带参考参数的子功能：它是如何工作的？

Question

I originally show the a program, which have a sub-function with non reference parameter like below:

#include<iostream>
using namespace std;
int inc(int x)
{
    x++;
    return x;
}
main()
{
    n=3;
    inc(n);
    cout<<n;
}

I think that about sub-function and may be it works like below:

when we call inc(n) , after that may be computer will create a var int m; and:

m=n; m++; and finally it is erased.

With sub-function with reference parameter :

#include<iostream>
using namespace std;
int inc(int &x)
{
    x++;
    return x;
}
main()
{
    n=3;
    inc(n);
    cout<<n;
}

I think:

when we call inc(n) , after that may be computer will create a var int &m=n ; m++; return m; and finally it is erased, but n is changed by n+1 .

Why am I asking? There are 2 reasons: 1) My fiend say that : there is not creation int &m , function inc(int &x) works such as : n run into and go out. 2) If there is int &m why everyone always say " inc (int &x) faster inc(int x) "?

Is my understanding of this correct? If not, can you tell me why?

Answer 1

From the compiler point of view, a reference is not far from a pointer. So in

int inc(int &x)
{
    x++;
    return x;
}
...
inc(n);

you only pass a reference (the address of the variable n ) to the function, and the callee variable is actually increased.

Whereas in

int inc(int x)
{
    x++;
    return x;
}
...
inc(n);

you pass by value and you are modifying a local copy.