Does void(*) in C++ mean anything?

Question

I'm trying understand C++ function pointer syntax. In Eclipse on Linux when I typed:

void(*);

It highlighted the statement with a message saying syntax error, but it let me compile it and the program ran. Then on Visual Studio I tried it and it won't compile, saying "Expected an expression". However what's strange is that when I do:

std::vector<void(*)> myVector;

It compiles fine on Visual Studio. Also on a couple of online compilers void(*); on its own works fine. I know that:

void (*)();

... is a function pointer and..

void();

... is a function signature, which is why you can do:

std::function<void()> func;

I'm having a lot of trouble understanding function pointer syntax.

Thanks.

Answer 1

Remember that parentheses can be used to change the precedence of certain things. That's why you have the parentheses around the asterisk in void (*)() because it's very different from void *().

In the case of void(*) the parentheses are such precedence-changing parentheses. But they are not needed. The type void(*) is void*, plain and simple.

The context where you use it is important though.