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

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: stackoverflow.com/q/5573310/274261

ArjunShankar
– ArjunShankar

2015-11-30 14:43:01 +00:00
Commented Nov 30, 2015 at 14:43 — ArjunShankar
– ArjunShankar, Commented Nov 30, 2015 at 14:43

Sourav Ghosh · Accepted Answer · 2015-11-30 14:43:40Z

4

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.

edited Nov 30, 2015 at 14:43

answered Nov 30, 2015 at 14:38

Sourav Ghosh

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