How Can I Get the Size of an Array Using a C Template Function?

How does this "size of array" template function work?

This template function, GetArrLength, is designed to determine the length of an array. It achieves this by leveraging the knowledge of the array's size and type, which are set as template parameters of type T and int size, respectively.

The function takes a reference to an array of type T and size size as an input parameter. This reference is declared using the syntax T(&)[size], which signifies that the parameter should be a reference to an array of type T with a constant size of size.

When called, the compiler attempts to deduce the template parameters based on the type and size of the array being passed in. For instance, if we call GetArrLength with an array of integers named a with a size of 10, the compiler will infer that T is int and size is 10, and the function will return the value 10.

However, there are certain limitations to the GetArrLength function implementation. Firstly, the use of a signed integer for the template parameter and return type can lead to issues with negative array sizes. For practical purposes, it's advisable to use an unsigned type, such as std::size_t, for both the template parameter and the return type to avoid any unexpected behavior.

Additionally, the function's result is not a constant expression, although the size of an array is inherently constant. To address this limitation, more robust solutions exist that can provide a constant result, such as the following code block:

template <std::size_t N>
struct type_of_size
{
    typedef char type[N];
};

template <typename T, std::size_t Size>
typename type_of_size<Size>::type& sizeof_array_helper(T(&&)[Size]);

#define sizeof_array(pArray) sizeof(sizeof_array_helper(pArray))

This technique generates a constant result by encoding the array's size into the size of a type using a template metaprogramming approach. It ensures that the result is not evaluated dynamically and remains a constant expression.

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