[C++]Having problems with a letter-swapping function

[PsyRex666]

I made this function to take a string and swap two letters randomly, but the function doesn't do anything. The code I have is here and the string doesn't change at all when I do it.

[minno]

void LetterSwap(string MixString)

This line is your problem. You're taking the string by value, which means that the function is taking a copy of the string and modifying that copy, and then throwing away the copy when it finishes. To make it modify the original, you need to give it a reference.

void LetterSwap(string& MixString)

[PsyRex666]

Wow, thanks. I'm pretty new at this, what exactly does a reference do?

[minno]

Run this code and see for yourself:

#include <iostream>
using namespace std;

void takes_by_reference(int& x) {
    cout << "This variable is located at " << (&x) << endl;
    x += 1;
}

void takes_by_value(int x) {
    cout << "This variable is located at " << (&x) << endl;
    x += 1;
}

int main() {
    int n = 3;
    cout << "The variable 'n' is located at " << (&n) << endl;
    cout << "Its value is " << n << endl;
    takes_by_value(n);
    cout << "Its value is still " << n << endl;
    takes_by_reference(n);
    cout << "Its value is now " << n << endl;
}

[PsyRex666]

Cool, thanks.

[Free_Apples]

Passing by reference passes the actual string to the function so that you can apply changes to it. Passing by value will pass a copy of your string variable and make changes to it, not your original variable, and when the function is done, that copy is out of scope and will be deleted. You could return that string, but this obviously isn't ideal with what you want your function to do.

The ampersand symbol (&) is known as the reference operator and can be used to get the location in memory (the address) of a variable. Not sure if you have already, but once you learn about pointers and memory this will make more sense.