N
Nelu
Arguments are always passed by value in C function calls, meaning that local copies of the values of the arguments are passed to the routines. Changes that you may make to the arguments in the function are made only to the local copies of the arguments. Passing by reference means you can change the value of the argument outside the scope of your function, which doesn't happen. If you have a function like: void foo(char *a) you cannot change the value of the argument, but you can change the content of the memory that the address points to. Pass-by-xxxx refers to the way the arguments are passed to the function and, in C, it's always by value. The fact that the function can change the value at the address your pointer points to is just a side-effect and has nothing to do with how the arguments are passed.
I don't remember exatly and I don't have a copy of the book right now, but I think K&R says that arguments are always passed by value in C function calls. No one will say anything about pass by reference because there's no such thing in C.
JAVA is a languae where everything's a reference. JAVA also passes the arguments to functions by value *only* (if you don't believe me check the documentation), but you can change the elements of an array that's passed to a function and it doesn't mean that array was passed by reference.
If I remember correctly in C++:
void foo(int &i) {
i=2;
}
....
int i=0;
foo(i)
=> i=2
type is int, you send int, int is changed in the function. That's pass by reference.
in C:
void foo(int *i) {
*i=2;
}
....
int i=0;
foo(&i);
=> i=2
type is int, you send a pointer to int, int is changed in the function. The functions doesn't tell you that it needs an int to change an int. It tells you it needs a pointer to int that it can't change.
Although you can consider &int a reference to an int, the function doesn't ask for a reference but for a type which is called "pointer to int" and what you send is the value of a pointer to int type.
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