I am getting ERROR when I am running this program at the line
static int b = a; //error : initializer element is not constant
Can not understand why?
#include <stdio.h>
// #include <setjmp.h>
int main()
{
int a = 5;
static int b = a;
return 0;
}
Apart from the other reasons stated in other answers here, please see the below statement in the Standard.
The C Standard says this in Point-4 (Section 6.7.8 Initialization):
All the expressions in an initializer for an object that has static storage duration
shall be constant expressions or string literals.
Additionally, as to what is a constant expression, it says in Section 6.6 Constant Expressions as below:
A constant expression can be evaluated during translation rather than runtime, and
accordingly may be used in any place that a constant may be.
In C (unlike C++), the initializer for any object with static storage duration - including function statics - must be constant expressions. In your example a is not a constant expression so the initialization is not valid.
C99 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.
Static variable is always global in the sense it is not on any thread's stack, and it is not important if its declaration is inside a function or not.
So the initialization of the global variable b is performed during program start-up, before any function (including main) gets called, i.e. no a exists at that time, because a is local variable which gets its memory place on stack after the function (here main) is called.
Hence you really cannot expect the compiler to accept it.
Following from Als's answer ...
// This is really crappy code but demonstrates the problem another way ....
#include <stdio.h>
int main(int argc, char *argv[])
{
static int b = argc ; // how can the compiler know
// what to assign at compile time?
return 0;
}
//comments have been standard C for the past 13 years, and they were widely supported before then. Even MSVC supports them.