I have a bubble-sort function that takes an array, a compare function, and a boolean that indicates if it should sorts the array upside-down. It is a template function that supports any data-type, and will deduce array size automatically.
When specifying the compare function, if I pass function pointer, the compiler will deduce the data type of array automatically, which is great. But if I pass a lambda instead, it will not deduce automatically. I have to specify the data type explicitly, or static_cast
the lambda as fnCompare_t<double>
.
What is the reason behind this? Because according to this post, as long as the lambda doesn't capture, it can be used like the plain-old function pointer, but it seems it is not always the case? How come it can be different in this case?
#include <iostream>
using namespace std;
template <typename T>
using fnCompare_t = int(*)(T const &, T const &);
template <typename T, size_t count>
inline void BubbleSort(
T(&array)[count],
fnCompare_t<T> fnCompare,
bool reverse)
{
cout << "TODO: Implement BubbleSort" << endl;
}
double doubleArray[] = {
22.3, 11.2, 33.21, 44.2, 91.2, 15.2, 77.1, 8.2
};
int CompareDouble(double const & a, double const & b)
{
return a > b ? 1 : a == b ? 0 : -1;
}
int main()
{
auto fnCompare = [](double const & a, double const & b) -> int {
return a > b ? 1 : a < b ? -1 : 0;
};
// compile OK:
BubbleSort(doubleArray, CompareDouble, false);
BubbleSort(doubleArray, static_cast<fnCompare_t<double>>(fnCompare), false);
BubbleSort<double>(doubleArray, fnCompare, false);
// compile error, could not deduce template argument:
//BubbleSort(doubleArray, fnCompare, false);
return 0;
}