C++11 auto declaration with and without pointer declarator

Question

What\'s the difference between the types of bar1 and bar2?

int foo = 10;
auto bar1 = &foo;
auto *bar2 = &foo;

Answer 1

In this specific example both bar1 and bar2 are the same. It's a matter of personal preference though I'd say that bar2 is easier to read.

However, this does not hold true for references as seen in this example:

#include 
using namespace std;

int main() {
    int k = 10;
    int& foo = k;
    auto bar = foo; //value of foo is copied and loses reference qualifier!
    bar = 5; //foo / k won't be 5
    cout << "bar : " << bar << " foo : " << foo << " k : " << k << endl;
    auto& ref = foo;
    ref = 5; // foo / k will be 5
    cout << "bar : " << bar << " foo : " << foo << " k : " << k;
    return 0;
}