What is the result and example of invoking this function using "Call by name" in java or c++?

Question

I just took a quiz for my programming languages class and came across this question:

Assuming call by name, what are the results of following code:

public class MyClass {

    static int i = 1;

    static float f(int x, int i){
        int s = 0;
        for(i = 0; i <3; i++){
            s = s + x;
        }
        return s;
    }

    static void g(){
        int[] a= {10,30,50};
        int[] b = {20,40,60};
        System.out.println(f(i, i));
        i = 1;
        System.out.println(f(a[i],i));
        i = 1;
       System.out.println(f((a[i] *b[i]), i));
    }  



 public static void main(String[]args){
     g();

 }   
}

The above code is exactly as presented on my quiz, but it was on paper at the time so I did the calculations manually and got the results 3.0, 90.0, and 3600.0. After running the code on my own computer, the results are the same as what I calculated.

However, the question was multiple choice and the closest option available was 3.0, 90.0, and 4400.0. This leads me to assume that the words "Call by name" change how the function f is called on the line System.out.println(f((a[i] *b[i]), i));

Can someone please explain how 4400.0 is possible in Java or C++?

That wouldn't even compile in C++, so maybe 4400.0 is a compiler error message. — juanchopanza, Dec 01 '17 at 22:40
You've only provided a snippet for the java portion of your question. Why make mention of c++ at all here? — François Andrieux, Dec 01 '17 at 22:40
I know it won't compile in C++, but using a call by name function in c++ as an example would suffice and help me understand. — Remixt, Dec 01 '17 at 22:41
Fine, I'll add an example in c++ as well then. I'm simply asking for an example in either language... — Remixt, Dec 01 '17 at 22:41
https://stackoverflow.com/questions/3331143/example-of-call-by-name — François Andrieux, Dec 01 '17 at 22:44
I love the plug in chug! Is it possible you misread what the function has in it? — Jake Freeman, Dec 01 '17 at 22:46
Neither Java nor C++ use call-by-name, so it's hardly surprising that running the code as Java would give the right answer. — Ken Wayne VanderLinde, Dec 01 '17 at 22:50
@JakeFreeman I still have a blank copy the quiz in front of me right now. — Remixt, Dec 01 '17 at 22:52
@Remixt -- Sorry, but what the heck is "call by name" with respect to C++? C++ uses call-by-value and call-by-reference in parameter passing. Nothing more, nothing less. — PaulMcKenzie, Dec 01 '17 at 23:21
@PaulMcKenzie The point is to imagine the language (Java or C++) with the only change being that it uses call-by-name instead of the normal semantics. It's not a watertight concept, but it's good enough for a quiz question. — Ken Wayne VanderLinde, Dec 01 '17 at 23:45

Ken Wayne VanderLinde · Accepted Answer · 2017-12-01T22:59:01.773

3

In call-by-name, the expressions used as arguments are substituted directly in the called function. They are not evaluated before calling the function, but are substituted at each place the parameter occurs in the function.

This means the last call to f is equivalent to this:

int s = 0;
for(i = 0; i <3; i++){
    s = s + (a[i] * b[i]);
}
return s;

Here is the full sequence of steps inside the last call to f (I hope this is clear):

s = 0
i = 0

s = s + (a[i] * b[i])
  = 0 + (a[0] * b[0])
  = 0 + (10 * 20)
  = 200

i = 1

s = s + (a[i] * b[i])
  = 200 + (a[1] * b[1])
  = 200 + (30 * 40)
  = 200 + 1200
  = 1400

i = 2

s = s + (a[i] * b[i])
  = 1400 + (a[2] * b[2])
  = 1400 + (50 * 60)
  = 1400 + 3000
  = 4400

So you can see that s is 4400 when f returns.

edited Dec 01 '17 at 22:59

answered Dec 01 '17 at 22:53

Ken Wayne VanderLinde

18,915
3
47
72

That would return the same value for all prints. – user1803551 Dec 01 '17 at 22:57
@user1803551 How do you figure that? – Ken Wayne VanderLinde Dec 01 '17 at 22:59
I ran your code, that's how. Now that you expanded on the answer it's clear that you are actually doing something different than what you wrote in respect to the code in the question. – user1803551 Dec 01 '17 at 23:04
As stated, the code snippet only applies to the **last call** to `f`, not to all the calls. So you can't have run my code for all the calls. – Ken Wayne VanderLinde Dec 01 '17 at 23:06
Best if you post the code that you ran and gave you these results. – user1803551 Dec 01 '17 at 23:10
I did it by hand. – Ken Wayne VanderLinde Dec 01 '17 at 23:12
Well, running it in Java, as you previously suggested, doesn't agree with your calculations. – user1803551 Dec 01 '17 at 23:13
Then you didn't actually run my code. Check here for an example that prints `s`: https://ideone.com/SmRn2j – Ken Wayne VanderLinde Dec 01 '17 at 23:19
As I said, the code you linked is not the same as replacing `f` in the last call. You replaced the structure of the code as well. – user1803551 Dec 01 '17 at 23:25
All you have to do is take the body of `f`, and replace each occurrence of `x` with `(a[i] * b[i])`. That's all I'm going to say on this - it's pretty straightforward. – Ken Wayne VanderLinde Dec 01 '17 at 23:31

What is the result and example of invoking this function using "Call by name" in java or c++?

1 Answers1