Re: Why is the return type of count_if() "signed" rather than "unsigned"?

From:
Saeed Amrollahi <amrollahi.saeed@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Tue, 22 Jun 2010 03:21:47 -0700 (PDT)
Message-ID:
<ba779804-2e77-4074-8933-f8eb95bf04ad@q12g2000yqj.googlegroups.com>
Hi

On Jun 22, 9:55 am, xmllmx <xml...@gmail.com> wrote:

As we know, count_if() will never return a negative number. So, it
seems evident that the return type of count_if() should be "unsigned
integral type" rather than "signed integral type".

However, to my surprise, the C++ standard should define the return
type is "signed integer type", which causes a lot of conceptual
confusions and annoying compiling warnings such as "signed/unsigned
mismatch".


According to C++ standard document, the return type of count/count_if
is: iterator_traits<InputIterator>::difference_type.
Because it's like the distance of two pointers
it's may be positive or negative. However count_if accepts Input
Iterator
so such difference is always positive.

Frankly, I don't think the following code issues a lot of annoying
warnings:

#include <algorithm>
#include <vector>
#include <iostream>
bool is_odd(int i) { return i % 2 != 0; }
int main()
{
    using namespace std;
    vector<int> v;
    for (int i = 0; i < 1000; ++i)
        v.push_back(i);

    int c = count_if(v.begin(), v.end(), is_odd);

    cout << c << '\n';
    return 0;
}

I compiled and ran using VS 2008 (VC++ 9).

What's the rationale for the C++ standard committee to do so ?

Thanks in advance!


Regards,
  -- Saeed Amrollahi

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