C++ Address-of Operator &

Question

[1]

char f1(char c) {
return c == 'z' ? 'a' : c + 1;
}
char f2(char &c) {
c = f1(c);
return c;
}
int main() {
char x = 'x';
cout << f2(x);
cout << f2(x);
cout << f2(x);
}

Output: yza

f1 function changes the value of its argument as well as f2 function.
Do we pass the x variable itself to f2 above or do we pass its address? I thought we should pass the address if we want our variable to change its value.
What is the difference between these two functions' declarations?

[2]

Imagine we have a similar function with such a declaration:
char f3(char *c); that changes its arguments' value as well.
What would be the result of these actions:
a) char x = 'x';
cout << f3(x);
b) char x = 'x';
cout << f3(&x);
?

How should I interpret receiving variables/pointers/addresses as arguments by functions?
Thanks in advance.

Answer 1

Emitted Assembly code in debugging mode on VSC++ (without involving cout object for the sake of simplicity)

char x = 'x';
mov byte ptr [x], 78h // Load the memory address indicated by label x to 78h (120 or 'x')

f2(x);
lea eax, [x] // Place the effective addres of x (pointer to mem location indicated by x) into accumulator register
push eax // Push the addres of x into program stack.
call f2 (11B2550h) // obvious enough, calling the f2(by address of x)

Live version for GCC: https://godbolt.org/g/Rdeeks

In case of f3(char *), calling as f3(x) is a compilation error. The legal form would be as

char f3(char *c) {
*c = f1(*c);
return *c;
}

int main() {
char x = 'x';
char *xp = &x;

// method 1
f3(&x);

// method 2
f3(xp);
}

Pointer and reference are pretty much identical in what they do. The big advantage of having a reference parameter is to passing large objects or containers witch otherwise cause deep copying and lead to performance penalty.