function not returning anything but its data member (char array) can be accessed in main function. How is this possible?

Question

#include <stdio.h>

void foo(char a[],char b[])
{
    a[0]='t';
    a[1]='e';
    a[2]='s';
    a[3]='t';
    a[4]='_';
    a[5]='a';
    a[6]=NULL;

    b[0]='t';
    b[1]='e';
    b[2]='s';
    b[3]='t';
    b[4]='_';
    b[5]='b';
    b[6]=NULL;

}

int main()
{
    char a[10],b[10];
    foo(a,b);
    printf("%s \n",a);    //outputs "test_a"
    printf("%s \n",b);    //outputs "test_b"
}

How does a[] and b[] char arrays in main() get its value, when foo() is neither using pointers nor returning any value when it is called through main()?

Correct me if I am wrong in saying "foo() is not using pointers".

See this: http://stackoverflow.com/q/5573310/274261 – ArjunShankar Nov 30 '15 at 14:43 — ArjunShankar, Nov 30 '15 at 14:43

Sourav Ghosh · Accepted Answer · 2015-11-30T14:43:40.000

Function foo() is actually using pointers, somehow. When arrays are passed to functions, it decays to the pointer to the first element of the array.

As it happens,

void foo(char a[],char b[])

and

void foo(char *a,char *b)

are perfectly interchangeable.

In your case, when called with parameters like foo(a,b);, from the function you can change the content of the memory area pointed by a and b. You don't need to return anything.

function not returning anything but its data member (char array) can be accessed in main function. How is this possible?

1 Answers1