what happens here: typedef int (*ptr) (void) in .h file C

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I have a piece of C code and don't understand what happens here:

typedef int (*ptr) (void *ptr2, const char *name);

What I do understand is the typedef int (*ptr) part, but what happens in the second()? I've seen some questions where it was the other way around: typedef void (*ptr) (int), is this similar or different (and how)? I'm not the best at C, so I thought maybe *ptr now points to a function where *ptr2 and *name are declared or *ptr now points to *ptr2 and *name?

It would be great if someone could explain this to me. Thanks in advance!

function-pointers declaration typedef implicit-conversion c
1 Answers

If you have for example a function declaration like

int f( void *ptr2, const char *name );

(as it is seen the function type is int( void *, const char * )) then a pointer to the function will look like

int ( *pf )( void *, const char * ) = f;

and the type of the pointer pf is int ( * )( void *, const char * ). That is the pointer pf now contains the address of the function f.

To introduce an alias for this function pointer type you can write

typedef int (*ptr) (void *ptr2, const char *name);

In this case the above declaration of the pointer pf will look like

ptr pf = f;

that is the declaration of the pointer is simplified.

Pay attention to that the function name used as an initializer of the pointer is implicitly converted to a pointer to the function. That is you could write

ptr pf = &f;

but due to the implicit conversion it is enough to write

ptr pf = f;
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