Why functions do not give integer overflow

Question

I have wrote functions in C++ and Pascal that give me n-th Fibbonacci number. As expected for large n-values(n>92,because even f(93) > 2^63+1) I was getting incorrect results.
But when I compared them for same n I would get same result in both languages.

This was opposed to my idea that I would get some random number.
I am wondering why I am getting same results and why I didn't get integer overflow in the first place.

Could somebody explain this to me?

Code:

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;

long long fibo(int n){
    long long a1,a2,pom;
    int i=1;
    a1 = 0; a2 = 1;
    while(i<=n){
        pom = a2;
        a2 = a1 + a2;
        a1 =  pom;
        i++;
    }
    return a1;
}

int main(){
    int n;
    cin >> n;
    cout << "Function: "<< setprecision(50) << fibo(n) << endl;
}

Program AddNums(output);
function fibo(n:integer):int64;
    var
        a1,a2,pom:int64;
        i:integer;
    begin
        a1:=0;a2:=1;i:=1;
        while(i<=n)do
            begin
                pom:= a2;
                a2:= a1 + a2;
                a1:= pom;
                inc(i);
            end;
        fibo:=a1;
    end;
var
    n:integer;
begin
     readln(n);
    writeln(fibo(n));
end.

What is integer overflow? I'm afraid there's no such thing in c++. — user0042
– user0042, Commented Dec 10, 2017 at 12:13
It sounds like you don't quite understand how integer overflow works – when an overflow happens, the result is incorrect but not random. — JJJ
– JJJ, Commented Dec 10, 2017 at 12:15
I get same results for the given C and Pascal. As the answers here will say, modulo arithmetic is not guaranteed for one or both of your compilers. Information about the compiler and platform helps in such questions. — Potatoswatter
– Potatoswatter, Commented Dec 10, 2017 at 12:23
Using modulo arithmetic you can calculate the result. Which one deviates? — Marco van de Voort
– Marco van de Voort, Commented Dec 10, 2017 at 12:52

Sergey Kalinichenko · Accepted Answer · 2017-12-10 12:21:14Z

7

A result that is undefined is not necessarily random. When you perform the same computation on the same platform using the same initial condition, you will arrive at the same result, even if it is incorrect.

In your case Pascal and C++ programs use the same underlying hardware with the same representation of int64 and long long, and they instruct that hardware to perform the same sequence of mathematical operations on numbers that start off the same. Hence, they do arrive at the same number, representing the end result of this sequence of operations.

The result is still undefined, because if you run the same computation on a different platform, or even on the same platform but with different compiler settings, you may get an entirely different incorrect result.

answered Dec 10, 2017 at 12:21

Sergey Kalinichenko

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3 Comments

Potatoswatter Over a year ago

"they instruct that hardware to perform the same sequence of mathematical operations on numbers that start off the same" — It would be very surprising if the compilers produced the same binary code. In all likelihood the results differ because the compilers took different approaches at some point.

Sergey Kalinichenko Over a year ago

@Potatoswatter I tried to word this very carefully to avoid giving a wrong impression that compilers from different languages would somehow arrive at identical binary code. The sequence of mathematical operations (additions of a1 and a2) would be the same, while the rest of the code (i.e. loop control, increments, and assignments) around the "payload" sequence of additions would almost certainly be different.

Potatoswatter Over a year ago

Different results can only come from different calculations, from different reasoning applied by the compiler. For example, in printing the check for a negative result could be inlined and removed.

Ahmed Salah · Accepted Answer · 2017-12-10 12:21:03Z

1

First of all it's (2^63 - 1) not (2^63 + 1).

when a number overflows it go for the next value in the closed numbers "loop".

Assume it starts with 0 and ends with 8, so if your var value is 8 and you ++ it so it will be 0.

answered Dec 10, 2017 at 12:21

Ahmed Salah

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