Definition of global variables using a non constant initializer

Question

#include <stdio.h>

int i=10;
int j=i;
int main()
{
    printf("%d",j);
}

I get an error stating that initialization element is not a constant? What is the reason behind this?

i is a variable, not a constant. And C does not allow non-constant initializers for global or static variables. C is different from C++ in this respect. — Jonathan Leffler
– Jonathan Leffler, Commented Dec 9, 2012 at 6:29
@user963472: In C(unlike C++) const declarations do not produce constant expressions. — Alok Save
– Alok Save, Commented Dec 9, 2012 at 6:39
@Vignesh_dino: const doesn't mean constant; it means read-only. (And const doesn't always imply constant in C++; consider const int r = rand();.) — Keith Thompson
– Keith Thompson, Commented Dec 9, 2012 at 7:36

Alok Save · Accepted Answer · 2012-12-09 06:28:35Z

13

What is the reason behind this?

C(unlike C++) does not allow initialization of global values with non constant values.

C99 Standard: Section 6.7.8:

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

answered Dec 9, 2012 at 6:28

Alok Save

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2 Comments

Vignesh_dino Over a year ago

I saw in the internet that variables with static storage are allocated at load time. Has this got anything to do with my above question.

Alok Save Over a year ago

@user963472: I do not understand your question. Can you please elaborate?

askmish · Accepted Answer · 2012-12-09 07:07:11Z

You could try using:

int i=10;
int j=0;

int main()
{
   j=i;//This should be the first statement in the main() and you achieve the same functionality as ur code
   return 0;
}

The only true C way is to initialize it at runtime. Although in C++ your code will work fine, without any compilation errors.

The C standard clearly prohibits initialization of global objects with non-constant values. The Section 6.7.8 of the C99 standard says:

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

The definition of an object with static storage duration is in section 6.2.4:

An object whose identifier is declared with external or internal linkage, or with the storage-class specifier static has static storage duration. Its lifetime is the entire execution of the program and its stored value is initialized only once, prior to program startup.

AnT stands with Russia · Accepted Answer · 2012-12-09 08:31:13Z

1

The idea behind this requirement is to have all static storage duration object initialized at compile time. The compiler prepares all static data in pre-initialized form so that it requires no additional initializing code at run time. I.e. when the compiled program is loaded, all such variables begin their lives in already initialized state.

In the first standardized version of C language (C89/90) this requirement also applied to aggregate initializers, even when they were used with local variables.

void foo(void)
{
  int a = 5;
  struct S { int x, y; } s = { a, a }; /* ERROR: initializer not constant */
}

Apparently the reason for that restriction was that the aggregate initializers were supposed to be built in advance in the pre-initialized data segment, just like global variables.

edited Dec 9, 2012 at 8:31

answered Dec 9, 2012 at 7:29

AnT stands with Russia

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Comments

sushant goel · Accepted Answer · 2012-12-11 14:46:03Z

0

Use this:-

int i=10,j=1;
int main()
{
  printf("%d",j);
}

Though it is a minor change but it will work

edited Dec 11, 2012 at 14:46

answered Dec 9, 2012 at 18:52

sushant goel

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