Length of string at compile time?

This is a discussion on Length of string at compile time? within the C Programming forums, part of the General Programming Boards category; Hi. Is there any way to get the length of a constant string at compile time, by following ISO C? ...

  1. #1
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    Length of string at compile time?

    Hi. Is there any way to get the length of a constant string at compile time, by following ISO C? I have this test code:

    Code:
    #include <string.h>
    
    int main()
    {
        const char *text_base = "text base";
        char text_full[strlen(text_base) + 10];
        return 0;
    }
    It compiles, but I get the following warning:

    Code:
    main.c:6: warning: ISO C90 forbids variable length array 'text_full'
    Is there any way fix this so that it follows the C standard?
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  2. #2
    spurious conceit MK27's Avatar
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    I believe it is allowed under C99 but not C90. You may have to use a compiler switch, for gcc:

    gcc --std=c99

    The other option is to malloc text_full instead.
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    That's it! Thanks!
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    It's also possible to declare text_base as an array and use sizeof text_base.

  5. #5
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    What is the practical difference between declaring text_base as an array and declaring it as a char*?
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  6. #6
    spurious conceit MK27's Avatar
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    Not much. It might be easier to circumvent the const qualifier with a cast that way, but that's kind of irrelevant. Using const provides enough of a warning.
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    You can do something like:
    Code:
    #define YOUR_STRING "text base"
    
    const char *text_base = YOUR_STRING;
    char text_full[ sizeof(YOUR_STRING)-1 + 10 ];

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    If you declare it an array, the compiler automatically knows its size.

  9. #9
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    Quote Originally Posted by Bayint Naung View Post
    You can do something like:
    Code:
    #define YOUR_STRING "text base"
    
    const char *text_base = YOUR_STRING;
    char text_full[ sizeof(YOUR_STRING)-1 + 10 ];
    I don't think that works. Since YOUR_STRING is #define'd as a string literal, you're really just taking the size of a string literal. But I'm pretty sure this is just the same as taking the size of a char* and always evaluates to the size of a pointer (i.e. 4 on a 32-bit system). Also, you never use text_base . . . .

    I would definitely use an array, because then you can take the size of it.
    Code:
    char str[] = "Hello, World!\n";
    char copy[sizeof(str) + 10];
    The difference between a const char * and a char array is this. With a const char * initialized to a string literal like "hello", the string "hello" gets put into a special segment of the program -- usually the read-only .text segment, on Intel-compatible platforms. Your const char * variable just points to this memory location. A char[] array, however, is allocated on the stack (if it is an auto variable, declared inside a function) or in a read-write static data segment (if it is a global or static variable); the string literal that you initialized it to is copied into the array. An array is mutable and a literal is not.
    dwk

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    and the hat of wrongness Salem's Avatar
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    sizeof on a string constant is a pretty neat way of looking at the problem
    Code:
    $ cat bar.c
    #include <stdio.h>
    int main ( ) {
        printf("%zd\n", sizeof("hello world") );
        return 0;
    }
    $ gcc bar.c
    $ ./a.out 
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    Frequently Quite Prolix dwks's Avatar
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    Interesting -- I didn't know sizeof() would return the length of a string literal. I guess that means string literals must have type char[N] instead of char* like I thought they did.
    dwk

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    sizeof() does not return length of string but total bytes (ie. including NUL byte for C string).

  13. #13
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    sizeof() does not return length of string but total bytes (ie. including NUL byte for C string).
    No it does not. The null is ignored:
    strlen

    Specifically:
    The strlen() function shall compute the number of bytes in the string to which s points, not including the terminating null byte.
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    Quote Originally Posted by dwks
    I guess that means string literals must have type char[N] instead of char* like I thought they did.
    Yes, a string literal is an array, not a pointer, though perhaps const char[N] comes closer.

    Quote Originally Posted by jeffcobb
    No it does not. The null is ignored:
    strlen
    strlen and sizeof are not the same thing
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    Quote Originally Posted by laserlight View Post
    Yes, a string literal is an array, not a pointer, though perhaps const char[N] comes closer.


    strlen and sizeof are not the same thing
    Thats what I get for replying before becoming fully caffeinated.
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