Passing expression as a parameter in Call by reference

Question

All,

When we are passing an expression as a parameter, how does the evaluation occur? Here is a small example. This is just a pseudocode kind of example:

f (x,y)
{
    y = y+1;
    x = x+y;
}
main()
{
    a = 2; b = 2;
    f(a+b, a)
    print a;
}

When accessing variable x in f, does it access the address of the temp variable which contains the result of a+b or will it access the individual addresses of a and b and then evaluate the value of a+b

Please help.

Regards, darkie15

Answer 1

Somewhat language dependent, but in C++

f(a+b, a)

evaluates a + b and and pushes the result of evaluation onto the stack and then passes references to this value to f(). This will only work if the first parameter is of f() is s const reference, as temporary objects like the result of a + b can only be bound to const references.