 returns undefined, per its name. Further calls of onceFunc will also return undefined, despite being called with an argument other than 1.
    – CertainPerformance
    9 hours ago






  • 2




    Please use typeof to compare with undefined
    – Pierre Arlaud
    8 hours ago






  • 1




    @PierreArlaud In any remotely reasonable code, it shouldn't be necessary, I think? Many global object can be redefined, that doesn't mean they shouldn't be used, that means not to reassign them.
    – CertainPerformance
    8 hours ago






  • 1




    @CertainPerformance typeof also does not trigger an exception if the variable is not declared. There is no scenario where not using typeof to compare with undefined is the better choice. My opinion on the matter is that using typeof is a good habbit to have, and "remotely reasonable codes" are no exceptions.
    – Pierre Arlaud
    8 hours ago






  • 2




    @PierreArlaud: In this case, getting an exception if the variable is not declared is desirable, since we're trying to test the value of a variable we know we just declared, and the exception is telling us that we made a typo in the variable name and the code is broken.
    – Ilmari Karonen
    5 hours ago





















6














You should only call the function and assign num if it's undefined, otherwise you overwrite it every time:






function once(func) {
let num;

function retFunc(x) {
num = (num === undefined) ? func(x) : num
return num;
}
return retFunc;
}

function addByTwo(input) {
return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6





Note that if the function you pass in returns undefined it will be called more than once. If you need to handle that case you can set a flag to indicate whether the cached value is valid.






share|improve this answer























  • wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
    – Yang K
    9 hours ago










  • @YangK mostly insomnia!
    – Mark Meyer
    9 hours ago










  • same here but not as smart or helpful as you lol
    – Yang K
    9 hours ago



















3














What you're missing is removing the original function after the first execution. You should modify your code in the following way:



function once(func) {
let num;
function retFunc(x){
if (func)
num = func(x);
func = null;
return num;
}
return retFunc;
}

function addByTwo(input){
return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6


This way you remove the function after first usage and you're just keeping the result.






share|improve this answer





















  • Also i'd like to say that this way is better than relying on the result, because function doesn't need to always return something. It may return undefined, but we still want to run it only once.
    – tswistak
    5 hours ago



















1














This is your problem



function retFunc(x){
num = func(x);
return num;
}


You're always calling func(x). Add an if condition to check if num is undefined before calling func(x).






share|improve this answer





















  • Hi by adding if condition to check whether num is undefined my code is working now.
    – Amit Kumar
    9 hours ago





















1














different approach: overwrite the call to func



no flags, check if result is undefined or anything required



function once(func) {
let num;

let internalFn = function(x) {
num = func(x);
internalFn = function(x) {
return num;
}
return num;
}

function retFunc(x){
return internalFn(x);
}
return retFunc;
}

function addByTwo(input){
return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6





share|improve this answer





















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    5 Answers
    5






    active

    oldest

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    5 Answers
    5






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    12














    You should only assign num = func(x) when num is undefined - that is, on the very first call of retFunc:






    function once(func) {
    let num;
    function retFunc(x){
    if (num === undefined) {
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    But this isn't a guaranteed general solution - what if the passed function (addByTwo in your example) results in undefined when called? Then, the === undefined check won't work. So, it might be better to set a flag or something similar, and reassign that flag the first time the callback is called:






    function once(func) {
    let num;
    let done = false;
    function retFunc(x){
    if (!done) {
    done = true;
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function returnsUndefinedOn1(input){
    return input === 1 ? undefined : input;
    }

    var onceFunc = once(returnsUndefinedOn1);

    console.log(onceFunc(1));
    console.log(onceFunc(10));
    console.log(onceFunc(9001));








    share|improve this answer

















    • 2




      That's the desired behavior - on an input of 1, the function returnsUndefinedOn1 returns undefined, per its name. Further calls of onceFunc will also return undefined, despite being called with an argument other than 1.
      – CertainPerformance
      9 hours ago






