Write a function that will swap two integers using refrences

Question

This is a particularly strange question, but i'm attempting to write a function that swaps the values of two integers without using references or '&'. I don't see how this is even possible. Here's what I have so far.

    void swap (int a, int b)
    {

    int temp;
    temp = a;
    a = b;
    b = temp;

    }

This, normally would be the way that I would do it, but since the integers don't permanently change, I have no idea as to how I would do this without referencing. Any suggestions?

Thanks!

You are passing by value. Meaning your swap function may swap inside but that will have no effect on the caller. — drescherjm, Nov 21 '17 at 17:45
Pointers are *technically* references (in that they reference something), but not in the same way as using `void swap (int &a, int &b)` would be, are they acceptable? — Nick is tired, Nov 21 '17 at 17:45
What is the entire content of the request? It doesn't make sense. — Passer By, Nov 21 '17 at 17:46
@JakeFromSteakFarm Then what is this actually for? I see no benefit in trying to do this when better methods are available — Nick is tired, Nov 21 '17 at 17:47
@JakeFromSteakFarm You can change your signature to `void swap (std::reference_wrapper a, std::reference_wrapper b)` to avoid the `&` or a pointer. — user0042, Nov 21 '17 at 17:51
@user0042 just because a reference gets wrapped doesn't mean its not a reference :P — Nick is tired, Nov 21 '17 at 17:56
@user0042 Unless we're ampersand-golfing what is the point of it? lol — miradulo, Nov 21 '17 at 17:57
@miradulo None of course. Silly questions deserve silly answers, that's all :-) ... — user0042, Nov 21 '17 at 17:58
You can't write your function without references or pointers. The compiler will pass *copies* of the variables to your function. Your function will change the copies but not the original variables. — Thomas Matthews, Nov 21 '17 at 18:48

score -2 · Answer 1 · answered Nov 21 '17 at 19:19

You should correct the question title. It says “using references”. That’s the opposite of what you mean, apparently.

Assuming that:

truly no references and pointers allowed, exluding any wrapper tricks
a runtime swap is what you want – as opposed to compile time template trickery
no classes, because that would trivial

I can think of one utterly horrible solution. Make your ints globals.

ridiculous_swap.hpp

#ifndef RIDICULOUS_SWAP_HPP
#define RIDICULOUS_SWAP_HPP

extern int first;
extern int second;

void swap_ints();

#endif // RIDICULOUS_SWAP_HPP

ridiculous_swap.cpp

int first = 0;
int second = 0;

void swap_ints()
{
    auto tmp = first;
    first = second;
    second = tmp;
}

main.cpp

#include "ridiculous_swap.hpp"
#include <iostream>

int main()
{
    first = 23;
    second = 42;
    std::cout << "first " << first << " second " << second << "\n";
    // prints: first 23 second 42

    swap_ints();
    std::cout << "first " << first << " second " << second << "\n";
    // prints: first 42 second 23
}

It’s not useful for anything, but it does swap two integers without using references or pointers.

score -2 · Answer 2 · answered Nov 21 '17 at 19:46

-2

There is this old trick to swap two integer-like variables without using a temporary variable:

void swap(int& a, int& b)
{
    a ^= b; 
    b ^= a; // b ^ (a ^b)  = a
    a ^= b; // (a ^ b) ^ a = b
}

answered Nov 21 '17 at 19:46

Michaël Roy

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Write a function that will swap two integers using refrences

2 Answers2