Weird, very weird case. Consider the code:
int n = 50;
auto p1 = new double[n][5]; //OK
auto p2 = new double[5][n]; //Error
main.cpp: In function ‘int main()’:
main.cpp:17:26: error: array size in new-expression must be constant
auto p2 = new double[5][n]; //Errormain.cpp:17:26: error: the value of ‘n’ is not usable in a constant expression
main.cpp:15:8: note: ‘int n’ is not const
Can anyone explain why do I get a compile error on the second one but the first one works perfectly?
With new double[n][5] you are allocating n values of type double[5].
With new double[5][n] you are allocating 5 variable-length arrays. And C++ doesn't have VLA's so that's invalid.
The solution, as ever, is to use std::vector:
std::vector<std::vector<double>> p2(5, std::vector<double>(n));
The above defines p2 to be a vector of vectors of double. It constructs p2 to have a size of 5 elements, where each element is initialized to a vector of n values.
Your problem is explained, incidentially with the exact example that you give (funny!?) on the new[] expression page on cppreference under the section "Explanation". See excerpt:
If type is an array type, all dimensions other than the first must be specified as positive integral constant expression (until C++14) converted constant expression of type std::size_t (since C++14), but the first dimension may be any expression convertible to std::size_t. This is the only way to directly create an array with size defined at runtime, such arrays are often referred to as dynamic arrays.
