CODEWITHSUNDEEP
c
user8669406
user8669406

Reputation: 21

Deciphering a Confusing Function Prototype

I have the following function prototype:

char *(*scan)(char *, char *, char *, char *, int , int);

scanleft() is a function and has the type of static char *.

When I try to compile, I found that there is a pointer type mismatch between scan and scanleft.

if ((subtype & 1) ^ zero) scan = scanleft; else scan = scanright;

What does the prototype of scan() mean?

Upvotes: 0

Views: 71

Answers (2)

paxdiablo
paxdiablo

Reputation: 881373

char *(*scan)(char *, char *, char *, char *, int , int);

This defines a function pointer scan, which points to a function that accepts four char * and two int parameters, and returns a char *.

That means you can do things like:

char *scanLeft(char *p1, char *p2, char *p3, char *p4, int p5, int p6) {
    // do something
}
char *scanRight(char *p1, char *p2, char *p3, char *p4, int p5, int p6) {
    // do something else
}
:
scan = scanLeft;  char *xyzzy = scan("a", "b", "c", "d", 271828, 314159);
scan = scanRight; char *plugh = scan("a", "b", "c", "d", 271828, 314159);

and the two calls via the scan function pointer will go to different functions.

Upvotes: 0

melpomene
melpomene

Reputation: 85767

scan is not a function; it has no prototype.

scan is a pointer to a function (taking a pointer to char, pointer to char, pointer to char, pointer to char, int, and int) and returning a pointer to char.

According to your description, scanleft isn't a function either; it's a pointer to char.

The only prototype in your question is this:

char *, char *, char *, char *, int , int

... which doesn't look particularly confusing to me. It's 6 straightforward parameters.

Upvotes: 2

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