Passing by Reference in C

Question

If C does not support passing a variable by reference, why does this work?

#include <stdio.h>

void f(int *j) {
  (*j)++;
}

int main() {
  int i = 20;
  int *p = &i;
  f(p);
  printf("i = %d\n", i);

  return 0;
}

Output:

$ gcc -std=c99 test.c
$ a.exe
i = 21

Accepted Answer

Because you're passing the value of the pointer to the method and then dereferencing it to get the integer that is pointed to.

Answer 2

That is not pass-by-reference, that is pass-by-value as others stated.

The C language is pass-by-value without exception. Passing a pointer as a parameter does not mean pass-by-reference.

The rule is the following:

A function is not able to change the actual parameters value.

(The above citation is actually from the book K&R)

Let's try to see the differences between scalar and pointer parameters of a function.

Must Read

Scalar variables

This short program shows pass-by-value using a scalar variable. param is called the formal parameter and variable at function invocation is called actual parameter. Note incrementing param in the function does not change variable.

#include <stdio.h>

void function(int param) {
    printf("I've received value %d\n", param);
    param++;
}

int main(void) {
    int variable = 111;

    function(variable);
    printf("variable %d\m", variable);
    return 0;
}

The result is

I've received value 111
variable=111

Passing by Reference in C

Output:

19 Answers

Must Read

Scalar variables

Maya Lin-Takahashi