CODEWITHSUNDEEP
c++arraystypedef
Mattias
Mattias

Reputation: 127

Dimensions of C-style arrays in typedefs

I have a question about the dimensions of C-style arrays in C++. In using declarations/typedefs the dimensions seems strange when using more than one dimension. For example:

using A1 = int[23]; //! << A1 = int[23]
using A2 = A1[4];   //! << A2 = int[4][23]

std::cout << std::is_same<int[23][4], A2>::value << std::endl; //false
std::cout << std::is_same<int[4][23], A2>::value << std::endl; //true

I thought that the type of A2 would be int[23][4] and not int[4][23]. The same behaviour is observed is observed in the snippet below:

template<typename T>
struct ArrayTest;

template<typename T, size_t N>
struct ArrayTest<T[N]>
{
    using type = T;
};

ArrayTest<int[23][2][45]>::type A3;  //! T is int[2][45], N is 23

In this example I thought the type would be int[23][2] and not int[2][45]. Does anyone know why the types are deduced like this? I have tried to find an explanation in the standard, but I guess I have not looked hard enough.

Upvotes: 4

Views: 73

Answers (1)

juanchopanza
juanchopanza

Reputation: 227608

Does anyone know why the types are deduced like this?

using A2 = A1[4];

A2 is a length 4 array of A1 objects. 

using A1 = int[23];

A1 is a length 23 array of int. So the type of A2 is length 4 array of length 23 int, or int[4][23].

Upvotes: 6

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