Doing exercise 6.22 in C Primer, a function swapping two pointers:

void swap(int*& a, int*& b) {
    auto temp = a;
    a = b;
    b = temp;
}

int main()
{
    int a = 15;
    int b = 4;
    int* p1 = &a;
    int* p2 = &b;
    cout << "p1 address: " << p1 << "; p1 value: " << *p1 << "\np2 address: " << p2 << "; p2 value: " << *p2 << endl;
    swap(p1, p2);
    cout << "p1 address: " << p1 << "; p1 value: " << *p1 << "\np2 address: " << p2 << "; p2 value: " << *p2 << endl;
}

Not sure if it's correct, but it works as expected, the values and addresses are swapped. I hover in MS Visual Studio over the a inside swap function:

It tells me the type is int*&, but changing the definition of temp to int*& temp = a breaks the program:

p1 address: 004FFE00; p1 value: 15
p2 address: 004FFDF4; p2 value: 4
p1 address: 004FFDF4; p1 value: 4
p2 address: 004FFDF4; p2 value: 4

What is the correct type here? What's the logic behind MS Visual Studio type tooltip?

CodePudding user response：

As mentioned in the comments, temp is of type int*, not int*&. If you want it to be a reference to a pointer, you can declare the type explicitly or use auto&. However, what you really want is int* anyway. Think about how you would write this if you were swapping two ints instead:

void swap(int& a, int& b) {
    int temp = a;
    a = b;
    b = temp;
}

Here, you want temp just to be an int because you just want it to store the value of a so you can copy it to b. There is no need for it to be a reference.

As to your other question, what is happening when temp is a reference?

• a = 15
• b = 4
• Call swap()
• temp references a
• a references b (so temp references b)
• b references temp (so b references b)
• Both a and b now reference the value 4

To see this in action, walk through the function in debug mode and examine the pointer addresses and values at each step.

CodePudding user response：

swap function can be changed like this.

template<typename T>
void swap(T& a, T& b) {
    auto temp = a;
    a = b;
    b = temp;
}

First your question is what type of int*&. in my function, your question is same what type of T&. T& mean is reference T. so int*& mean is reference int*.

this is why P2 is not changed when use int*&. but if you use auto, temp type is int*. this is why it works so well.