当C ++中的变量不在全局范围内并在自定义函数中使用时，如何设置它的值？

Question

The variables are not accepting the values I'm entering in my C++ program. I must avoid global variables and use only local variables. And the function returns nothing, so I have used "void" instead of "int" type. Same thing happening when I use strings or any type of custom function. Here is the example to explain my problem:

#include <iostream>


void sum (int a, int b, int c);

int main (void)
{
    int a = 0, b = 0, c = 0;

    sum (a, b, c);

    std::cout << a << b << c;

    return 0;
}


void sum (int a, int b, int c) // It doesn't have to be the same variable name :)
{
    std::cout << "Enter value of a:\n";
    std::cin  >> a;
    std::cout << "Enter value of b:\n";
    std::cin  >> b;
    std::cout << "Enter value of c:\n";
    std::cin  >> c;

    a = b+c;
}

Answer 1

通过引用传递 ：

void sum (int &a, int &b, int &c)

Answer 2

You can use pass by reference (or by pointer, for the educational purpose):

void sum (int& a, int& b, int& c);
void sum (int* a, int* b, int* c);

int main (void)
{
   int a = 0, b = 0, c = 0;
   sum (a, b, c);
   std::cout << a << b << c;

   a = 0, b = 0, c = 0;
   sum (&a, &b, &c);
   std::cout << a << b << c;

   return 0;
}


void sum (int& a, int& b, int& c)
{
   std::cout << "Enter value of a:\n";
   std::cin  >> a;
   std::cout << "Enter value of b:\n";
   std::cin  >> b;
   std::cout << "Enter value of c:\n";
   std::cin  >> c;
   a = b+c;
}

void sum (int* a, int* b, int* c)
{
   std::cout << "Enter value of a:\n";
   std::cin  >> *a;
   std::cout << "Enter value of b:\n";
   std::cin  >> *b;
   std::cout << "Enter value of c:\n";
   std::cin  >> *c;
   *a = *b + *c;
}

Answer 3

Arguments can be passed by value or by reference to a function.
When you pass an argument by a value (which is what you are doing), a separate copy is created of the variable and is stored in a different memory location.
When you pass by reference (using a pointer), same memory location is referenced. Basically in your code you are creating a separate copy of the variable referenced by the same name and modifying that copy and expecting the changes in the original. The solution is to use pointers.