c++ signatures, pointers

Answer 1

An ampersand in a type declaration indicates a reference type.

int i = 4;
int& refi = i;  // reference to i
int* ptri = &i; // pointer to i

refi = 6;  // modifies original 'i', no explicit dereferencing necessary
*ptri = 6; // modifies through the pointer

References have many similarities with pointers, but they're easier to use and less error-prone if address arithmetic is not needed. Also, unlike pointers, references can't be rebound to 'point' to another object after their initialization. Just ask google for references vs. pointers in C++.

Answer 2

T * f(T & identifier);
This is a function which takes a reference to T and returns a pointer to T.

T & f(T & identifier);
This is a function which takes a reference to T and returns a reference to T.

T f(T & identifier);
This one takes a reference to a T and returns a copy of a T.

void f(T * identifier);
This one takes a pointer to a T and returns nothing.

void f(T & identifier);
This one takes a reference to a T and returns nothing.

void f(T identifier);
This one takes a T by value (copies) and returns nothing.

References behave almost exactly like pointers except a reference is never going to be set to NULL, and a reference is implicitly created and dereferenced for you so you don't need to deal with pointer syntax while calling the function or inside the function.