Passing by reference in C

Question

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

#include 

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

int main() {
  int i = 20;         


        

Answer 1

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;
}