    • 2




      Please use typeof to compare with undefined
      – Pierre Arlaud
      8 hours ago






    • 1




      @PierreArlaud In any remotely reasonable code, it shouldn't be necessary, I think? Many global object can be redefined, that doesn't mean they shouldn't be used, that means not to reassign them.
      – CertainPerformance
      8 hours ago






    • 1




      @CertainPerformance typeof also does not trigger an exception if the variable is not declared. There is no scenario where not using typeof to compare with undefined is the better choice. My opinion on the matter is that using typeof is a good habbit to have, and "remotely reasonable codes" are no exceptions.
      – Pierre Arlaud
      8 hours ago






    • 2




      @PierreArlaud: In this case, getting an exception if the variable is not declared is desirable, since we're trying to test the value of a variable we know we just declared, and the exception is telling us that we made a typo in the variable name and the code is broken.
      – Ilmari Karonen
      5 hours ago


















    12














    You should only assign num = func(x) when num is undefined - that is, on the very first call of retFunc:






    function once(func) {
    let num;
    function retFunc(x){
    if (num === undefined) {
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    But this isn't a guaranteed general solution - what if the passed function (addByTwo in your example) results in undefined when called? Then, the === undefined check won't work. So, it might be better to set a flag or something similar, and reassign that flag the first time the callback is called:






    function once(func) {
    let num;
    let done = false;
    function retFunc(x){
    if (!done) {
    done = true;
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function returnsUndefinedOn1(input){
    return input === 1 ? undefined : input;
    }

    var onceFunc = once(returnsUndefinedOn1);

    console.log(onceFunc(1));
    console.log(onceFunc(10));
    console.log(onceFunc(9001));








    share|improve this answer

















    • 2




      That's the desired behavior - on an input of 1, the function returnsUndefinedOn1 returns undefined, per its name. Further calls of onceFunc will also return undefined, despite being called with an argument other than 1.
      – CertainPerformance
      9 hours ago






    • 2




      Please use typeof to compare with undefined
      – Pierre Arlaud
      8 hours ago






    • 1




      @PierreArlaud In any remotely reasonable code, it shouldn't be necessary, I think? Many global object can be redefined, that doesn't mean they shouldn't be used, that means not to reassign them.
      – CertainPerformance
      8 hours ago






    • 1




      @CertainPerformance typeof also does not trigger an exception if the variable is not declared. There is no scenario where not using typeof to compare with undefined is the better choice. My opinion on the matter is that using typeof is a good habbit to have, and "remotely reasonable codes" are no exceptions.
      – Pierre Arlaud
      8 hours ago






    • 2




      @PierreArlaud: In this case, getting an exception if the variable is not declared is desirable, since we're trying to test the value of a variable we know we just declared, and the exception is telling us that we made a typo in the variable name and the code is broken.
      – Ilmari Karonen
      5 hours ago
















    12












    12








    12






    You should only assign num = func(x) when num is undefined - that is, on the very first call of retFunc:






    function once(func) {
    let num;
    function retFunc(x){
    if (num === undefined) {
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    But this isn't a guaranteed general solution - what if the passed function (addByTwo in your example) results in undefined when called? Then, the === undefined check won't work. So, it might be better to set a flag or something similar, and reassign that flag the first time the callback is called:






    function once(func) {
    let num;
    let done = false;
    function retFunc(x){
    if (!done) {
    done = true;
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function returnsUndefinedOn1(input){
    return input === 1 ? undefined : input;
    }

    var onceFunc = once(returnsUndefinedOn1);

    console.log(onceFunc(1));
    console.log(onceFunc(10));
    console.log(onceFunc(9001));








    share|improve this answer












    You should only assign num = func(x) when num is undefined - that is, on the very first call of retFunc:






    function once(func) {
    let num;
    function retFunc(x){
    if (num === undefined) {
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    But this isn't a guaranteed general solution - what if the passed function (addByTwo in your example) results in undefined when called? Then, the === undefined check won't work. So, it might be better to set a flag or something similar, and reassign that flag the first time the callback is called:






