Reputation: 25
i find this in Pointers on C
int f[](); /* this one is illegal */
and:
int (* f [])(); /* this one legal. */
i really want know what's the usage of the second one.
thank you.
Upvotes: 2
Views: 4873
Reputation: 51326
int f[](); /* this one is illegal */
That's trying to declare an array of functions, which is impossible.
int (* f [])(); /* this one NOT legal, despite what the OP's post says. */
That's trying to declare an array of function pointers, which would be perfectly legal (and sensible) if the array size were specified, e.g.:
int (* f [42])(); /* this one legal. */
EDIT: The type int (* f [])()
can be used as a function parameter type, because for function parameter types, array-to-pointer conversion takes place immediately, meaning we don't ever need to specify the dimension of the innermost array of a (possibly multidimensional) array:
void some_func(int (* f [])()); /* This is also legal. */
Upvotes: 1
Reputation: 9234
int f[]();
// this is illegal because of you can't create array of functions . It's illegal in C
But second is legal
int (* f [])();
It says that f is an array of function pointers returning int
and taking unspecified number of arguments
Upvotes: 1
Reputation: 106483
The second example is quite valid, if you use initialization block. For example:
#include <stdio.h>
int x = 0;
int a() { return x++ + 1; }
int b() { return x++ + 2; }
int c() { return x++ + 3; }
int main()
{
int (* abc[])() = {&a, &b, &c};
int i = 0,
l = sizeof(abc)/sizeof(abc[0]);
for (; i < l; i++) {
printf("Give me a %d for %d!\n", (*abc[i])(), i);
}
return 0;
}
Upvotes: 2
Reputation: 11954
I'm not sure if the second example is legal, since the size of the function array is not known, but what it is supposed to be is an array of function pointers, and here is a possible example of usage if the size would be known:
int a()
{
return 0;
}
int main(int argc ,char** argv)
{
int (* f [1])();
f[0] = a;
}
Upvotes: 1