how to pass value of array from one function to another in c

Everyone has to learn how to use arrays, strings, pointers and functions -- and how to pass arrays and pointers to functions and handle function returns.

When you declare a variable (including an array type), the lifetime for the variable exists only within the scope it is declared unless memory for the variable is allocated dynamically or it is declared as static or otherwise has static storage duration.

Why? Storage for variables declared within a function is created on the function stack which is destroyed (released for reuse) on the function return. (main() is itself a function). Now a function can always return a value of its own type, but it cannot declare a an array of type with automatic storage within the function and return a pointer to the array.

char str[]="something";    /* violates that rule.... */

You have a few options, but they rely on understanding the storage duration for "something" and exactly what it is. let's look at what it appears you are attempting to do:

/* retuning pointer to string literal */
const char *string1 ()
{
    const char *s = "something";

    return s;
}

There are a few subtleties above. First "something" is a string literal and it has static storage duration for the life of the program. So you are assigning the address of "something" that is created in read-only memory that will survive the return of string1. Since a function can always return it's own type (in this case const char * -- chosen for that purpose), you can return the address to "something".

To use s in string2, you must pass s as an argument to string2, or assign the return of string1 to a global pointer (discouraged). Since s points to read-only memory, you should declare the parameter passed to string2 as const char * to signal it cannot be modified. Something similar to:

/* simply use string 's' in string2 */
void string2 (const char *s)
{
    printf ("in string2(), s : %s\n", s);
}

Putting it altogether, you could do something like the following:

#include <stdio.h>

/* retuning pointer to string literal */
const char *string1 ()
{
    const char *s = "something";

    return s;
}

/* simply use string 's' in string2 */
void string2 (const char *s)
{
    printf ("in string2(), s : %s\n", s);
}

int main (void) {

    /* call string 2 passing string1 as the parameter */
    string2 (string1 ());

    return 0;
}

Example Use/Output

$ ./bin/literalreturn
in string2(), s : something

note: you can only get away with returning s in string1 because of the string literal static storage duration. Otherwise, you would need to pass s as a parameter to string1 or dynamically allocate storage for s within string1 and return a pointer to the starting address for the new block of memory.

If you were passing a character array to string1, then you would need to do something similar to the following:

#include <stdio.h>
#include <string.h>

#define MAXC 32     /* constant for array size */

/* anytime you are passing an array, pass the size. */
char *string1 (char *s, size_t max)
{
    if (max < 2) {
        fprintf (stderr, "error: insufficient max size.\n");
        return NULL;
    }

    if (strlen (s) < max - 1)
        strcpy (s, "something");

    return s;
}

/* simply use string 's' */
void string2 (const char *s)
{
    printf ("in string2(), s : %s\n", s);
}

int main (void) {

    char buf[MAXC] = "";  /* declare character array of MAXC chars */

    /* call string 2 passing string1 passing array and size */
    string2 (string1 (buf, sizeof buf));

    return 0;
}

(Since s isn't changed in string2 it is still properly passed as const char *s which allows certain compiler optimizations for your code.)

Look things over. You must understand the subtleties of each example as these concepts underlie much of C. Let me know if you have further questions.