int& and int difference

[DarkParadox]

What is the difference between int& and int inside a function?
Example:
void x(int a,int b)
and
void x(int& a,int& b)
Please explain how it works

[mbozzi]

http://www.cplusplus.com/doc/tutorial/functions/

[Flaze07]

Taken from that article

Arguments by reference do not require a copy. The function operates directly on (aliases of) the strings passed as arguments, and, at most, it might mean the transfer of certain pointers to the function. In this regard, the version of concatenate taking references is more efficient than the version taking values, since it does not need to copy expensive-to-copy strings.

Hope yer not too lazy to read it

[Handy Andy]

Hello DarkParadox,

After all that good reading it boils down to pass by value or pass by reference.

In the first example void x(int a,int b) This is pass by reference and "a" and "b" represent a copy of the original variables used to call the function. As a copy when the function ends so do the copies and the original calling variables know nothing of any changes that the function made. A simple program to illustrate this:

#include <iostream>

void x(int a, int b);

int main()
{
	int row{ 0 };
	int col{ 0 };
	char end{ ' ' };

	x(row, col);

	std::cout << "\n row = " << row << "\n col = " << col << std::endl;
	
	//std::cin >> end;
	std::cin.get();
	return 0;
}

void x(int a, int b)
{
	a++;
	b++;
}

Notice how "row" and "col" do not change.

The example void x(int& a,int& b) is pass by reference. In this case "a" and "b" become new names for "row" and "col" and any changes in in the function to "a" and "b" are reflected back in main for the variables "row" and "col".

#include <iostream>

void x(int& a, int& b);

int main()
{
	int row{ 0 };
	int col{ 0 };
	char end{ ' ' };

	x(row, col);

	std::cout << "\n row = " << row << "\n col = " << col << std::endl;

	std::cin >> end;

	return 0;
}

void x(int& a, int& b)
{
	a++;
	b++;
}

Notice how "row" and "col" have changed in main.

Advantages to pass by reference are that sometimes it is easier than making a copy of something large, less overhead than making the copy, and you have direct access to the variable that was passed in. Since a function can only return one of anything passing variables by reference allows you to change more than one variable with the function and when the function ends changes are made to the original variables before the local variables of the function are lost.

Should you not want to change the original variables just make them "const" in the function parameters.

That cover most of the major points, but I feel I have missed some of the minor points.

Hope that helps,

Andy

P.S. be sure to run the programs.

[keskiverto]

Current C++ has also rvalue references: void x( T&& a, T&& b );

[Flaze07]

Nice explanation Andy...
I hope DarkParadox will Understand because I told him with the same method as you do and he didnt understand... (DarkParadox is my friend)

[Handy Andy]

@Flaze07,

Your welcome.

Andy