afunction(int & value) {within the function, value is actually the external variable and any changes to value affect the external variable.
// 8_2.cpp :In example 8_2, testfunction is called twice, on variables a and b. The result is
//
#include "stdafx.h"
#include <iostream>
void testfunction(int & value) {
value++;
}
int main(int argc, char* argv[])
{
int a=10;
int b=20;
testfunction(a) ;
testfunction(b) ;
std::cout << "a = " << a << " b = "<< b << std::endl;
return 0;
}
a = 11 b = 21Showing that a and b were both changed by the calls to testfunction.
Const Parameters
If you put the word const before the type then any attempt to change the parameter will cause a compilation error. You've indicated to the compiler that it is a const, so the compiler will prevent it being altered. But as we've just discovered a way to alter variables passed in to functions, why try to prevent that?The Answer : for single variables like ints or even floats, there is little point in making them const and reference. Might as well just pass them by value. . For larger types such as structs and arrays though it makes a lot of sense, because if passed by value they have to be copied. Pass by reference doesn't make a copy, it just passes in a pointer. This is far faster to execute.
On the next page : Learn about references and return values.
