[Solved] C++ primer plus list:10.9 The “top” pointer problem

Question

The ampersand in front of &top->topval(...) returns the address of the object returned from topval. This is because the member access operator -> has higher precedence than the address-of operator. So when we assume that Stock::topval returns a reference to another Stock object, we can get the address of that one and bind it to Stock *top.

Accepted Answer

The ampersand in front of &top->topval(...) returns the address of the object returned from topval. This is because the member access operator -> has higher precedence than the address-of operator. So when we assume that Stock::topval returns a reference to another Stock object, we can get the address of that one and bind it to Stock *top.