Function to cache its argument's return value
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 memoization
add a comment |
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 memoization
1
Thenum
is already a good first step, but how do you remember that it was called a first time already? You always overwritenum = func(x)
– Bergi
yesterday
FYI, this is called memoization
– Barmar
22 hours ago
add a comment |
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 memoization
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 memoization
javascript memoization
edited 22 hours ago
Barmar
420k35244344
420k35244344
asked yesterday
Amit KumarAmit Kumar
1108
1108
1
Thenum
is already a good first step, but how do you remember that it was called a first time already? You always overwritenum = func(x)
– Bergi
yesterday
FYI, this is called memoization
– Barmar
22 hours ago
add a comment |
1
Thenum
is already a good first step, but how do you remember that it was called a first time already? You always overwritenum = func(x)
– Bergi
yesterday
FYI, this is called memoization
– Barmar
22 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
yesterday
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
yesterday
FYI, this is called memoization
– Barmar
22 hours ago
FYI, this is called memoization
– Barmar
22 hours ago
add a comment |
5 Answers
5
active
oldest
votes
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));
2
That's the desired behavior - on an input of1
, the functionreturnsUndefinedOn1
returnsundefined
, per its name. Further calls ofonceFunc
will also returnundefined
, despite being called with an argument other than1
.
– CertainPerformance
yesterday
2
Please usetypeof
to compare withundefined
– Pierre Arlaud
yesterday
2
@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
yesterday
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
yesterday
3
@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
yesterday
|
show 1 more comment
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.
wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
– Yang K
yesterday
@YangK mostly insomnia!
– Mark Meyer
yesterday
same here but not as smart or helpful as you lol
– Yang K
yesterday
add a comment |
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.
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
yesterday
add a comment |
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)
.
Hi by adding if condition to check whether num is undefined my code is working now.
– Amit Kumar
yesterday
add a comment |
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
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
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));
2
That's the desired behavior - on an input of1
, the functionreturnsUndefinedOn1
returnsundefined
, per its name. Further calls ofonceFunc
will also returnundefined
, despite being called with an argument other than1
.
– CertainPerformance
yesterday
2
Please usetypeof
to compare withundefined
– Pierre Arlaud
yesterday
2
@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
yesterday
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
yesterday
3
@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
yesterday
|
show 1 more comment
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));
2
That's the desired behavior - on an input of1
, the functionreturnsUndefinedOn1
returnsundefined
, per its name. Further calls ofonceFunc
will also returnundefined
, despite being called with an argument other than1
.
– CertainPerformance
yesterday
2
Please usetypeof
to compare withundefined
– Pierre Arlaud
yesterday
2
@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
yesterday
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
yesterday
3
@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
yesterday
|
show 1 more comment
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));
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));
answered yesterday
CertainPerformanceCertainPerformance
77.8k143865
77.8k143865
2
That's the desired behavior - on an input of1
, the functionreturnsUndefinedOn1
returnsundefined
, per its name. Further calls ofonceFunc
will also returnundefined
, despite being called with an argument other than1
.
– CertainPerformance
yesterday
2
Please usetypeof
to compare withundefined
– Pierre Arlaud
yesterday
2
@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
yesterday
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
yesterday
3
@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
yesterday
|
show 1 more comment
2
That's the desired behavior - on an input of1
, the functionreturnsUndefinedOn1
returnsundefined
, per its name. Further calls ofonceFunc
will also returnundefined
, despite being called with an argument other than1
.
– CertainPerformance
yesterday
2
Please usetypeof
to compare withundefined
– Pierre Arlaud
yesterday
2
@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
yesterday
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
yesterday
3
@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
yesterday
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
yesterday
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
yesterday
2
2
Please use
typeof
to compare with undefined
– Pierre Arlaud
yesterday
Please use
typeof
to compare with undefined
– Pierre Arlaud
yesterday
2
2
@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
yesterday
@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
yesterday
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
yesterday
@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
yesterday
3
3
@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
yesterday
@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
yesterday
|
show 1 more comment
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.
wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
– Yang K
yesterday
@YangK mostly insomnia!
– Mark Meyer
yesterday
same here but not as smart or helpful as you lol
– Yang K
yesterday
add a comment |
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.
wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
– Yang K
yesterday
@YangK mostly insomnia!
– Mark Meyer
yesterday
same here but not as smart or helpful as you lol
– Yang K
yesterday
add a comment |
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.
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
edited yesterday
answered yesterday
Mark MeyerMark Meyer
37.1k33059
37.1k33059
wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
– Yang K
yesterday
@YangK mostly insomnia!
– Mark Meyer
yesterday
same here but not as smart or helpful as you lol
– Yang K
yesterday
add a comment |
wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
– Yang K
yesterday
@YangK mostly insomnia!
– Mark Meyer
yesterday
same here but not as smart or helpful as you lol
– Yang K
yesterday
wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
– Yang K
yesterday
wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep
– Yang K
yesterday
@YangK mostly insomnia!
– Mark Meyer
yesterday
@YangK mostly insomnia!
– Mark Meyer
yesterday
same here but not as smart or helpful as you lol
– Yang K
yesterday
same here but not as smart or helpful as you lol
– Yang K
yesterday
add a comment |
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.
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
yesterday
add a comment |
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.
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
yesterday
add a comment |
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.
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.
answered yesterday
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
yesterday
add a comment |
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
yesterday
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
yesterday
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
yesterday
add a comment |
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)
.
Hi by adding if condition to check whether num is undefined my code is working now.
– Amit Kumar
yesterday
add a comment |
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)
.
Hi by adding if condition to check whether num is undefined my code is working now.
– Amit Kumar
yesterday
add a comment |
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)
.
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)
.
answered yesterday
asleepysamuraiasleepysamurai
729614
729614
Hi by adding if condition to check whether num is undefined my code is working now.
– Amit Kumar
yesterday
add a comment |
Hi by adding if condition to check whether num is undefined my code is working now.
– Amit Kumar
yesterday
Hi by adding if condition to check whether num is undefined my code is working now.
– Amit Kumar
yesterday
Hi by adding if condition to check whether num is undefined my code is working now.
– Amit Kumar
yesterday
add a comment |
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
add a comment |
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
add a comment |
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
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
answered yesterday
LexLex
190119
190119
add a comment |
add a comment |
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1
The
num
is already a good first step, but how do you remember that it was called a first time already? You always overwritenum = func(x)
– Bergi
yesterday
FYI, this is called memoization
– Barmar
22 hours ago