Question

#include<stdio.h>
#include<string.h>
char *y;
y=(char *)malloc(40); // gives an error here
int main()
{
    strcpy(y,"hello world");
}

---

error: conflicting types for 'y'
error: previous declaration of 'y' was here
warning: initialization makes integer from pointer without a cast
error: initializer element is not constant
warning: data definition has no type or storage class
warning: passing arg 1 of `strcpy' makes pointer from integer without cast

Now the real question is, can’t we make the dynamic memory allocation globally? Why does it show an error when I use malloc globally? And the code works with no error if I put malloc statement inside the main function or some other function. Why is this so?

#include<stdio.h>
#include<string.h>
char *y;
int main()
{
    y=(char *)malloc(40); 
    strcpy(y,"hello world");
}

Answer 1

You can’t execute code outside of functions. The only thing you can do at global scope is declaring variables (and initialize them with compile-time constants).

malloc is a function call, so that’s invalid outside a function.

If you initialize a global pointer variable with malloc from your main (or any other function really), it will be available to all other functions where that variable is in scope (in your example, all functions within the file that contains main).

(Note that global variables should be avoided when possible.)