How to sort in descending order?

[HazySmoke)345]

By default, sort() sorts everthing in ascending order. That's not what I want. First, I tried this:

code:

#include <cstdio> #include <algorithm> using namespace std; int comp(int a, int b){         return b-a; } int main(){         int a[] = {3,6,1,7,1,7,4,2,5,2}, i;         sort(a, a+10, comp);         for (i = 0; i < 10; ++i) printf ("%d ", a[i]); return 0;}

The output turned out to be: 2 5 2 4 7 1 7 1 6 3 - not very appealing.
Then I tried this:

code:

#include <cstdio> #include <algorithm> using namespace std; int main(){         int a[] = {3,6,1,7,1,7,4,2,5,2}, i;         sort(a, a+10, operator>);         for (i = 0; i < 10; ++i) printf ("%d ", a[i]); return 0;}

The compiler gaved me an error...
testing.cpp:7: error: no matching function for call to `sort(int[10], int*, <unknown type>)'

Apparantly it doesn't recognize what 'operator>' is.

What am I doing wrong?

[PaulButler]

c++:

int comp(int a, int b){    return b-a; }

Shouldn't this be:

c++:

int comp(int a, int b){    return b < a; }

(or b > a?)

[HazySmoke)345]

Thanks, you're absolutely right. I guess the compare function works differently in C++ than in C. 8 bits.

I still have no clue why wouldn't operator> work, though. Doesn't C++ treat it like any other functions?

[OneOffDriveByPoster]

You can try std::greater<T> in <functional>.

HazySmoke)345 @ Tue Aug 07, 2007 9:16 pm wrote:

I still have no clue why wouldn't operator> work, though. Doesn't C++ treat it like any other functions?

Yes, but you aren't dealing with operator> when you do ( 0 > 0 ). You cannot do operator>(0, 0).

[PaulButler]

Is there not a way to make an infix operator into a function? In OCaml you just wrap it in brackets (eg., 5 + 3 is the same as (+) 5 3). Maybe C++ has something similar?

[wtd]

O'Caml also lacks operator overloading, which adds some complexity to the issue.

The way one does this in C++ is to create a function object.

code:

template <typename T> struct gt {     bool operator()(T a, T b)     {         return a > b;     } };