What does `*&` in a function declaration mean?

Question

I wrote a function along the lines of this:

void myFunc(myStruct *&out) {     out = new myStruct;     out->field1 = 1;     out->field2 = 2; } 

Now in a calling function, I might write something like this:

myStruct *data; myFunc(data); 

which will fill all the fields in data. If I omit the '&' in the declaration, this will not work. (Or rather, it will work only locally in the function but won't change anything in the caller)

Could someone explain to me what this '*&' actually does? It looks weird and I just can't make much sense of it.

Answer 1

The & symbol in a C++ variable declaration means it's a reference.

It happens to be a reference to a pointer, which explains the semantics you're seeing; the called function can change the pointer in the calling context, since it has a reference to it.

So, to reiterate, the "operative symbol" here is not *&, that combination in itself doesn't mean a whole lot. The * is part of the type myStruct *, i.e. "pointer to myStruct", and the & makes it a reference, so you'd read it as "out is a reference to a pointer to myStruct".

The original programmer could have helped, in my opinion, by writing it as:

void myFunc(myStruct * &out) 

or even (not my personal style, but of course still valid):

void myFunc(myStruct* &out) 

Of course, there are many other opinions about style. :)

Answer 2

In C and C++, & means call by reference; you allow the function to change the variable. In this case your variable is a pointer to myStruct type. In this case the function allocates a new memory block and assigns this to your pointer 'data'.

In the past (say K&R) this had to be done by passing a pointer, in this case a pointer-to-pointer or **. The reference operator allows for more readable code, and stronger type checking.