Error “initializer element is not constant” when trying to initialize variable with const

Question

I get an error on line 6 (initialize my_foo to foo_init) of the following program and I\'m not sure I understand why.

typedef struct foo_t {
    int a, b, c;         


        

Answer 1

This is a bit old, but I ran into a similar issue. You can do this if you use a pointer:

#include <stdio.h>
typedef struct foo_t  {
    int a; int b; int c;
} foo_t;
static const foo_t s_FooInit = { .a=1, .b=2, .c=3 };
// or a pointer
static const foo_t *const s_pFooInit = (&(const foo_t){ .a=2, .b=4, .c=6 });
int main (int argc, char **argv) {
    const foo_t *const f1 = &s_FooInit;
    const foo_t *const f2 = s_pFooInit;
    printf("Foo1 = %d, %d, %d\n", f1->a, f1->b, f1->c);
    printf("Foo2 = %d, %d, %d\n", f2->a, f2->b, f2->c);
    return 0;
}

Answer 2

gcc 7.4.0 can not compile codes as below:

#include <stdio.h>
const char * const str1 = "str1";
const char * str2 = str1;
int main() {
    printf("%s - %s\n", str1, str2);
    return 0;
}

constchar.c:3:21: error: initializer element is not constant const char * str2 = str1;

In fact, a "const char *" string is not a compile-time constant, so it can't be an initializer. But a "const char * const" string is a compile-time constant, it should be able to be an initializer. I think this is a small drawback of CLang.

A function name is of course a compile-time constant.So this code works:

void func(void)
{
    printf("func\n");
}
typedef void (*func_type)(void);
func_type f = func;
int main() {
    f();
    return 0;
}

Answer 3

In C language, objects with static storage duration have to be initialized with constant expressions, or with aggregate initializers containing constant expressions.

A "large" object is never a constant expression in C, even if the object is declared as const.

Moreover, in C language, the term "constant" refers to literal constants (like 1, 'a', 0xFF and so on), enum members, and results of such operators as sizeof. Const-qualified objects (of any type) are not constants in C language terminology. They cannot be used in initializers of objects with static storage duration, regardless of their type.

For example, this is NOT a constant

const int N = 5; /* `N` is not a constant in C */

The above N would be a constant in C++, but it is not a constant in C. So, if you try doing

static int j = N; /* ERROR */

you will get the same error: an attempt to initialize a static object with a non-constant.

This is the reason why, in C language, we predominantly use #define to declare named constants, and also resort to #define to create named aggregate initializers.

Answer 4

Just for illustration by compare and contrast The code is from http://www.geeksforgeeks.org/g-fact-80/ /The code fails in gcc and passes in g++/

#include<stdio.h>
int initializer(void)
{
    return 50;
}

int main()
{
    int j;
    for (j=0;j<10;j++)
    {
        static int i = initializer();
        /*The variable i is only initialized to one*/
        printf(" value of i = %d ", i);
        i++;
    }
    return 0;
}

Answer 5

It's a limitation of the language. In section 6.7.8/4:

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

In section 6.6, the spec defines what must considered a constant expression. No where does it state that a const variable must be considered a constant expression. It is legal for a compiler to extend this (6.6/10 - An implementation may accept other forms of constant expressions) but that would limit portability.

If you can change my_foo so it does not have static storage, you would be okay:

int main()
{
    foo_t my_foo = foo_init;
    return 0;
}