在 foreach 循环中将变量声明为 i vs &i

Question

Why do these loops give the same output:

#include<iostream>
#include<vector>
using namespace std;
int main()
{
    vector<int> ar = {2, 3 ,4};
    for(auto i: ar) //this line changes in the next loop
        cout<<i<<" ";

    cout<<"\n";

    for(auto &i: ar) //i changed to &i
        cout<<i<<" ";
}

They both give the same output:

2 3 4

2 3 4

When declaring the foreach loop variable, shouldn't adding ampersand make the variable take the references of the values in the array, and printing i make it print the references. What is happening here?

By print the references I meant something like this code prints:

for(auto i: ar)
    cout<<&i<<" ";

Output:

0x61fdbc 0x61fdbc 0x61fdb

Answer 1

The ampersand operator has many purposes. These two of the many are hard to recognize, namely the & infront of a variable in a declaration ( such as int& i ( or int &i , same thing ) ), and the & infront of a variable not in a declaration, such as cout << &i .

Try these and you will get a better understanding.

for (auto i : ar)
    cout << i << " "; // 2 3 4 // element of ar
    
for (auto &i : ar)
    cout << i << " "; // 2 3 4 // element of ar

for (auto i : ar)
    cout << &i << " "; // address of local variable i (probably same address)
    
for (auto &i: ar)
    cout << &i << " "; // address of elements of ar (increasing addresses)

Answer 2

The results are the same because in the first loop you copy the variable's value in a new variable i and print its value. (Additional RAM allocated)

In the second loop you access the value of the current element from the memory by assigning its address to i . (No additional RAM allocated)

On the other side:

cout<<&i<<" ";

causes printing the address of i .