CODEWITHSUNDEEP
c++
mirza.o258
mirza.o258

Reputation: 45

Predicate function with counter

How to make predicate function that counts number of iterations in sort algorithm ? I guess I'm supposed to make a functions that returns lambda becausee of: auto p1 = predicate_with_counter(predicate1, counter); .But I have no idea how to do it.In text of problem it says that main function should remain as it is, so changing main is not allowed

#include <vector>
#include <iostream>
#include <algorithm>

// predicate_with_counter() ???

bool predicate1(double a, double b){
   return a > b;
}

int main(){
   using namespace std;
   size_t counter;
   vector<double> v1{2,4,0,-1,5};
   auto p1 = predicate_with_counter(predicate1, counter);
   sort(begin(v1), end(v1), p1);
   cout << "Number of comparisons in sorting of v1 : " << counter << endl;

   return 0;
}

Upvotes: 1

Views: 123

Answers (2)

UnholySheep
UnholySheep

Reputation: 4096

Based on the requirements discussed in the comments I will present 2 solutions. The first one does not use lambdas and will work in all versions of C++

struct predicate_with_counter {
     // typedef for the function to be passed as argument
     typedef bool(*predicate)(double, double);
     predicate_with_counter(predicate pred, size_t& counter) :
         pred_(pred), counter_(counter) {
             // Initialize counter to 0 since it's uninitialized in main
             counter_ = 0;
         }

     // Overload function call operator so it can be used in `sort`
     bool operator()(double a, double b){
         ++counter_;
         return pred_(a,b);
     }

     predicate pred_;
     size_t& counter_;
};

The other option, using lambdas and C++14 return type deduction

auto predicate_with_counter(bool(*predicate)(double a, double b), size_t& counter) {
    // Initialize counter to 0 since it's uninitialized in main
    counter = 0;
    return [predicate, &counter](double a, double b) {
        ++counter;
        return predicate(a,b);
    };
}

While the second option is a lot shorter it is (subjectively) harder to understand. Pick whichever one you prefer

Upvotes: 2

Michał Turek
Michał Turek

Reputation: 721

That's how I solved your problem using lambda expression:

#include <vector>
#include <iostream>
#include <algorithm>

int main() {
    using namespace std;
    size_t counter=0;
    vector<double> v1{ 2,4,0,-1,5 };
    sort(begin(v1), end(v1), [&counter](double a, double b)
        {
            counter++;
            return a > b;
        });
    cout << "Number of comparisons in sorting of v1 : " << counter << endl;

    return 0;
}

Answering your question, if you want to make it as a function, you can use auto:

auto count_comparisons = [&counter](double a, double b)
    {
        counter++;
        return a > b;
    };
 sort(begin(v1), end(v1),count_comparisons );

Upvotes: 3

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