value of parameter cannot be used as a constant c++

Question

void merge(vector<Flight>& data, int low, int high, int mid, string criteria)
{
int i, j, k, temp[high - low + 1];
...

The error that comes up is "the value of parameter "high" (declared at line 100) cannot be used as a constant". I haven't managed to find an appropriate answer to this question online.

You are trying to use `high` as array length in a declaration. As the compiler tells you, that's not possible. — user0042, Nov 28 '17 at 09:01

score 4 · Accepted Answer · answered Nov 28 '17 at 09:01

4

high - low + 1 needs to be a compile time evaluable constant expression in C++. (C++ does not support variable length arrays.)

And that isn't, so the compiler issues a diagnostic.

The simple solution is to use a std::vector<int> as the type for temp.

answered Nov 28 '17 at 09:01

Bathsheba

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like this: `vector * temp = new vector[high - low + 1];`? – asymmetryFan Nov 28 '17 at 09:17
Almost, but drop the pointer type. There's no need. – Bathsheba Nov 28 '17 at 09:17
if I drop the pointer type, I get: "no suitable constructor exists to convert from "std::vector> *" to "std::vector>"". – asymmetryFan Nov 28 '17 at 09:19
The vector type needs to be an `int`, in order to match what you want for `temp`. i.e. `std::vector temp(high - low + 1);` – Bathsheba Nov 28 '17 at 09:20
I made a mistake in the OP. I've changed the type of temp, because I'm going to transfer something from it to `vector data` later on in the code, which means that I need it to be a `vector` instead of a `vector`. – asymmetryFan Nov 28 '17 at 09:22
`std::vector temp(high - low + 1);` will work, so long as `Flight` has the required constructors. – Bathsheba Nov 28 '17 at 09:33

value of parameter cannot be used as a constant c++

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