Why can't I create an array with size determined by a global variable?

Question

Why does the array a not get initialized by global variable size?

#include

int size = 5;

int main()
{
    int a[si         


        

Answer 1

The compiler needs to know the size of the array while declaring it. Because the size of an array doesn't change after its declaration. If you put the size of the array in a variable, you can imagine that the value of that variable will change when the program is executed. In this case, the compiler will be forced to allocate extra memory to this array. In this case, this is not possible because the array is a static data structure allocated on the stack. I hope that this will help.

Answer 2

In C99, 6.7.8/3:

The type of the entity to be initialized shall be an array of unknown size or an object type that is not a variable length array type.

6.6/2:

A constant expression can be evaluated during translation rather than runtime

6.6/6:

An integer constant expression shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants, sizeof expressions whose results are integer constants, and floating constants that are the immediate operands of casts.

6.7.5.2/4:

If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type; otherwise, the array type is a variable length array type.

a has variable length array type, because size is not an integer constant expression. Thus, it cannot have an initializer list.

In C90, there are no VLAs, so the code is illegal for that reason.

In C++ there are also no VLAs, but you could make size a const int. That's because in C++ you can use const int variables in ICEs. In C you can't.

Presumably you didn't intend a to have variable length, so what you need is:

#define size 5

If you actually did intend a to have variable length, I suppose you could do something like this:

int a[size];
int initlen = size;
if (initlen > 5) initlen = 5;
memcpy(a, (int[]){1,2,3,4,5}, initlen*sizeof(int));

Or maybe:

int a[size];
for (int i = 0; i < size && i < 5; ++i) {
    a[i] = i+1;
}

It's difficult to say, though, what "should" happen here in the case where size != 5. It doesn't really make sense to specify a fixed-size initial value for a variable-length array.

Answer 3

size is a variable, and C does not allow you to declare (edit: C99 allows you to declare them, just not initialize them like you are doing) arrays with variable size like that. If you want to create an array whose size is a variable, use malloc or make the size a constant.

Answer 4

It looks like that your compiler is not C99 Compliant...speaking of which, which compiler are you using? If it's gcc you need to pass the switch '-std=c99'.... if you are using a pre-C99 compiler, that statement is illegal, if that's the case, do this:

int main() { 
   int a[5]={1,2,3,4,5}; 
   printf("%d",a[0]); 
   return 0; 
}

In pre-C99 standard compilers, use a constant instead of a variable.

Edit: You can find out more about the C99 standard here... and here....

Answer 5

#include<stdio.h> 

/* int size=5; */
#define size 5 /* use this instead*/
/*OR*/
int a[size]={1,2,3,4,5};  /* this*/

int main() 
{ 
    int a[size]={1,2,3,4,5}; 
    printf("%d",a[0]); 
    return 0; 
} 

int size means that size is a variable and C does not allow variablesize arrays.

I am using VS2008 where using

const int size=5;  

allows

int a[size]={1,2,3,4,5}; 

Answer 6

You don't need to tell the compiler what size the array is if you're giving an initializer. The compiler will figure out the size based on how many elements you're initializing it with.

int a[] = {1,2,3,4,5};

Then you can even let the compiler tell you the size by getting the total size of the array in bytes sizeof(a) and dividing it by the size of one element sizeof(a[0]):

int size = sizeof(a) / sizeof(a[0]);