Why do I need an ampersand in a function parameter?

Please tell me what role the ampersand plays in the parameter of this function because the program works correctly both with and without it

#include <iostream>
#include <vector>

using namespace std;
typedef vector<int>  T_summands;  

void  print_summands(int n_start,const T_summands& summands)
{
    cout << n_start << "=";
    for(T_summands::const_iterator  summands_it = summands.begin(); summands_it != summands.end(); ++summands_it)
    {
        cout << *summands_it << (summands_it != summands.end() - 1 ? " + " : "\n");
    }
}
Author: Harry, 2020-02-27
Source

1 answers

For understanding - well, consider that when passing by reference, you're just passing a pointer, but you don't have to dereference it all the time.

Type

void f(int * x) { *x = *x * 2; }
...
int y = 4;
f(&y);

The same via the link - 

void f(int& x) { x = x * 2; }
...
int y = 4;
f(y);

This is, of course, an approximation, but it is suitable for understanding.

When you pass without reference, by value - a copy is created. Something like this:

void f(int x) { x = x * 2; }

It's as if (inaccurate! just for understanding!!)

void f(int * y) 
{
    int x;
    x = *y;

    x = x * 2;
}

At the "household" level-the value passed by is not changes what is passed by reference-changes the passed variable itself.

void f(int& x) { x = x * 2; }
...
int y = 4;
f(y);  // Теперь y == 8


void f(int x) { x = x * 2; }
...
int y = 4;
f(y);  // y == 4...
 3
Author: Harry, 2020-02-27 15:04:59