Question

As I asked and answered in this post. I have the following example code.

#include <stdio.h>

char foo()    { return 'a'; }
char bar()    { return 'b'; }
char blurga() { return 'c'; }
char bletch() { return 'd'; }

char (*gfunclist[])() = {foo, bar, blurga, bletch};

char (*(*x())[])()
{
  static char (*funclist[4])() = {foo, bar, blurga, bletch};
  return funclist;
}

int main() 
{
  printf("%c\n",gfunclist[0]());

  char (**fs)();
  fs = x();
  printf("%c\n",fs[1]()); 
}

My questions are

• Why the

  return funclist (with "warning: return from incompatible pointer type")

  and

  return &funclist

  both works?

• I get warning at the line 21 (fs = x();) of

  warning: assignment from incompatible pointer type

  . How to remove this warning?

ADDED

With AndreyT’s help. I could get the following code that doesn’t have the warnings.

#include <stdio.h>

char foo()    { return 'a'; }
char bar()    { return 'b'; }
char blurga() { return 'c'; }
char bletch() { return 'd'; }

char (*gfunclist[])() = {foo, bar, blurga, bletch};

char (*(*x())[])()
{
  static char (*funclist[4])() = {foo, bar, blurga, bletch};
  return &funclist;
}

int main() 
{
  printf("%c\n",gfunclist[0]());

  char (*(*fs)[4])();
  fs = x();
  printf("%c\n",(*fs)[1]()); 
}

And this is less messy code with the help from peoro.

typedef char (*funptr)();

funptr gfunclist[] = {foo, bar, blurga, bletch};

funptr* x()
{
  static funptr funclist[4] = {foo, bar, blurga, bletch};
  return funclist;
}

int main() 
{
  printf("%c\n",gfunclist[0]());

  funptr *fs;
  fs = x();
  printf("%c\n",fs[1]()); 
}

Answer 1

You have to decide whether you are using C or C++. These languages are significantly different in their treatment of the situations like yours.

In C++ a “pointer to an [] array” (i.e an array of unspecificed size) is a completely different type from a “pointer to an [N] array” (i.e. an array of specified size). This immediately means that your code has no chance to compile as C++ code. It is not a “warning”, it is an error. If you want your code to compile as C++, you need to specfiy the exact array size in the function return type

char (*(*x())[4])() // <- see the explicit 4 here?
{
  static char (*funclist[4])() = {foo, bar, blurga, bletch};
  return &funclist;
}

And, of course, you have to return &funclist, since you are declaring your function as returning a pointer to an array.

In main declaring the receiving pointer as char (**fs)() makes no sense whatsoever. The function is returning a pointer to an array, not a pointer to a pointer. You need to declare your fs as

char (*(*fs)[4])(); // <- pointer to an array

i.e. as having pointer-to-array type (note the similarity to the function declaration). And in order to call the function through such a pointer you have to do

printf("%c\n", (*fs)[1]()); 

In C language the explicit array size in the pointer-to-array declarations can be omitted, since in C “pointer to an [] array” type is compatible with “pointer to an [N] array” type, but the other points still stand. However, even in C it might make more sense to specify that size explicitly.

---

Alternatively, you can stop using pointer-to-array type and instead use pointer-to-pointer type. In that case your function should be defined as follows

char (**x())()
{
  static char (*funclist[4])() = {foo, bar, blurga, bletch};
  return funclist; // <- no `&` here
}

and in main you’ll work with it as follows

char (**fs)();
fs = x();
printf("%c\n", fs[1]()); 

Note, that this main is the same as what you had in your original post. In other words, your original code is a bizarre fusion of two different absolutely incompatible techniques. You have to decide which one you want to use and stick to it.