    function once(func) {
    let num;
    let done = false;
    function retFunc(x){
    if (!done) {
    done = true;
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function returnsUndefinedOn1(input){
    return input === 1 ? undefined : input;
    }

    var onceFunc = once(returnsUndefinedOn1);

    console.log(onceFunc(1));
    console.log(onceFunc(10));
    console.log(onceFunc(9001));








    function once(func) {
    let num;
    function retFunc(x){
    if (num === undefined) {
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    function once(func) {
    let num;
    function retFunc(x){
    if (num === undefined) {
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    function once(func) {
    let num;
    let done = false;
    function retFunc(x){
    if (!done) {
    done = true;
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function returnsUndefinedOn1(input){
    return input === 1 ? undefined : input;
    }

    var onceFunc = once(returnsUndefinedOn1);

    console.log(onceFunc(1));
    console.log(onceFunc(10));
    console.log(onceFunc(9001));





    function once(func) {
    let num;
    let done = false;
    function retFunc(x){
    if (!done) {
    done = true;
    num = func(x);
    }
    return num;
    }
    return retFunc;
    }

    function returnsUndefinedOn1(input){
    return input === 1 ? undefined : input;
    }

    var onceFunc = once(returnsUndefinedOn1);

    console.log(onceFunc(1));
    console.log(onceFunc(10));
    console.log(onceFunc(9001));






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 9 hours ago









    CertainPerformanceCertainPerformance

    77.6k143863




    77.6k143863








    • 2




      That's the desired behavior - on an input of 1, the function returnsUndefinedOn1 returns undefined, per its name. Further calls of onceFunc will also return undefined, despite being called with an argument other than 1.
      – CertainPerformance
      9 hours ago






    • 2




      Please use typeof to compare with undefined
      – Pierre Arlaud
      8 hours ago






    • 1




      @PierreArlaud In any remotely reasonable code, it shouldn't be necessary, I think? Many global object can be redefined, that doesn't mean they shouldn't be used, that means not to reassign them.
      – CertainPerformance
      8 hours ago






    • 1




      @CertainPerformance typeof also does not trigger an exception if the variable is not declared. There is no scenario where not using typeof to compare with undefined is the better choice. My opinion on the matter is that using typeof is a good habbit to have, and "remotely reasonable codes" are no exceptions.
      – Pierre Arlaud
      8 hours ago






    • 2




      @PierreArlaud: In this case, getting an exception if the variable is not declared is desirable, since we're trying to test the value of a variable we know we just declared, and the exception is telling us that we made a typo in the variable name and the code is broken.
      – Ilmari Karonen
      5 hours ago
















    • 2




      That's the desired behavior - on an input of 1, the function returnsUndefinedOn1 returns undefined, per its name. Further calls of onceFunc will also return undefined, despite being called with an argument other than 1.
      – CertainPerformance
      9 hours ago






    • 2




      Please use typeof to compare with undefined
      – Pierre Arlaud
      8 hours ago






    • 1




      @PierreArlaud In any remotely reasonable code, it shouldn't be necessary, I think? Many global object can be redefined, that doesn't mean they shouldn't be used, that means not to reassign them.
      – CertainPerformance
      8 hours ago






    • 1




      @CertainPerformance typeof also does not trigger an exception if the variable is not declared. There is no scenario where not using typeof to compare with undefined is the better choice. My opinion on the matter is that using typeof is a good habbit to have, and "remotely reasonable codes" are no exceptions.
      – Pierre Arlaud
      8 hours ago






    • 2




      @PierreArlaud: In this case, getting an exception if the variable is not declared is desirable, since we're trying to test the value of a variable we know we just declared, and the exception is telling us that we made a typo in the variable name and the code is broken.
      – Ilmari Karonen
      5 hours ago










