std::sort and compare-function with template does not work

Question

I want to sort a vector of arbitrary types, so I wrote the following code:

#include <iostream> #include <vector> #include <algorithm> using namespace std; template<class T> bool compare(T a, T b) { return a < b; } int main() { vector<int> v; v.push_back(3); v.push_back(4); v.push_back(2); v.push_back(1); sort(v.begin(), v.end(), compare); for (size_t i = 0; i < v.size(); i++) { cout << v.at(i) << " "; } return 0; }

This code is not compiling, with a error message like:

..\src\Test.cpp:22:34: error: no matching function for call to 'sort(std::vector<int>::iterator, std::vector<int>::iterator, <unresolved overloaded function type>)' ..\src\Test.cpp:22:34: note: candidates are: ... and more

When I implement the compare-function with concrete types, it works. Can someone tell me how to do that with a template-compare-function?

Borgleader · Accepted Answer · 2015-01-16 13:22:49Z

8

You need to specify which specialization you want:

sort(v.begin(), v.end(), compare<int>);

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answered Jan 16, 2015 at 13:22

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Comments

Jonathan Wakely · Accepted Answer · 2015-01-16 13:26:39Z

compare is not a function, it's a function template that can be used generate functions, such as compare<int> and compare<long>

So to pass a function to sort you need to name a specialization of the function template:

sort(v.begin(), v.end(), compare<int>);

Alternatively, create a function object and pass that:

struct Compare { template<typename T> bool operator()(T a, Tb) const { return a < b; } }; sort(v.begin(), v.end(), Compare());

This function object has a member function template that can compare any types (like your compare) but you don't need to refer to a specific specialization when passing it to sort, you pass a temporary of type Compare and inside the sort algorithm the compiler will select the right specialization of the function template.

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