Re: compilation error with merge algorithm

From:
Barry <dhb2000@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Mon, 26 May 2008 16:55:20 +0800
Message-ID:
<g1dtts$mii$1@news.cn99.com>
Barry wrote:

subramanian100in@yahoo.com, India wrote:

Consider the following program:

#include <cstdlib>
#include <iostream>
#include <set>
#include <map>
#include <algorithm>
#include <iterator>
#include <utility>

using namespace std;

int main()
{
        typedef pair<int, int> p_type;

        multiset<p_type> c1;

        c1.insert(make_pair(0, 2));
        c1.insert(make_pair(0, 0));
        c1.insert(make_pair(0, 1));
        c1.insert(make_pair(2, 3));

        multimap<int, int> c3;

        c3.insert(make_pair(5, 0));
        c3.insert(make_pair(0, 0));
        c3.insert(make_pair(3, 2));
        c3.insert(make_pair(0, -1));
        c3.insert(make_pair(1, -1));

        multimap<int, int> dest;

        merge(c1.begin(), c1.end(), c3.begin(), c3.end(),
inserter(dest, dest.begin()));

        return EXIT_SUCCESS;
}

Suppose the above program is named z.cpp and is compiled as
g++ -std=c++98 -pedantic -Wall -Wextra z.cpp

Then I get compilation error for the line
merge(...)

z.cpp:32: instantiated from here
/usr/lib/gcc/i386-redhat-linux/3.4.3/../../../../include/c++/3.4.3/
bits/stl_algo.h:2979: error: no match for 'operator<' in '(&__first2)-

std::_Rb_tree_iterator<_Tp>::operator* [with _Tp = std::pair<const

int, int>]() < (&__first1)-

std::_Rb_tree_const_iterator<_Tp>::operator* [with _Tp =

main()::p_type]()'

Kindly help me to fix this compilation error.


typeof "multiset<p_type>::value_type" == p_type == pair<int, int> ... Type1
typeof "multimap<int, int>::value_type" == pair<const int, int> ... Type2

Type1. Type2 are two different specialization of template class pair

so there are no specialization of
template <class T1, class T2>
bool operator<(const pair<T1, T2>& x, const pair<T1, T2>& y);

while the 5-arg merge needs operator<
take a look at the implementation of merge.


"merge" also requires the two value_type to be the same type.

--
Best Regards
Barry

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