If I have a declaration like this:
int foo1 (int foo2 (int a));
How can I implement this foo1 function? Like,
int foo1 (int foo2 (int a))
{
// How can I use foo2 here, which is the argument?
}
And how do I call the foo1 function in main? Like:
foo1(/* ??? */);
When you declare a function parameter as a function, the compiler automatically adjusts its type to "pointer to function".
int foo1 (int foo2 (int a))
is exactly the same as
int foo1 (int (*foo2)(int a))
(This is similar to how declaring a function parameter as an array (e.g. int foo2[123]) automatically makes it a pointer instead (e.g. int *foo2).)
As for how you can use foo2: You can call it (e.g. foo2(42)) or you can dereference it (*foo2), which (as usual with functions) immediately decays back to a pointer again (which you can then call (e.g. (*foo2)(42)) or dereference again (**foo2), which immediately decays back to a pointer, which ...).
To call foo1, you need to pass it a function pointer. If you don't have an existing function pointer around, you can define a new function (outside of main), such as:
int bar(int x) {
printf("hello from bar, called with %d\n", x);
return 2 * x;
}
Then you can do
foo1(&bar); // pass a pointer to bar to foo1
or equivalently
foo1(bar); // functions automatically decay to pointers anyway
int foo2[123] it's not exactly the same as a pointer. There is a difference with sizeof.
Perhaps, this simple example could help you :
#include <stdio.h>
int foo1 (int foo2 (int),int i);
int sub_one (int);
int add_one (int);
int main() {
int i=10,j;
j=foo1(sub_one,i);
printf("%d\n",j);
j=foo1(add_one,i);
printf("%d\n",j);
}
int sub_one (int i) {
return i-1;
}
int add_one (int i) {
return i+1;
}
int foo1 (int foo2 (int),int i) {
return foo2(i);
}
Have a look at the following code, it shows how to call functions they way you wanted to call.
#include <stdio.h>
/* Declaration of foo1 . It receives a specific function pointer foo2 and an integer. */
int foo1 (int (*foo2)(int), int a);
int cube(int number)
{
return (number * number * number);
}
int square(int number)
{
return (number * number);
}
int foo1 (int (*foo2)(int), int a)
{
int ret;
/* Call the foo2 function here. */
ret = foo2(a);
printf("Result is: %d\r\n", ret);
return (ret);
}
int main()
{
int a = 3;
foo1(square, a);
foo1(cube, a);
return 0;
}
