C++ what is the difference of type * and type *& in function

Question

I have two function signatures in C++

void printArray(int* arrayPtr);
void printArray(int*& arrayPtr);

I understand the 1st function. It says the function takes in a arrayPtr argument which is of type that's a pointer pointing to an integer.

Both function signature works, but I have a hard time understanding the 2nd signature(*&) and what benefits it offers?

Ask yourself, " What if I want the function to change the address the pointer's pointing at?" — user4581301, Jun 09 '18 at 03:53
Thinking on this a bit more, we may have to back up a bit. Are you familiar with references in C++? — user4581301, Jun 09 '18 at 03:57

score 2 · Accepted Answer · answered Jun 09 '18 at 04:01

2

It's exactly the same as type versus type&; the first is a value and the second is a reference. The fact that type is a pointer doesn't change that.

answered Jun 09 '18 at 04:01

Pete Becker

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I can't imagine there's no existing answer to what a reference to a pointer is – xaxxon Jun 09 '18 at 04:02
This is a pretty clear answer. Putting `&` at the end of any type has a very consistent effect. – Drew Dormann Jun 09 '18 at 04:04

C++ what is the difference of type * and type *& in function

1 Answers1