Question

In the C program below, I don’t understand why buf[0] = ‘A’ after I call foo. Isn’t foo doing pass-by-value?

#include <stdio.h>
#include <stdlib.h>

    void foo(char buf[])
    {
      buf[0] = 'A';
    }

    int main(int argc, char *argv[])
    {
      char buf[10];

      buf[0] = 'B';
      printf("before foo | buf[0] = %c\n", buf[0]);
      foo(buf);
      printf("after foo | buf[0] = %c\n", buf[0]);

      system("PAUSE"); 
      return 0;
      }

output:

before foo | buf[0] = 'B' 
after foo | buf[0] = 'A'

Answer 1

void foo(char buf[])

is the same as

void foo(char* buf)

When you call it, foo(buf), you pass a pointer by value, so a copy of the pointer is made.

The copy of the pointer points to the same object as the original pointer (or, in this case, to the initial element of the array).

C does not have pass by reference semantics in the sense that C++ has pass by reference semantics. Everything in C is passed by value. Pointers are used to get pass by reference semantics.