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 

Answer 1

In C, Pass-by-reference is simulated by passing the address of a variable (a pointer) and dereferencing that address within the function to read or write the actual variable. This will be referred to as "C style pass-by-reference."

Source: www-cs-students.stanford.edu

Answer 2

Your example works because you are passing the address of your variable to a function that manipulates its value with the dereference operator.

While C does not support reference data types, you can still simulate passing-by-reference by explicitly passing pointer values, as in your example.

The C++ reference data type is less powerful but considered safer than the pointer type inherited from C. This would be your example, adapted to use C++ references:

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

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

  return 0;
}

Answer 3

Because there is no pass-by-reference in the above code. Using pointers (such as void func(int* p)) is pass-by-address. This is pass-by-reference in C++ (won't work in C):

void func(int& ref) {ref = 4;}

...
int a;
func(a);
// a is 4 now

Answer 4

'Pass by reference' (by using pointers) has been in C from the beginning. Why do you think it's not?

Answer 5

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 6

Because you're passing a pointer(memory address) to the variable p into the function f. In other words you are passing a pointer not a reference.

Answer 7

You're passing a pointer(address location) by value.

It's like saying "here's the place with the data I want you to update."

Answer 8

You're not passing an int by reference, you're passing a pointer-to-an-int by value. Different syntax, same meaning.

Answer 9

In C, to pass by reference you use the address-of operator & which should be used against a variable, but in your case, since you have used the pointer variable p, you do not need to prefix it with the address-of operator. It would have been true if you used &i as the parameter: f(&i).

You can also add this, to dereference p and see how that value matches i:

printf("p=%d \n",*p);

Answer 10

p is a pointer variable. Its value is the address of i. When you call f, you pass the value of p, which is the address of i.

Answer 11

No pass-by-reference in C, but p "refers" to i, and you pass p by value.