C Programming - Pointers - Discussion

I would like to add one more thing.
#include<stdio.h>
int *check( int, int);
int main()
{
int *c;
c = check(10, 20);
printf("%d\n", c);
return 0;
}
int *check(int i, int j)
{
int *p, *q;
p = &i;
q = &j;
if(i >= 45)
return (p);
else
return (q);
}


This removes the 'static' keyword, then also will not give the correct output as the pointer p is local to the function and on the return of control from the function it gets destroyed - returning a garbage value.

Static variables compiler stores in static/global (data section) as per storage class.

Here we are having pointers present in the code section pointing to static variables which are present in other storage sections giving error.

Also, we cannot use static variables as function parameters in C.

@Avinash.

* means that's this function return pointer to integer value.

#include<stdio.h>
int *check( int, int);
int main()
{
int *c;
c = check(10, 20);
printf("%d\n", *c);
return 0;
}
int *check( int i, int j)
{
int *p, *q;
p = &i;
q = &j;
if(i >= 45)
return (p);
else
return (q);
}

Here we get the output is 20.

Can anyone tell me what is the reason why we can not pass static variables in a function?

Can any one explain that, what happens when the parameters are non-static?

Static variables can not be declared inside method. These are always declared inside the main method.

@ M@C:

The static variable is initialized to "1" as global variable and in local block variable is initialized to "10". Since both are same variables when the local block finishes and comes to the function call it will take value of global variable and it is printed as output.

I am using visual c++, in that I can able to pass static variable as the arguments, it is showing only warning.

Because static variable can not be change, It is initialized at one time. So answer is 1.