pointer to an array of pointer to functions

Question

I would like f to point to an array of f_pointer_type. So what did i miss here ? I'm getting an error 'lvalue required as left operand of assignment'. What does it mean ?

#include <stdio.h>
typedef char* (*f_pointer_type)();
char * retHello()
{
    return "hello";
}
int main()
{
    char (* ( *f())[])(); //Declare f to be a pointer to an array of pointers to function that gets nothing and return a char.
    f_pointer_type funcs[3]; //an array of pointers to function that gets nothing and returns a pointer to char
    f = &funcs; //ERROR : lvalue required as left operand of assignment
    return 0;
}

Why don't you use typedef in f variable definition? f_pointer_type** f; - — Alex F
– Alex F, Commented Aug 7, 2013 at 11:50
I guess f is interpreted as some crazy function forward declaration. — Alex F
– Alex F, Commented Aug 7, 2013 at 11:51

Some programmer dude · Accepted Answer · 2013-08-07 12:07:20Z

3

If you read about the clockwise/spiral rule you will see that f is actually a function returning a pointer to an array of pointers to functions returning char*. In other words it's a function prototype and not a variable declaration.

Try this instead:

f_pointer_type (*f)[];

edited Aug 7, 2013 at 12:07

answered Aug 7, 2013 at 11:56

Some programmer dude

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Lundin · Accepted Answer · 2013-08-07 11:58:44Z

1

An array of function pointers is written as char* (*f[])(). You have the parenthesis wrong.

Though it is of course better to always use typedef as in the f_pointer_type funcs[3] example.

answered Aug 7, 2013 at 11:58

Lundin

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Lidong Guo · Accepted Answer · 2013-08-07 11:58:54Z

1

You define f as a function pointer. f is pointer to a function who has void params and return char *

answered Aug 7, 2013 at 11:58

Lidong Guo

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