Is it possible to make a clamp function shorter than a ternary in JS?Let's create random number...
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Is it possible to make a clamp function shorter than a ternary in JS?
Let's create random number genratorBase-2 integer logarithm of 64-bit unsigned integerInteger square root of integerTweetable Mathematical ArtShortest Minmod FunctionNon-repeating random numbersCompute the Mertens functionMagic popcount numbersBe as evil as possibleTernary Triangles
$begingroup$
Imagine this short function to clamp a number between 0 and 255:
c = n => n > 0 ? n < 255 ? n : 255 : 0
Is this the shortest possible version of a clamp function with JavaScript (without ES.Next features)?
P.S: Not sure if it's relevant but, the 0 and 255 are not random, the idea is to clamp a number as an 8-bit unsigned integer.
code-golf math tips javascript
asked yesterday
Ricardo AmaralRicardo Amaral
1986
New contributor
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 2 more comments
$begingroup$
Imagine this short function to clamp a number between 0 and 255:
c = n => n > 0 ? n < 255 ? n : 255 : 0
Is this the shortest possible version of a clamp function with JavaScript (without ES.Next features)?
P.S: Not sure if it's relevant but, the 0 and 255 are not random, the idea is to clamp a number as an 8-bit unsigned integer.
code-golf math tips javascript
asked yesterday
Ricardo AmaralRicardo Amaral
1986
New contributor
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
Hi and welcome to PPCG! Just to be clear, any answer you receive here will not necessarily be a good idea to use in anything except for code golfing. Aside from that, if you care about what version / environment it has to work in you might want to specify it.
$endgroup$
– FryAmTheEggman
yesterday
1
$begingroup$
Oh, I'm well aware. I've updated the question a bit. Thank you :)
$endgroup$
– Ricardo Amaral
yesterday
2
$begingroup$
I don't know JS, but one way to clamp is to sort[0,n,255]and take the middle element -- might that be shorter?
$endgroup$
– xnor
yesterday
1
$begingroup$
@xnor Unfortunately, the JSsort()method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)
$endgroup$
– Arnauld
yesterday
5
$begingroup$
@Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
$endgroup$
– xnor
yesterday
|
show 2 more comments
$begingroup$
Imagine this short function to clamp a number between 0 and 255:
c = n => n > 0 ? n < 255 ? n : 255 : 0
Is this the shortest possible version of a clamp function with JavaScript (without ES.Next features)?
P.S: Not sure if it's relevant but, the 0 and 255 are not random, the idea is to clamp a number as an 8-bit unsigned integer.
code-golf math tips javascript
asked yesterday
Ricardo AmaralRicardo Amaral
1986
New contributor
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Imagine this short function to clamp a number between 0 and 255:
c = n => n > 0 ? n < 255 ? n : 255 : 0
Is this the shortest possible version of a clamp function with JavaScript (without ES.Next features)?
P.S: Not sure if it's relevant but, the 0 and 255 are not random, the idea is to clamp a number as an 8-bit unsigned integer.
code-golf math tips javascript
code-golf math tips javascript
asked yesterday
Ricardo AmaralRicardo Amaral
1986
New contributor
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Ricardo AmaralRicardo Amaral
1986
New contributor
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Ricardo Amaral
asked yesterday
Ricardo AmaralRicardo Amaral
1986
New contributor
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Ricardo AmaralRicardo Amaral
1986
asked yesterday
Ricardo AmaralRicardo Amaral
1986
1986
New contributor
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ricardo Amaral is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
Hi and welcome to PPCG! Just to be clear, any answer you receive here will not necessarily be a good idea to use in anything except for code golfing. Aside from that, if you care about what version / environment it has to work in you might want to specify it.
$endgroup$
– FryAmTheEggman
yesterday
1
$begingroup$
Oh, I'm well aware. I've updated the question a bit. Thank you :)
$endgroup$
– Ricardo Amaral
yesterday
2
$begingroup$
I don't know JS, but one way to clamp is to sort[0,n,255]and take the middle element -- might that be shorter?
$endgroup$
– xnor
yesterday
1
$begingroup$
@xnor Unfortunately, the JSsort()method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)
$endgroup$
– Arnauld
yesterday
5
$begingroup$
@Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
$endgroup$
– xnor
yesterday
|
show 2 more comments
2
$begingroup$
Hi and welcome to PPCG! Just to be clear, any answer you receive here will not necessarily be a good idea to use in anything except for code golfing. Aside from that, if you care about what version / environment it has to work in you might want to specify it.
$endgroup$
– FryAmTheEggman
yesterday
1
$begingroup$
Oh, I'm well aware. I've updated the question a bit. Thank you :)
$endgroup$
– Ricardo Amaral
yesterday
2
$begingroup$
I don't know JS, but one way to clamp is to sort[0,n,255]and take the middle element -- might that be shorter?
