What is the returned value? - pass-by-reference

Question

In a language that passes parameters by reference, given the following function:
int function g(x, y) {
x = x + 1;
y = y + 2;
return x + y;
}
If i = 3, and g(i,i) is called, what is value returned? I thought it is 9, is this correct?

Answer 1

If it's pass-by-reference (your original question was C but C doesn't have pass-by-reference and the question has changed since then anyway, so I'll answer generically), it's probably the case that x and y will simply modify the variables that are passed in for them. That's what a reference is, after all.
In this case, they're both a reference to the same variable i, so your sequence is likely to be:
i = i + 1; // i becomes 4.
i = i + 2; // i becomes 6.
return i + i; // return i + i, or 12.
You can see this in operation with the following C (using pointers to emulate pass-by-reference):
pax$ cat qq.c
#include <stdio.h>
int g(int *x, int *y) {
*x = *x + 1;
*y = *y + 2;
return *x + *y;
}
int main (void) {
int i = 3;
int rv = g (&i, &i);
printf ("Returned: %d\n", rv);
return 0;
}
pax$ gcc -o qq qq.c ; ./qq
Returned: 12
Your result of 9 seems to be assuming that the references are distinct from one another, such as in the following code:
#include <stdio.h>
int g(int *x, int *y) {
*x = *x + 1;
*y = *y + 2;
return *x + *y;
}
int main (void) {
int i1 = 3, i2 = 3;
int rv = g (&i1, &i2);
printf ("Returned: %d\n", rv);
return 0;
}
(this does output 9) but that's not usually the case with reference types.