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;
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.
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
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
We change the piece of code slightly. param
is a pointer now.
#include
void function2(int *param) {
printf("I've received value %d\n", *param);
(*param)++;
}
int main(void) {
int variable = 111;
function2(&variable);
printf("variable %d\n", 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.
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
void function2(int *param) {
printf("param's address %d\n", param);
param = NULL;
}
int main(void) {
int variable = 111;
int *ptr = &variable;
function2(ptr);
printf("ptr's address %d\n", 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.