Passing by reference in C

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 = %dn", i);

  return 0;
}

Output

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

---

因为您将指针的值传递给方法，然后将其解引用以获取指向的整数。

---

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.

---

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

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 %dn", param);
    param++;
}

int main(void) {
    int variable = 111;

    function(variable);
    printf("variable %dm", variable);
    return 0;
}

The result is

I've received value 111
variable=111

Illusion of pass-by-reference

We change the piece of code slightly. param is a pointer now.

#include <stdio.h>

void function2(int *param) {
    printf("I've received value %dn", *param);
    (*param)++;
}

int main(void) {
    int variable = 111;

    function2(&variable);
    printf("variable %dn", variable);
    return 0;
}

The result is

I've received value 111
variable=112

That makes you believe that the parameter was passed by reference. It was not. It was passed by value, the param value being an address. The int type value was incremented, and that is the side effect that make us think that it was a pass-by-reference function call.

Pointers - passed-by-value

How can we show/prove that fact? Well, maybe we can try the first example of Scalar variables, but instead of scalar we use addresses (pointers). Let's see if that can help.

#include <stdio.h>

void function2(int *param) {
    printf("param's address %dn", param);
    param = NULL;
}

int main(void) {
    int variable = 111;
    int *ptr = &variable;

    function2(ptr);
    printf("ptr's address %dn", ptr);
    return 0;
}

The result will be that the two addresses are equal (don't worry about the exact value).

Example result:

param's address -1846583468
ptr's address -1846583468

In my opinion this proves clearly that pointers are passed-by-value. Otherwise ptr would be NULL after function invocation.

---

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

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