error: initializer element is not a compile-time constant

Question

I have been looking for answers but could not find anything to make this code run. I get av[1] highlighted by the compiler in the main function when declaring:

static char const *str = av[1];

Here is the code I tried to run with gcc:

#include <stdio.h>
#include <stdlib.h>

char    *ft_strjoin(char const *s1, char const *s2);

void    fct(char **av)
{
    static char const *str = av[1];
    str = ft_strjoin(av[1], av[1]);
    printf("%s\n", str);
}

int main(int ac, char **av)
{
    fct(&av[1]);
    fct(&av[1]);
    fct(&av[1]);
    fct(&av[1]);
    fct(&av[1]);
    fct(&av[1]);
}

I found this interesting but I still don't get it and don't know how to run this code.

Possible dup: [Error “initializer element is not constant” when trying to initialize variable with const](http://stackoverflow.com/questions/3025050/error-initializer-element-is-not-constant-when-trying-to-initialize-variable-w) — P.P, Dec 24 '15 at 09:21
I was asking this because I am working on a project. The function I need to create has to be prototyped like this: int get_next_line(int const fd, char ** line); I need to get the buffer from any kind of file director and return the pointer to the beginning of each line (of a text file for instance). I wrote a code storing each of the lines in a linked list so its not the problem. To be clearer, the function can be inserted in a loop and it will change at every loop (until EOF) the pointer to the beginning of the line. How would could the static variable look like in my code? — ziKmouT, Dec 24 '15 at 09:42

score 7 · Answer 1 · edited Dec 24 '15 at 11:18

7

Quoting C11, §6.7.9, Initialization

All the expressions in an initializer for an object that has static or thread storage duration shall be constant expressions or string literals.

In your code,

static char const *str = av[1];

av[1] is not a compile time constant value (i.e., not a constant expression). Hence the error.

You need to remove static from str to avoid the issue.

edited Dec 24 '15 at 11:18

fuz

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answered Dec 24 '15 at 09:20

Sourav Ghosh

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score 2 · Answer 2 · answered Dec 24 '15 at 09:15

2

static variables need to be initialised with a compile time constants (constant literals). av[1] will be calculated at runtime and that's why you are getting the error message.

answered Dec 24 '15 at 09:15

haccks

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score 0 · Answer 3 · answered Dec 24 '15 at 09:34

You could simulate that behaviour by writing:

static const char *str;
static bool already;
if ( !already )
{
    str = av[1];
    ++already;
}

However this would be redundant compared to the solution of:

const char *str;

because you immediately overwrite that value anyway with the return value of your function.

(Also you pass the same argument in every call, so even if you used str, it still doesn't need to be static).

error: initializer element is not a compile-time constant

3 Answers3

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