Length of a UTF-8 byte sequence

Question

Determine the length of a UTF-8 byte sequence given its first byte. The following table shows which ranges map to each possible length:

  Range    Length
---------  ------
0x00-0x7F    1
0xC2-0xDF    2
0xE0-0xEF    3
0xF0-0xF4    4

_{Notes on gaps in the table: 0x80-0xBF are continuation bytes, 0xC0-0xC1 would start an overlong, invalid sequence, 0xF5-0xFF would result in a codepoint beyond the Unicode maximum.}

Write a program or function that takes the first byte of a UTF-8 byte sequence as input and outputs or returns the length of the sequence. I/O is flexible. For example, the input can be a number, an 8-bit character or a one-character string. You can assume that the first byte is part of a valid sequence and falls into one of the ranges above.

This is code golf. The shortest answer in bytes wins.

Test cases

0x00 => 1
0x41 => 1
0x7F => 1
0xC2 => 2
0xDF => 2
0xE0 => 3
0xEF => 3
0xF0 => 4
0xF4 => 4

Answer 1

Forth, 6 bytes

x-size

see https://forth-standard.org/standard/xchar/X-SIZE

Input and output follows a standard Forth model:

Input

Memory address + length (i.e. 1) of a single-byte UTF-8 "string".

Output

UTF-8 sequence length in bytes.

Sample Code

Store 0xF0 in a memory cell, and invoke x-size:

variable v
0xF0 v !
v 1 x-size

Check the result:

.s <1> 4  ok

Answer 2

Z80Golf, 19 14 bytes

00000000: 2f6f 3e10 37ed 6a3d 30fb ee07 c03c       /o>.7.j=0....<

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-5 bytes thanks to @Bubbler

Example with input 0x41-Try it online! Assembly

Example with input 0xC2-Try it online!

Example with input 0xE0-Try it online!

Example with input 0xF4-Try it online!

Assembly:

;input: register a
;output: register a
byte_count:			;calculate 7^(log2(255^a))||1
	cpl			;xor 255
	ld l,a
	log2:
		ld	a,16
		scf
	log2loop:
		adc	hl,hl
		dec	a
		jr	nc,log2loop
	xor 7
	ret nz
	inc a

Try it online!

Answer 3

C (gcc), 39 bytes

t(char x){x=(__builtin_clz(~x)-24)%7u;}

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Answer 4

Jelly,  8  7 bytes

+⁹BIITḢ

A monadic link accepting the byte as an integer.

Try it online! Or see all inputs evaluated.

If an input of a list of the 8 bits were acceptable then the method is only 6 bytes: 1;IITḢ, however it has been deemed as talking flexible I/O too far.

How?

+⁹BIITḢ - Link: integer       e.g.: 127 (7f)            223 (df)            239 (ef)            244 (f4)
 ⁹      - literal 256
+       - add                       383                 479                 495                 500
  B     - to a list of bits         [1,0,1,1,1,1,1,1,1] [1,1,1,0,1,1,1,1,1] [1,1,1,1,0,1,1,1,1] [1,1,1,1,1,0,1,0,0]
   I    - increments                [-1,1,0,0,0,0,0,0]  [0,0,-1,1,0,0,0,0]  [0,0,0,-1,1,0,0,0]  [0,0,0,0,-1,1,-1,0]
    I   - increments                [2,-1,0,0,0,0,0]    [0,-1,2,-1,0,0,0]   [0,0,-1,2,-1,0,0]   [0,0,0,-1,2,-2,1]
     T  - truthy indices            [1,2]               [2,3,4]             [3,4,5]             [4,5,6,7]
      Ḣ - head                      1                   2                   3                   4

Answer 5

Haskell, 28 bytes

f x=sum[1|y<-"Áßï",x>y]+1

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Answer 6

Jelly, 8 7 bytes

»Ø⁷Ba\S

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How it works

»Ø⁷Ba\S  Main link. Argument: n (integer)

 Ø⁷      Yield 128.
»        Take the maximum of n and 128.
   B     Yield the array of binary digits.
    a\   Cumulatively reduce by AND, replacing 1's after the first 0 with 0's.
      S  Take the sum.

Answer 7

Python 2, 28 bytes

lambda x:1.4**(x/16-11)//1+1

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Answer 8

JavaScript (Node.js), 24 bytes

x=>7^Math.log2(255^x)||1

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Answer 9

Ruby, 27 23 bytes

->x{2+x[7]+(x/16<=>14)}

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Answer 10

Charcoal, 12 bytes

Ｉ⌕⍘⌈⟦Ｎ¹²⁸⟧²0

Try it online! Link is to verbose version of code. Explanation:

     Ｎ          Input number
      ¹²⁸       Literal 128
   ⌈⟦    ⟧      Take the maximum
  ⍘       ²     Convert to base 2 as a string
 ⌕         0    Find the position of the first `0`
Ｉ               Cast to string
                Implicitly print

Answer 11

05AB1E, 8 7 bytes

žy‚àb0k

Port of @Neil's Charcoal answer.
-1 byte thanks to @Grimy.

Input as integer.

Try it online or verify all test cases.

Explanation:

žy       # Push 128
  ‚      # Pair it with the (implicit) input-integer
   à     # Take the maximum of this pair (128 and input)
    b    # Convert it to a binary-string
     0k  # Get the 0-based first index of a "0" in this binary-string
         # (and output it implicitly as result)

Answer 12

Jelly, 7 bytes

»Ø⁷Bi0’

Port of my 05AB1E answer.

Try it online or verify all test cases.

Explanation:

 Ø⁷        # Push 128
»          # Take the max of 128 and the input
   B       # Convert it to binary
    i0     # Get the 1-indexed first index of a 0
      ’    # Decrease it by 1 to make it 0-indexed (and output it implicitly)

Answer 13

Perl 6, 18 bytes

{7-msb(255-$_)||1}

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Port of user202729's JavaScript answer. Alternatives with WhateverCode:

(255-*).msb*6%34%7
-(255-*).msb%6%5+1

Answer 14

x86 Assembly, 11 bytes

00000000 <f>:
   0:   f6 d1                   not    %cl
   2:   0f bd c1                bsr    %ecx,%eax
   5:   34 07                   xor    $0x7,%al
   7:   75 01                   jne    a <l1>
   9:   40                      inc    %eax
0000000a <l1>:
   a:   c3                      ret

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Port of user202729's JavaScript answer. Uses fastcall conventions.

Answer 15

Labyrinth, 35 bytes

? 28& 16/ )!@!
:_1 ";_ _3&""2
   @1

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Unwrapped version of the code:

?:_128&1!@
      ;
      _16/_3&2!@
            )
            !
            @

Answer 16

C, 31 bytes

f(x){return(x-160>>20-x/16)+2;}

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27 bytes with gcc (-O0)

f(x){x=(x-160>>20-x/16)+2;}

Alternatives, 31 and 33 bytes

f(x){return(10>>15-x/16)+7>>2;}
f(x){return x/128-(-3>>15-x/16);}

I found these expressions when playing around with the Aha! superoptimizer a few years ago.