How to properly add hex escapes into a string-literal?

c c99
22,875

Solution 1

Use 3 octal digits:

char problem[] = "abc\022e";

or split your string:

char problem[] = "abc\x12" "e";

Why these work:

  • Unlike hex escapes, standard defines 3 digits as maximum amount for octal escape.

    6.4.4.4 Character constants

    ...

    octal-escape-sequence:
        \ octal-digit
        \ octal-digit octal-digit
        \ octal-digit octal-digit octal-digit
    

    ...

    hexadecimal-escape-sequence:
        \x hexadecimal-digit
        hexadecimal-escape-sequence hexadecimal-digit
    
  • String literal concatenation is defined as a later translation phase than literal escape character conversion.

    5.1.1.2 Translation phases

    ...

    1. Each source character set member and escape sequence in character constants and string literals is converted to the corresponding member of the execution character set; if there is no corresponding member, it is converted to an implementation- defined member other than the null (wide) character. 8)

    2. Adjacent string literal tokens are concatenated.

Solution 2

Since string literals are concateneated early on in the compilation process, but after the escaped-character conversion, you can just use:

char problem[] = "abc\x12" "e";

though you may prefer full separation for readability:

char problem[] = "abc" "\x12" "e";

For the language lawyers amongst us, this is covered in C11 5.1.1.2 Translation phases (my emphasis):

  1. Each source character set member and escape sequence in character constants and string literals is converted to the corresponding member of the execution character set; if there is no corresponding member, it is converted to an implementation-defined member other than the null (wide) character.

  2. Adjacent string literal tokens are concatenated.

Solution 3

Why I'm asking? When you want to build UTF-8 string as constant, you have to use hex values of character is larger than ASCII table can hold.

Well, no. You don't have to. As of C11, you can prefix your string constant with u8, which tells the compiler that the character literal is in UTF-8.

char solution[] = u8"no need to use hex-codes á駵";

(Same thing is supported by C++11 as well, by the way)

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tilz0R
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tilz0R

Embedded firmware developer with 5+ years experience with master of electrical engineering.

Updated on July 08, 2022

Comments

  • tilz0R
    tilz0R almost 2 years

    When you have string in C, you can add direct hex code inside.

    char str[] = "abcde"; // 'a', 'b', 'c', 'd', 'e', 0x00
    char str2[] = "abc\x12\x34"; // 'a', 'b', 'c', 0x12, 0x34, 0x00
    

    Both examples have 6 bytes in memory. Now the problem exists if you want to add value [a-fA-F0-9] after hex entry.

    //I want: 'a', 'b', 'c', 0x12, 'e', 0x00
    //Error, hex is too big because last e is treated as part of hex thus becoming 0x12e
    char problem[] = "abc\x12e";
    

    Possible solution is to replace after definition.

    //This will work, bad idea
    char solution[6] = "abcde";
    solution[3] = 0x12;
    

    This can work, but it will fail, if you put it as const.

    //This will not work
    const char solution[6] = "abcde";
    solution[3] = 0x12; //Compilation error!
    

    How to properly insert e after \x12 without triggering error?


    Why I'm asking? When you want to build UTF-8 string as constant, you have to use hex values of character if it is larger than ASCII table can hold.