Evaluation of variable through pass-by-name

Question

I am learning parameter passing.

I've been working on a question about parameter passing, but I don't understand one problem.

I searched the Internet, but there was no example of a parameter being expression.

If it's a expression, can I know how to evaluate pass-by-name?

The problem is as follows.

function func(int a, int b, int c)
  begin
    a := b + c;
    b := c + 1; 
    print a, b, c;
 end
function main
  begin
    int x := 10; 
    int y := 5; 
    ink z := 15;
    func(x, y, y + z);
    print x, y, z;
  end

score 0 · Answer 1 · answered Nov 24 '19 at 06:26

From Wikipedia: (note that "call by name" and "pass by name" mean the same thing.)

Call by name is an evaluation strategy where the arguments to a function are not evaluated before the function is called—rather, they are substituted directly into the function body (using capture-avoiding substitution) and then left to be evaluated whenever they appear in the function. If an argument is not used in the function body, the argument is never evaluated; if it is used several times, it is re-evaluated each time it appears.

So for your function, we should make the substitutions a → x, b → y and c → y + z accordingly:

  begin
    x := y + (y + z);
    y := (y + z) + 1; 
    print x, y, (y + z);
  end

I've put brackets around the places where c is substituted, to make clear that when the expression is "copy/pasted" into the function, it doesn't change the precedence of the other operations; for example, 3 * c would be equivalent to 3 * (x + y), not 3 * x + y.

The fact that c is replaced by the expression y + z rather than a simple variable, doesn't create any problems here, because c never appeared on the left side of an assignment statement.

Evaluation of variable through pass-by-name

1 Answers1