I need to make a generic function which will allocate an array that has same elements as vector in the main function.
This generic function should accept pointers/iterators to the beginning and end of the vector.
#include <iostream>
#include <new>
#include <vector>
template <typename type>
type *MakeArray(type::iterator start, type::iterator after_end) {
int n = 0;
while (start < after_end) {
start ;
n ;
}
start -= n;
type *arr = nullptr;
arr = new type[n];
throw std::bad_alloc("Not enough memory!");
while (start < after_end) {
*arr = *start;
start ;
arr ;
}
delete[] arr;
arr-=n;
return arr;
}
int main() {
int n=5;
std::vector<double>a{1,2,3,4,5};
double *arr = nullptr;
try {
arr = MakeArray(a.begin(),a.end());
} catch (std::bad_alloc e) {
std::cout << "Exception: " << e.what();
}
delete[] arr;
return 0;
}
ERRORS:
line 5:
expected ‘)’ before ‘after_end’
expected ‘;’ before ‘{’ token
line 30:
missing template arguments before ‘(’ token
I don't see any reason why should I get these errors. Could you help me to fix my code? Iterators and dynamic allocation are new to me.
CodePudding user response:
You can use the iterator type itself as the template arguments and then extract the underlying type of the contained elements in the function.
Also, you shouldn't be using delete[] arr;
in your function because: (a) at that point, it no longer points to the memory allocated by the new
call; (b) if you do, you won't be able to use it in the calling module.
There are also some significant other simplifications and improvements you can make to your function, which I have shown in the below code:
template <typename it_type>
auto* MakeArray(it_type start, it_type after_end)
{
using basetype = typename std::decay< decltype(*start) >::type; // Type of contained objects
size_t size = static_cast<size_t>(std::distance(start, after_end)); // Quick calculation of size
basetype* arr = new basetype[size]; // This will automatically throw (bad_alloc) if it fails
std::copy(start, after_end, arr); // Quicker/easier way to do the data copy
return arr;
}