$endgroup$
– xnor
yesterday
1
$begingroup$
@xnor Unfortunately, the JSsort()method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)
$endgroup$
– Arnauld
yesterday
5
$begingroup$
@Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
$endgroup$
– xnor
yesterday
2
2
$begingroup$
Hi and welcome to PPCG! Just to be clear, any answer you receive here will not necessarily be a good idea to use in anything except for code golfing. Aside from that, if you care about what version / environment it has to work in you might want to specify it.
$endgroup$
– FryAmTheEggman
yesterday
$begingroup$
Hi and welcome to PPCG! Just to be clear, any answer you receive here will not necessarily be a good idea to use in anything except for code golfing. Aside from that, if you care about what version / environment it has to work in you might want to specify it.
$endgroup$
– FryAmTheEggman
yesterday
1
1
$begingroup$
Oh, I'm well aware. I've updated the question a bit. Thank you :)
$endgroup$
– Ricardo Amaral
yesterday
$begingroup$
Oh, I'm well aware. I've updated the question a bit. Thank you :)
$endgroup$
– Ricardo Amaral
yesterday
2
2
$begingroup$
I don't know JS, but one way to clamp is to sort
[0,n,255] and take the middle element -- might that be shorter?$endgroup$
– xnor
yesterday
$begingroup$
I don't know JS, but one way to clamp is to sort
[0,n,255] and take the middle element -- might that be shorter?$endgroup$
– xnor
yesterday
1
1
$begingroup$
@xnor Unfortunately, the JS
sort() method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)$endgroup$
– Arnauld
yesterday
$begingroup$
@xnor Unfortunately, the JS
sort() method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)$endgroup$
– Arnauld
yesterday
5
5
$begingroup$
@Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
$endgroup$
– xnor
yesterday
$begingroup$
@Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
$endgroup$
– xnor
yesterday
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
20 bytes
For reference, this is the original version without whitespace and without naming the function:
n=>n>0?n<255?n:255:0
Try it online!
19 bytes
We can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.
n=>n<0?0:n>>8?255:n
Try it online!
19 bytes
This one returns $false$ instead of $0$ but works for $nge 2^{32}$.
n=>n>255?255:n>0&&n
Try it online!
18 bytes
By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.
n=>n>>8?n>0&&255:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
n > 0 && // return false if n is negative
255 // or 255 otherwise
: // else:
n // return n unchanged
(This is a fixed revision of the code proposed by @ValueInk in the comments.)
17 bytes
We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:
n=>n>>8?-n>>>24:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
-n >>> 24 // non-arithmetic right-shift of -n by 24 positions
: // else:
n // return n unchanged
answered yesterday
ArnauldArnauld
77.9k695326
$endgroup$
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
yesterday
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will be truthy for any negative input.
$endgroup$
– Arnauld
yesterday
$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago
add a comment |
Your Answer
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votes
$begingroup$
20 bytes
For reference, this is the original version without whitespace and without naming the function:
n=>n>0?n<255?n:255:0
Try it online!
19 bytes
We can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.
n=>n<0?0:n>>8?255:n
Try it online!
19 bytes
This one returns $false$ instead of $0$ but works for $nge 2^{32}$.
n=>n>255?255:n>0&&n
Try it online!
18 bytes
By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.
n=>n>>8?n>0&&255:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
n > 0 && // return false if n is negative
255 // or 255 otherwise
: // else:
n // return n unchanged
(This is a fixed revision of the code proposed by @ValueInk in the comments.)
17 bytes
We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:
n=>n>>8?-n>>>24:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
-n >>> 24 // non-arithmetic right-shift of -n by 24 positions
: // else:
n // return n unchanged
answered yesterday
ArnauldArnauld
77.9k695326
$endgroup$
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
yesterday
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will be truthy for any negative input.
$endgroup$
– Arnauld
yesterday
$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago
add a comment |
$begingroup$
20 bytes
For reference, this is the original version without whitespace and without naming the function:
n=>n>0?n<255?n:255:0
Try it online!
19 bytes
We can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.
n=>n<0?0:n>>8?255:n
Try it online!
19 bytes
This one returns $false$ instead of $0$ but works for $nge 2^{32}$.
n=>n>255?255:n>0&&n
Try it online!
18 bytes
By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.
n=>n>>8?n>0&&255:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
n > 0 && // return false if n is negative
255 // or 255 otherwise
: // else:
n // return n unchanged
(This is a fixed revision of the code proposed by @ValueInk in the comments.)
17 bytes
We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:
n=>n>>8?-n>>>24:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
-n >>> 24 // non-arithmetic right-shift of -n by 24 positions
: // else:
n // return n unchanged
answered yesterday
ArnauldArnauld
77.9k695326
$endgroup$
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
yesterday
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will be truthy for any negative input.
