Interpretation of a pointer variable declaration

Question

1) [see here](https://cdecl.org/?q=int+*%28cmp%29%28char*%2C+char*%29%3B) 2) just writing style. — Blaze, Mar 09 '20 at 09:56
1. `int *(cmp)(char*, char*);` is simply `int *cmp(char*, char*);`: `cmp` is a function that takes two arguments and both of type `char *` and returns a pointer to `int`. 2. No difference. It's matter of choice. — haccks, Mar 09 '20 at 10:02
In the second I prefer `char *ptr2;` because in `char* ptr2, ptr3;` the `ptr3` is not a pointer, it is `char ptr3;` — Weather Vane, Mar 09 '20 at 10:10
Since it's called `cmp` (compare?), I'd be inclined to think the signature is actually `int (*cmp)(char*, char*);`, i.e. a pointer to a function which returns an `int`, as a parameter for a `qsort`-style *comparison* function (as [explained here](https://cdecl.org/?q=int+%28*cmp%29%28char*%2C+char*%29%3B)). [This thread](https://stackoverflow.com/q/27284185/69809) might be useful if you want to see how to write such a function. — vgru, Mar 09 '20 at 10:30
@WeatherVane Simply don't write multiple declarations on a single line - doing so is a well-known safety hazard. Not just for pointers; I've also seen things like `int a,b,c = 0;` when the programmer meant to set all items to zero, not just `c`. — Lundin, Mar 09 '20 at 10:36
@Groo Good remark, though in that case it should have been `int (*cmp)(const void*, const void*)`. — Lundin, Mar 09 '20 at 10:37

Vlad from Moscow · Accepted Answer · 2020-03-09T10:28:01.900

this

int *(cmp)(char*, char*);

is a declaration of a function that has the return type int * and two parameters of the type char *.

You may enclose a declarator in parentheses. So the above function declaration can be also rewritten like

int * ( (cmp)(char*, char*) );

The both declarations are equivalent to

int * cmp(char*, char*);

A declaration of a pointer to such a function will look like

int * ( *p_cmp )(char*, char*) = cmp;

There is no difference between these declarations

char* ptr1;

char *ptr1;

char * ptr1;

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