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Is it possible to make a clamp function shorter than a ternary in JS?

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20

1

$begingroup$

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

share|improve this question

edited 2 days ago

Ricardo Amaral

asked 2 days ago

Ricardo AmaralRicardo Amaral

2036

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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I'm well aware. I've updated the question a bit. Thank you :)
  $endgroup$
  – Ricardo Amaral
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  @Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
  $endgroup$
  – xnor
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

20

1

$begingroup$

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

share|improve this question

edited 2 days ago

Ricardo Amaral

asked 2 days ago

Ricardo AmaralRicardo Amaral

2036

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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, I'm well aware. I've updated the question a bit. Thank you :)
  $endgroup$
  – Ricardo Amaral
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  @Arnauld Wow, that's pretty silly. But it looks like it would be longer even if the sort was numerical.
  $endgroup$
  – xnor
  2 days ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

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

share|improve this question

edited 2 days ago

Ricardo Amaral

asked 2 days ago

Ricardo AmaralRicardo Amaral

2036

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.

$endgroup$

c = n => n > 0 ? n < 255 ? n : 255 : 0

share|improve this question

edited 2 days ago

Ricardo Amaral

asked 2 days ago

Ricardo AmaralRicardo Amaral

2036

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.

share|improve this question

edited 2 days ago

Ricardo Amaral

asked 2 days ago

Ricardo AmaralRicardo Amaral

2036

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 2 days ago

Ricardo AmaralRicardo Amaral

2036

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 2 days ago

Ricardo AmaralRicardo Amaral

2036

1 Answer
1

active

oldest

votes

21

$begingroup$

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 $256-2^32le n<2^32$ 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

share|improve this answer

edited yesterday

answered 2 days ago

ArnauldArnauld

78k795326

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go 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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk If you don't test $n<0$ beforhand, n>>8 will be truthy for any negative input.
  $endgroup$
  – Arnauld
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice, thank you so much!
  $endgroup$
  – Ricardo Amaral
  yesterday
  

  
  
  
  

add a comment | 

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21

$begingroup$

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 $256-2^32le n<2^32$ 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

share|improve this answer

edited yesterday

answered 2 days ago

ArnauldArnauld

78k795326

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go 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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk If you don't test $n<0$ beforhand, n>>8 will be truthy for any negative input.
  $endgroup$
  – Arnauld
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice, thank you so much!
  $endgroup$
  – Ricardo Amaral
  yesterday
  

  
  
  
  

add a comment | 

21

$begingroup$

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 $256-2^32le n<2^32$ 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

share|improve this answer

edited yesterday

answered 2 days ago

ArnauldArnauld

78k795326

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why stop there? If you're extremely liberal with what counts as a 0 (as JavaScript tends to do) you can always go 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
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ValueInk If you don't test $n<0$ beforhand, n>>8 will be truthy for any negative input.
  $endgroup$
  – Arnauld
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Very nice, thank you so much!
  $endgroup$
  – Ricardo Amaral
  yesterday
  

  
  
  
  

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 $256-2^32le n<2^32$ 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

share|improve this answer

edited yesterday

answered 2 days ago

ArnauldArnauld

78k795326

$endgroup$

n=>n>0?n<255?n:255:0

n=>n<0?0:n>>8?255:n

n=>n>255?255:n>0&&n

n=>n>>8?n>0&&255:n

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

n=>n>>8?-n>>>24:n

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

share|improve this answer

edited yesterday

answered 2 days ago

ArnauldArnauld

78k795326

answered 2 days ago

ArnauldArnauld

78k795326

answered 2 days ago

ArnauldArnauld

78k795326