从C ++中具有非恒定数组大小的函数返回数组指针

Question

I built a function to generate a random 2-d array in C++. I was hoping that I could set the size of the array at compile time, so I included variables for the number of rows and columns in the array. However, when I try to compile the function, I get an error about the storage size of the array is not constant. This seems to do with the the static keyword that I have to add to the array definition, so that I can return the pointer from the function. I was not sure if there is a way around this error? Any suggestions.

double * generate_random_array(int rows, int cols, double lower_, double upper_){

static double test_array[rows][cols];

for (int i = 0; i < sizeof test_array / sizeof test_array[0]; i++) {
    for (int j = 0; j < sizeof test_array[0] / sizeof(double); j++) {
        test_array[i][j] = generate_random_numbers(lower_, upper_);
    }
}
return(test_array);

}

Answer 1

You can make generate_random_array a template , enforcing rows and cols to be known at compile-time:

template <int rows, int cols>
double* generate_random_array(double lower_, double upper_)
{
    /* ... as before ... */
}

Example usage:

generate_random_array<5, 10>(51.4, 66.0);

---

Nevertheless, you should use std::array instead of C-style arrays. If you want resizable arrays, then you should use std::vector instead.

std::array example:

template <int rows, int cols>
auto generate_random_array(double lower_, double upper_)
{
    const auto idx = [](int x, int y){ return y * rows + x; };
    std::array<double, rows * cols> result;

    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < cols; j++) {
            result[idx(i, j)] = generate_random_numbers(lower_, upper_);
        }
    }

    return result;
}

Example usage:

auto test_array = generate_random_array<5, 10>(11.0, 66.33);

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