Why is it allowed to initialize static variable with non-const here?
I read this one . The first answer by @Andrei T says that
A "large" object is never a constant expression in C, even if the object is declared const. Objects constructed using constants (of any type) are not constants in C 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 to do
static int j = N; /* ERROR */
you will get the same error: trying to initialize a static object with a volatile
I agree with his answer. I also tried a simple example like gcc 4.8.2 and 4.9.2 and it gives compiler errors as I expected:
#include <stdio.h>
int main(void)
{
const int a=5;
static int b=a;
printf("%d",b);
}
But when I tried it on ideone.com
, it compiles and works fine and gives the expected output. See the demo here . Also, in 13.12 IDE codeblocks (gcc 4.7.1) this program works fine. So is this a compiler error or a gcc extension? What combination of compiler options ideone
are used under the hood? So how and why does it compile to ideone
? What is the reason?
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This is because ideon probably calls gcc
with an option -O
(optimization level 1). This applies to older versions too gcc
(mine is 4.4.7):
$ gcc -ansi main.c
main.c: In function ‘main’:
main.c:6: error: initializer element is not constant
$ gcc -ansi -O main.c
$ echo $?
0
Interestingly, with -pedantic
it works correctly again and a diagnostic message is present (tested only with 4.4.7,see Keith's comment sub>):
gcc -ansi -pedantic -O main.c
main.c: In function ‘main’:
main.c:6: error: initializer element is not constant
$ echo $?
1
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