Using a type parameter and a pointer to the same type parameter in a function template
Posted
by Darel
on Stack Overflow
See other posts from Stack Overflow
or by Darel
Published on 2010-04-26T01:56:06Z
Indexed on
2010/04/26
2:03 UTC
Read the original article
Hit count: 390
Hello,
I've written a template function to determine the median of any vector or array of any type that can be sorted with sort. The function and a small test program are below:
#include <algorithm>
#include <vector>
#include <iostream>
using namespace::std;
template <class T, class X>
void median(T vec, size_t size, X& ret)
{
sort(vec, vec + size);
size_t mid = size/2;
ret = size % 2 == 0 ? (vec[mid] + vec[mid-1]) / 2 : vec[mid];
}
int main()
{
vector<double> v;
v.push_back(2); v.push_back(8);
v.push_back(7); v.push_back(4);
v.push_back(9);
double a[5] = {2, 8, 7, 4, 9};
double r;
median(v.begin(), v.size(), r);
cout << r << endl;
median(a, 5, r);
cout << r << endl;
return 0;
}
As you can see, the median function takes a pointer as an argument, T vec. Also in the argument list is a reference variable X ret, which is modified by the function to store the computed median value.
However I don't find this a very elegant solution. T vec will always be a pointer to the same type as X ret. My initial attempts to write median had a header like this:
template<class T>
T median(T *vec, size_t size)
{
sort(vec, vec + size);
size_t mid = size/2;
return size % 2 == 0 ? (vec[mid] + vec[mid-1]) / 2 : vec[mid];
}
I also tried:
template<class T, class X>
X median(T vec, size_t size)
{
sort(vec, vec + size);
size_t mid = size/2;
return size % 2 == 0 ? (vec[mid] + vec[mid-1]) / 2 : vec[mid];
}
I couldn't get either of these to work. My question is, can anyone show me a working implementation of either of my alternatives?
Thanks for looking!
© Stack Overflow or respective owner