Why can't some types (with array notation) be used as return type in C without typedef?

Question

While messing around with the type syntax, I noticed this is legal:

typedef int *((* T)[10]);

T fun(){
  return 0;
};

int main(int argc, char * argv[]){
  //int c = fun(); // (1)
  return 0;
}

...And if you uncomment (1) , then you get an error message of this kind (GCC / Clang): " error: cannot initialize a variable of type 'int' with an rvalue of type 'T' (aka 'int *((*)[10])') " (Normal so far). Notice however the " aka " that points out the type is an alias of int *((*)[10]) and not simply int ***

However, It seems impossible to declare a function with this type without using a typedef:

int *((*)[10]) fun(){ // The compiler does not approve
  return 0;
};

int *((* fun2)[10]) (){ // The compiler does not approve either
  return 0;
};

int main(int argc, char * argv[]){
  //int c = fun(); // (1)
  return 0;
}

...Then I was wondering why? (the question is for the C language, but it looks like it's the same for C++)

Answer 1

This type:

typedef int *((* T)[10]);

Is a pointer to an array of size 10 whose members are of type int * . This is not the same as an int *** .

As for creating a function that returns this type, you would need this:

int *(*fun())[10] {
  return 0;
};

But using a typedef makes this much clearer.

Answer 2

int *((*fun())[10]) {
  return 0;
};

... Yup. You should probably stick to the typedef for the sake of readability:)

Answer 3

The original

typedef int *((* T)[10])

can shed the outer parens:

typedef int *(* T)[10]

Or aligned with dbush's function:

typedef int *(*   T   )[10]
        int *(* fun() )[10]