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:

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#include <string.h>

int main()
{
    const char *text_base = "text base";
    char text_full[strlen(text_base) + 10];
    return 0;
}

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It compiles, but I get the following warning:

Code:

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main.c:6: warning: ISO C90 forbids variable length array 'text_full'

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Is there any way fix this so that it follows the C standard?

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.

That's it! Thanks!

It's also possible to declare text_base as an array and use sizeof text_base.

What is the practical difference between declaring text_base as an array and declaring it as a char*?

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.

You can do something like:

Code:

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#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.

Quote:

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Originally Posted by Bayint Naung

You can do something like:

Code:

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#define YOUR_STRING "text base"

const char *text_base = YOUR_STRING;
char text_full[ sizeof(YOUR_STRING)-1 + 10 ];

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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:

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char str[] = "Hello, World!\n";
char copy[sizeof(str) + 10];

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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.

sizeof on a string constant is a pretty neat way of looking at the problem

Code:

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$ cat bar.c
#include <stdio.h>
int main ( ) {
    printf("%zd\n", sizeof("hello world") );
    return 0;
}
$ gcc bar.c
$ ./a.out
12

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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. :)

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

Quote:

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

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No it does not. The null is ignored:
strlen

Specifically:

Quote:

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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:

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Originally Posted by dwks

I guess that means string literals must have type char[N] instead of char* like I thought they did.

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Yes, a string literal is an array, not a pointer, though perhaps const char[N] comes closer.

Quote:

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Originally Posted by jeffcobb

No it does not. The null is ignored:
strlen

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strlen and sizeof are not the same thing ;)

Quote:

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Originally Posted by laserlight

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 ;)

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Thats what I get for replying before becoming fully caffeinated. :)