c++ - 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. :)