Interpretation of a pointer variable declaration

Question

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
– haccks, Commented Mar 9, 2020 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
– Weather Vane, Commented Mar 9, 2020 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). This thread might be useful if you want to see how to write such a function. — vgru
– vgru, Commented Mar 9, 2020 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
– Lundin, Commented Mar 9, 2020 at 10:36

Vlad from Moscow · Accepted Answer · 2020-03-09 10:28:01Z

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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