$endgroup$
– Arnauld
yesterday
$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago
add a comment |
$begingroup$
20 bytes
For reference, this is the original version without whitespace and without naming the function:
n=>n>0?n<255?n:255:0
Try it online!
19 bytes
We can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.
n=>n<0?0:n>>8?255:n
Try it online!
19 bytes
This one returns $false$ instead of $0$ but works for $nge 2^{32}$.
n=>n>255?255:n>0&&n
Try it online!
18 bytes
By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.
n=>n>>8?n>0&&255:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
n > 0 && // return false if n is negative
255 // or 255 otherwise
: // else:
n // return n unchanged
(This is a fixed revision of the code proposed by @ValueInk in the comments.)
17 bytes
We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:
n=>n>>8?-n>>>24:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
-n >>> 24 // non-arithmetic right-shift of -n by 24 positions
: // else:
n // return n unchanged
answered yesterday
ArnauldArnauld
77.9k695326
$endgroup$
20 bytes
For reference, this is the original version without whitespace and without naming the function:
n=>n>0?n<255?n:255:0
Try it online!
19 bytes
We can save a byte by inverting the logic of the ternary tests and using n>>8 to test whether $n$ is greater than $255$. Because of the bitwise operation, this will however fail for $nge 2^{32}$.
n=>n<0?0:n>>8?255:n
Try it online!
19 bytes
This one returns $false$ instead of $0$ but works for $nge 2^{32}$.
n=>n>255?255:n>0&&n
Try it online!
18 bytes
By combining both versions above, we end up with a function that works for $256-2^{32}le n<2^{32}$ and returns $false$ for $n<0$.
n=>n>>8?n>0&&255:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
n > 0 && // return false if n is negative
255 // or 255 otherwise
: // else:
n // return n unchanged
(This is a fixed revision of the code proposed by @ValueInk in the comments.)
17 bytes
We can go a step further by limiting the valid input range to $-2^{24}< nle 2^{24}$:
n=>n>>8?-n>>>24:n
Try it online!
Commented
n => // n = input number
n >> 8 ? // if n is greater than 255 or n is negative:
-n >>> 24 // non-arithmetic right-shift of -n by 24 positions
: // else:
n // return n unchanged
answered yesterday
ArnauldArnauld
77.9k695326
edited 21 hours ago
answered yesterday
ArnauldArnauld
77.9k695326
answered yesterday
ArnauldArnauld
77.9k695326
answered yesterday
ArnauldArnauld
77.9k695326
77.9k695326
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
yesterday
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will be truthy for any negative input.
$endgroup$
– Arnauld
yesterday
$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago
add a comment |
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go forn=>n>>8?255:n>0&&nfor 18 bytes, sincefalsecan be coerced to0and this will make all negative numbers evaluate tofalse
$endgroup$
– Value Ink
yesterday
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,n>>8will be truthy for any negative input.
$endgroup$
– Arnauld
yesterday
$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go for
n=>n>>8?255:n>0&&n for 18 bytes, since false can be coerced to 0 and this will make all negative numbers evaluate to false$endgroup$
– Value Ink
yesterday
$begingroup$
Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go for
n=>n>>8?255:n>0&&n for 18 bytes, since false can be coerced to 0 and this will make all negative numbers evaluate to false$endgroup$
– Value Ink
yesterday
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,
n>>8 will be truthy for any negative input.$endgroup$
– Arnauld
yesterday
$begingroup$
@ValueInk If you don't test $n<0$ beforhand,
n>>8 will be truthy for any negative input.$endgroup$
– Arnauld
yesterday
$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago
$begingroup$
Very nice, thank you so much!
$endgroup$
– Ricardo Amaral
20 hours ago
add a comment |
Ricardo Amaral is a new contributor. Be nice, and check out our Code of Conduct.
Ricardo Amaral is a new contributor. Be nice, and check out our Code of Conduct.
Ricardo Amaral is a new contributor. Be nice, and check out our Code of Conduct.
Ricardo Amaral is a new contributor. Be nice, and check out our Code of Conduct.
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2
$begingroup$
Hi and welcome to PPCG! Just to be clear, any answer you receive here will not necessarily be a good idea to use in anything except for code golfing. Aside from that, if you care about what version / environment it has to work in you might want to specify it.
$endgroup$
– FryAmTheEggman
yesterday
1
$begingroup$
Oh, I'm well aware. I've updated the question a bit. Thank you :)
$endgroup$
– Ricardo Amaral
yesterday
2
$begingroup$
I don't know JS, but one way to clamp is to sort
[0,n,255]and take the middle element -- might that be shorter?$endgroup$
– xnor
yesterday
1
$begingroup$
@xnor Unfortunately, the JS
sort()method uses a lexicographical comparison by default, so that would require an explicit callback. (Something like that.)$endgroup$
– Arnauld
yesterday
5
$begingroup$
@Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
$endgroup$
– xnor
yesterday