CODEWITHSUNDEEP
c++c
hl037_
hl037_

Reputation: 3887

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

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++)

Upvotes: 2

Views: 82

Answers (3)

user15943018
user15943018

Reputation:

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]

Upvotes: 0

dbush
dbush

Reputation: 224072

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.

Upvotes: 3

Quentin
Quentin

Reputation: 63124

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

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

Upvotes: 2

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