What is the syntax in this typedef statement doing?

Question

I ran across a line of code that looks like the following:

typedef Foo* (*CREATE_BAR)(uint32_t);

How exactly does this work? What is happening in this code?

It's at times like these when http://cdecl.ridiculousfish.com/ is helpful... Of course you'd have to already know that `Foo` was a type and swap it for a primitive type, like: [`int* (*CREATE_BAR)(uint32_t)`](http://cdecl.ridiculousfish.com/?q=int*+%28*CREATE_BAR%29%28uint32_t%29) — Jonathan Mee, Jun 14 '16 at 21:37

score 3 · Accepted Answer · answered Jun 14 '16 at 21:30

3

It's a function pointer type named CREATE_BAR which accepts a uint32_t argument and returns a Foo*. It could hold a pointer to any such function.

answered Jun 14 '16 at 21:30

score 2 · Answer 2 · answered Jun 14 '16 at 21:30

2

It is a type for a pointer on function returning Foo*, and taking uint32_t

In c++11, it would be

using CREATE_BAR = Foo* (*)(uint32_t);

answered Jun 14 '16 at 21:30

Jarod42

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What is the syntax in this typedef statement doing?

2 Answers2