Is void *function() a pointer to function or a function returning a void*?

The function has the return type void *.

void *function();

So I always prefer in such cases to separate the symbol * from the function name like

void * function();

And as Jarod42 pointed to in a comment you can rewrite the function declaration in C++ using the trailing return type like

auto function() -> void *;

If you want to declare a pointer to function then you should write

void ( *function )();

where the return type is void Or

void * ( *function )();

where the return type void *.

Or a pointer to function that returns pointer to function

void * ( *( *function )() )();

It is a function returning a pointer to void.

Think of your declaration this way:

void *(function());

This would be a function returning void (or nothing):

void (*function2)();

Think of the above declaration this way:

void ((*function2)());

A much easier way to write these is to use typedefs:

typedef void *function_returning_void_pointer();
typedef void function_returning_nothing();

function_returning_void_pointer function;
function_returning_nothing *function2;

This generally eliminates the confusion around function pointers and is much easier to read.

Whenever I'm unsure about C syntax issues, I like to use the cdecl utility (online version) to interpret for me. It translates between C syntax and English.

For example, I input your example of void *foo() and it returned

declare foo as function returning pointer to void

To see what the other syntax would look like, I input declare foo as pointer to function returning void and it returned

void (*foo)()

This gets particularly useful when you have multiple levels of typecasts, stars, or brackets in a single expression.