    2




    2




    That's the desired behavior - on an input of 1, the function returnsUndefinedOn1 returns undefined, per its name. Further calls of onceFunc will also return undefined, despite being called with an argument other than 1.
    – CertainPerformance
    9 hours ago




    That's the desired behavior - on an input of 1, the function returnsUndefinedOn1 returns undefined, per its name. Further calls of onceFunc will also return undefined, despite being called with an argument other than 1.
    – CertainPerformance
    9 hours ago




    2




    2




    Please use typeof to compare with undefined
    – Pierre Arlaud
    8 hours ago




    Please use typeof to compare with undefined
    – Pierre Arlaud
    8 hours ago




    1




    1




    @PierreArlaud In any remotely reasonable code, it shouldn't be necessary, I think? Many global object can be redefined, that doesn't mean they shouldn't be used, that means not to reassign them.
    – CertainPerformance
    8 hours ago




    @PierreArlaud In any remotely reasonable code, it shouldn't be necessary, I think? Many global object can be redefined, that doesn't mean they shouldn't be used, that means not to reassign them.
    – CertainPerformance
    8 hours ago




    1




    1




    @CertainPerformance typeof also does not trigger an exception if the variable is not declared. There is no scenario where not using typeof to compare with undefined is the better choice. My opinion on the matter is that using typeof is a good habbit to have, and "remotely reasonable codes" are no exceptions.
    – Pierre Arlaud
    8 hours ago




    @CertainPerformance typeof also does not trigger an exception if the variable is not declared. There is no scenario where not using typeof to compare with undefined is the better choice. My opinion on the matter is that using typeof is a good habbit to have, and "remotely reasonable codes" are no exceptions.
    – Pierre Arlaud
    8 hours ago




    2




    2




    @PierreArlaud: In this case, getting an exception if the variable is not declared is desirable, since we're trying to test the value of a variable we know we just declared, and the exception is telling us that we made a typo in the variable name and the code is broken.
    – Ilmari Karonen
    5 hours ago






    @PierreArlaud: In this case, getting an exception if the variable is not declared is desirable, since we're trying to test the value of a variable we know we just declared, and the exception is telling us that we made a typo in the variable name and the code is broken.
    – Ilmari Karonen
    5 hours ago















    6














    You should only call the function and assign num if it's undefined, otherwise you overwrite it every time:






    function once(func) {
    let num;

    function retFunc(x) {
    num = (num === undefined) ? func(x) : num
    return num;
    }
    return retFunc;
    }

    function addByTwo(input) {
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    Note that if the function you pass in returns undefined it will be called more than once. If you need to handle that case you can set a flag to indicate whether the cached value is valid.






    share|improve this answer























    • wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
      – Yang K
      9 hours ago










    • @YangK mostly insomnia!
      – Mark Meyer
      9 hours ago










    • same here but not as smart or helpful as you lol
      – Yang K
      9 hours ago
















    6














    You should only call the function and assign num if it's undefined, otherwise you overwrite it every time:






    function once(func) {
    let num;

    function retFunc(x) {
    num = (num === undefined) ? func(x) : num
    return num;
    }
    return retFunc;
    }

    function addByTwo(input) {
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    Note that if the function you pass in returns undefined it will be called more than once. If you need to handle that case you can set a flag to indicate whether the cached value is valid.






    share|improve this answer























    • wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
      – Yang K
      9 hours ago










    • @YangK mostly insomnia!
      – Mark Meyer
      9 hours ago










    • same here but not as smart or helpful as you lol
      – Yang K
      9 hours ago














    6












    6








    6






    You should only call the function and assign num if it's undefined, otherwise you overwrite it every time:






    function once(func) {
    let num;

    function retFunc(x) {
    num = (num === undefined) ? func(x) : num
    return num;
    }
    return retFunc;
    }

    function addByTwo(input) {
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    Note that if the function you pass in returns undefined it will be called more than once. If you need to handle that case you can set a flag to indicate whether the cached value is valid.






