How to use a function argument of a function in the function implementation?

Question

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(/* ??? */);

You mean you want to pass a function pointer as an argument to a function? See [Function pointer as an argument](https://stackoverflow.com/questions/1789807/function-pointer-as-an-argument) — lurker, Mar 05 '19 at 17:04
What kind of syntax is this even? It seems to compile, but it's not a function pointer. — mkrieger1, Mar 05 '19 at 17:05
@lurker but in my declaration, there is no a pointer. I have already see that topic. — javac, Mar 05 '19 at 17:06
@melpomene , I can't see a pointer `*`, how could you say that? — javac, Mar 05 '19 at 17:09

score 8 · Accepted Answer · answered Mar 05 '19 at 17:09

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

If you have a Parameter `int foo2[123]` it's not exactly the same as a pointer. There is a difference with `sizeof`. — harper, Mar 05 '19 at 17:53
(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).) I don't think this is a technically correct statement. Though it may appear . — Mazhar, Mar 05 '19 at 17:55

score 0 · Answer 2 · answered Mar 05 '19 at 17:24

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);
}

Mazhar · Answer 3 · 2019-03-05T17:48:28.820

-1

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;
}

edited Mar 05 '19 at 17:48

answered Mar 05 '19 at 17:26

Mazhar

575
5
20

A code dump is not an answer, especially if the code doesn't even contain `int foo1 (int foo2 (int a));` or address OP's question. – melpomene Mar 05 '19 at 17:30
@melpomene Would be please remove that vote ? Only if you think it's not wrong answer. Otherwise, let it be there. – Mazhar Mar 05 '19 at 17:50

How to use a function argument of a function in the function implementation?

3 Answers3