创建一个C ++模板函数，该函数将返回特定大小的std :: array

Question

I am creating a function titled linspase with C++17 with the following input structure linspace(double upper, double lower, int size) . The function should create size evenly spaced numbers between lower and upper . Programming this function is easy; however, I want it to create an std::array<double, size> where the function will determine the array size from the function call and pass back the data type std::array<double, size> . I know templates are the only way to do this, but I am not sure how to build the template. In a general pseudo code I think it looks something like this.

template <typedef T, size_t size>
T linspace(double lower, double upper, int num)
{
    /* ... Function meat goes here to create arr of size 
           num of evenly space numbers between lower and upper
    */     
    return arr
}

However, I know this template declaration is not right and I am not sure what it is supposed to look like. To be clear, I want it to return an std:;array of a specific size, not an std::vector .

Answer 1

If you (correctly) pass the array size as a template parameter, you don't need it as one of the function arguments, so:

template <size_t size>
auto linspace(double lower, double upper) -> std::array<int, size>
{
    std::array<int, size> arr{};
    //populate the array ...
    return arr;
}

Since you're using c++14, you can get rid of the return value altogether, and have a prototype like:

template <size_t size>
auto linspace(double lower, double upper)

Then you can use it like this:

auto arr = linspace<1>(0, 1);

for(auto a : arr)
{
    std::cout << a << std::endl;
}