How does function overloading work in the following case?

The following code compiles and runs but I'm not sure what exactly is going on at a lower level. Doesn't a reference just store the address of the object being refer开发者_Python百科enced? If so, both test functions are receiving an address as a parameter? Or is the C++ implementation able to differentiate between these types in some other way?

int main() {
    int i = 1;
    cout << test(i) << endl;
}

char test(int &i) {
    return 'a';
}

char test(int *i) {
    return 'b';
}

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As int& and int* are distinct types and i can be treated as a int& but not as a int*, overload resolution is absolutely unambiguous here.

It doesn't matter at this point that references are just a somewhat cloaked kind of pointer. From a language point of view they are distinct types.

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References in C++ are more akin to an alias than a pointer. A reference is not a seperate variable in itself, but it is a new "name" for an exisiting variable. In your example the first test would get called because you are passing an integer to the function. A pointer is a seperate variable that holds the address of another variable so for the second function to be called you would have to call test with a pointer. Like so.. test(&i); While a tad confusing the operator & gets the address of a variable while a variable declared with an & like int &i declares a reference.

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you code only matches with char test(int&i) since you are passing an int& to the function and that can not be converted to int*