    share|improve this answer














    You should only call the function and assign num if it's undefined, otherwise you overwrite it every time:






    function once(func) {
    let num;

    function retFunc(x) {
    num = (num === undefined) ? func(x) : num
    return num;
    }
    return retFunc;
    }

    function addByTwo(input) {
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    Note that if the function you pass in returns undefined it will be called more than once. If you need to handle that case you can set a flag to indicate whether the cached value is valid.






    function once(func) {
    let num;

    function retFunc(x) {
    num = (num === undefined) ? func(x) : num
    return num;
    }
    return retFunc;
    }

    function addByTwo(input) {
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    function once(func) {
    let num;

    function retFunc(x) {
    num = (num === undefined) ? func(x) : num
    return num;
    }
    return retFunc;
    }

    function addByTwo(input) {
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 9 hours ago

























    answered 9 hours ago









    Mark MeyerMark Meyer

    36.9k33059




    36.9k33059












    • wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
      – Yang K
      9 hours ago










    • @YangK mostly insomnia!
      – Mark Meyer
      9 hours ago










    • same here but not as smart or helpful as you lol
      – Yang K
      9 hours ago


















    • wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
      – Yang K
      9 hours ago










    • @YangK mostly insomnia!
      – Mark Meyer
      9 hours ago










    • same here but not as smart or helpful as you lol
      – Yang K
      9 hours ago
















    wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
    – Yang K
    9 hours ago




    wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
    – Yang K
    9 hours ago












    @YangK mostly insomnia!
    – Mark Meyer
    9 hours ago




    @YangK mostly insomnia!
    – Mark Meyer
    9 hours ago












    same here but not as smart or helpful as you lol
    – Yang K
    9 hours ago




    same here but not as smart or helpful as you lol
    – Yang K
    9 hours ago











    3














    What you're missing is removing the original function after the first execution. You should modify your code in the following way:



    function once(func) {
    let num;
    function retFunc(x){
    if (func)
    num = func(x);
    func = null;
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6


    This way you remove the function after first usage and you're just keeping the result.






    share|improve this answer





















    • Also i'd like to say that this way is better than relying on the result, because function doesn't need to always return something. It may return undefined, but we still want to run it only once.
      – tswistak
      5 hours ago
















    3














    What you're missing is removing the original function after the first execution. You should modify your code in the following way:



    function once(func) {
    let num;
    function retFunc(x){
    if (func)
    num = func(x);
    func = null;
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6


    This way you remove the function after first usage and you're just keeping the result.






    share|improve this answer





















    • Also i'd like to say that this way is better than relying on the result, because function doesn't need to always return something. It may return undefined, but we still want to run it only once.
      – tswistak
      5 hours ago














    3












    3








    3






    What you're missing is removing the original function after the first execution. You should modify your code in the following way:



    function once(func) {
    let num;
    function retFunc(x){
    if (func)
    num = func(x);
    func = null;
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6


    This way you remove the function after first usage and you're just keeping the result.






    share|improve this answer












    What you're missing is removing the original function after the first execution. You should modify your code in the following way:



    function once(func) {
    let num;
    function retFunc(x){
    if (func)
    num = func(x);
    func = null;
    return num;
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6


    This way you remove the function after first usage and you're just keeping the result.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 9 hours ago









    tswistaktswistak

    4112




    4112












    • Also i'd like to say that this way is better than relying on the result, because function doesn't need to always return something. It may return undefined, but we still want to run it only once.
      – tswistak
      5 hours ago


















    • Also i'd like to say that this way is better than relying on the result, because function doesn't need to always return something. It may return undefined, but we still want to run it only once.
      – tswistak
      5 hours ago
















    Also i'd like to say that this way is better than relying on the result, because function doesn't need to always return something. It may return undefined, but we still want to run it only once.
    – tswistak
    5 hours ago




    Also i'd like to say that this way is better than relying on the result, because function doesn't need to always return something. It may return undefined, but we still want to run it only once.
    – tswistak
    5 hours ago











