CODEWITHSUNDEEP
c++
Priyam Rajvanshi
Priyam Rajvanshi

Reputation: 49

error: invalid operands of types '<unresolved overloaded function type>' and 'int' to binary 'operator>' if(min > max)

I am getting this error while compiling this program. This questions is to find max of min's window size. Here is my code:

    #include <iostream>

    using namespace std;

    int main() {

            int n;
            cin>>n;
            int a[n];
            for(int i=0; i<n; i++)
            {
                cin>>a[i];//to take input as array
            }
            for(int k=1; k<=n; k++)
            {
                int max=0;
                for(int i=0; i<n-k; i++)
                {
                    int min=a[i];
                    for(int j=0; j<k; j++)
                    {
                        if(a[i+j]<min)
                        {
                            min=a[i+j];
                        }
                    }
                }
               if(min > max)
               {
                   max=min;
               }
                cout<<max<<" ";
            }

        return 0;
    }

Upvotes: 3

Views: 5482

Answers (1)

Bo Persson
Bo Persson

Reputation: 92261

You are hiding (shadowing) the std::min and std::max functions by the local declarations of int max and int min. However, this shadowing only happens in the local scope, where these variables are declared.

When you leave the scope, the outer name becomes visible again:

        int max=0;                  // <-- Here max hides std::max
        for(int i=0; i<n-k; i++)
        {
            int min=a[i];          // <-- Here min hides std::min
            for(int j=0; j<k; j++)
            {
                if(a[i+j]<min)
                {
                    min=a[i+j];
                }
            }
        }                       // <-- But here the local min goes away again

       if(min > max)            // <-- And here min is std::min, a function

So the compiler believes that min > max is to compare an int to a function template. And it cannot see how to do that.

This is one of the reasons why using using namespace std; is considered not a good idea.

Upvotes: 5

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