confusion about using std::less and std::greater with std::sort
c, c++, sorting, std, stl
Solution
`std::sort` behaves like that because it's based on the idea of a strict weak ordering, which is (usually) defined in terms of the `<` operator.
As to your question; it currently seems to be "I wrote a C function that behaves differently to `std::sort`. Why is it different?". The answer is: because you wrote a different function!
Problem
In C, sort usually implements as in the following example: ``` #include <stdio.h> void Sort( int* arr, int n, bool(*cmp)(int,int) ) { for( int i=0; i<n-1; i++ ) { for( int j=i+1; j<n; j++ ) { if( cmp(arr[i], arr[j]) ) swap( arr[i], arr[j] ); } } } int ascending( int a, int b ) { return a > b; } // greater int descending( int a, int b ) { return a < b; } // less void main() { int arr[10] = { 1,3,5,7,9,2,4,6,8,10 }; // ascending Sort( arr, 10, ascending ); for( int i=0; i<10; i++ ) printf( "%d ", arr[i] ); printf( "\n" ); // descending Sort( arr, 10, descending ); for( int i=0; i<10; i++ ) printf( "%d ", arr[i] ); printf( "\n" ); } ``` So I wrote some source as in the following example, expecting same result: ``` #include <iostream> #include <algorithm> // for sort #include <functional> // for less & greater using namespace std; bool gt( int a, int b ) { return a > b; } // greater bool ls( int a, int b ) { return a < b; } // less void main() { int x[10] = { 1,3,5,7,9,2,4,6,8,10 }; // ascending but descending sort( x, x+10, gt ); for( int i=0; i<10; i++ ) cout << x[i] << " "; cout << endl; // descending but ascending sort( x, x+10, ls ); for( int i=0; i<10; i++ ) cout << x[i] << " "; cout << endl; greater<int> g; // a > b less<int> l; // a < b // ascending but descending sort( x, x+10, g ); for( int i=0; i<10; i++ ) cout << x[i] << " "; cout << endl; // descending but ascending sort( x, x+10, l ); for( int i=0; i<10; i++ ) cout << x[i] << " "; cout << endl; } ``` But my expectation was not correct. Why does not sort in STL work like sort in C?