    1














    This is your problem



    function retFunc(x){
    num = func(x);
    return num;
    }


    You're always calling func(x). Add an if condition to check if num is undefined before calling func(x).






    share|improve this answer





















    • Hi by adding if condition to check whether num is undefined my code is working now.
      – Amit Kumar
      9 hours ago


















    1














    This is your problem



    function retFunc(x){
    num = func(x);
    return num;
    }


    You're always calling func(x). Add an if condition to check if num is undefined before calling func(x).






    share|improve this answer





















    • Hi by adding if condition to check whether num is undefined my code is working now.
      – Amit Kumar
      9 hours ago
















    1












    1








    1






    This is your problem



    function retFunc(x){
    num = func(x);
    return num;
    }


    You're always calling func(x). Add an if condition to check if num is undefined before calling func(x).






    share|improve this answer












    This is your problem



    function retFunc(x){
    num = func(x);
    return num;
    }


    You're always calling func(x). Add an if condition to check if num is undefined before calling func(x).







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 9 hours ago









    asleepysamuraiasleepysamurai

    719614




    719614












    • Hi by adding if condition to check whether num is undefined my code is working now.
      – Amit Kumar
      9 hours ago




















    • Hi by adding if condition to check whether num is undefined my code is working now.
      – Amit Kumar
      9 hours ago


















    Hi by adding if condition to check whether num is undefined my code is working now.
    – Amit Kumar
    9 hours ago






    Hi by adding if condition to check whether num is undefined my code is working now.
    – Amit Kumar
    9 hours ago













    1














    different approach: overwrite the call to func



    no flags, check if result is undefined or anything required



    function once(func) {
    let num;

    let internalFn = function(x) {
    num = func(x);
    internalFn = function(x) {
    return num;
    }
    return num;
    }

    function retFunc(x){
    return internalFn(x);
    }
    return retFunc;
    }

    function addByTwo(input){
    return input + 2;
    }

    var onceFunc = once(addByTwo);

    console.log(onceFunc(4)); //should log 6
    console.log(onceFunc(10)); //should log 6
    console.log(onceFunc(9001)); //should log 6





    share|improve this answer


























      1














      different approach: overwrite the call to func



      no flags, check if result is undefined or anything required



      function once(func) {
      let num;

      let internalFn = function(x) {
      num = func(x);
      internalFn = function(x) {
      return num;
      }
      return num;
      }

      function retFunc(x){
      return internalFn(x);
      }
      return retFunc;
      }

      function addByTwo(input){
      return input + 2;
      }

      var onceFunc = once(addByTwo);

      console.log(onceFunc(4)); //should log 6
      console.log(onceFunc(10)); //should log 6
      console.log(onceFunc(9001)); //should log 6





      share|improve this answer
























        1












        1








        1






        different approach: overwrite the call to func



        no flags, check if result is undefined or anything required



        function once(func) {
        let num;

        let internalFn = function(x) {
        num = func(x);
        internalFn = function(x) {
        return num;
        }
        return num;
        }

        function retFunc(x){
        return internalFn(x);
        }
        return retFunc;
        }

        function addByTwo(input){
        return input + 2;
        }

        var onceFunc = once(addByTwo);

        console.log(onceFunc(4)); //should log 6
        console.log(onceFunc(10)); //should log 6
        console.log(onceFunc(9001)); //should log 6





        share|improve this answer












        different approach: overwrite the call to func



        no flags, check if result is undefined or anything required



        function once(func) {
        let num;

        let internalFn = function(x) {
        num = func(x);
        internalFn = function(x) {
        return num;
        }
        return num;
        }

        function retFunc(x){
        return internalFn(x);
        }
        return retFunc;
        }

        function addByTwo(input){
        return input + 2;
        }

        var onceFunc = once(addByTwo);

        console.log(onceFunc(4)); //should log 6
        console.log(onceFunc(10)); //should log 6
        console.log(onceFunc(9001)); //should log 6






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 4 hours ago









        LexLex

        190119




        190119






























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