Question

Consider the following code C++ like program:

int i, j, arr[5]; //arr is an array starting at index 0
void exchange(int x, int y) {
     int temp:= x;
     x:= y;
     y:= temp;
}
main(){
  for (j = 0; j < 5; j++) arr[j]:= j;
  i:= 1;
  exchange(i, arr[i+1]);
  output(i, arr[2]);   //print i and arr[2]
}

What is the output of the code if both parameters in function swapping are passed by:

a- value?

b- reference?

c- value-result?

Answer 1

a) output: 1 2

That's because in call by value, parameter supplied by the caller is copied into the callee's formal parameter and no modification happens to callers parameters.

b) output: 2 1

That's because, in call by reference, parameters supplied by the caller are referenced and can be modified in the callee.

c) output: 2 1

In case of 'call by value-result' actual parameter supplied by the caller is copied into the callee's formal parameter; Then when the function is run the formal parameter could be modified anywhere inside the function body but only when the function exits the modified formal parameter is then copied back to the caller.

So, in this case, is it similar to pass by reference, the only difference is that the values are modified only after the function exits.

C++ Does not support Pass-by-value-result, however, it can be simulated. To do so you create a copy of the variables, pass them by reference to your function, and then set your original values to the temporary values.