l-value specifies const object while using std::map - c++

Question

I'm trying to use std::map like in example below:
#include <map>
#include <algorithm>
int wmain(int argc, wchar_t* argv[])
{
typedef std::map<int, std::wstring> TestMap;
TestMap testMap;
testMap.insert(std::make_pair(0, L"null"));
testMap.insert(std::make_pair(1, L"one"));
testMap.erase(std::remove_if(testMap.begin(), testMap.end(), [&](const TestMap::value_type& val){ return !val.second.compare(L"one"); }), testMap.end());
return 0;
}
And my compiler (VS2010) gives me following message:
>c:\program files\microsoft visual studio 10.0\vc\include\utility(260): error C2166: l-value specifies const object
1> c:\program files\microsoft visual studio 10.0\vc\include\utility(259) : while compiling class template member function 'std::pair<_Ty1,_Ty2> &std::pair<_Ty1,_Ty2>::operator =(std::pair<_Ty1,_Ty2> &&)'
1> with
1> [
1> _Ty1=const int,
1> _Ty2=std::wstring
1> ]
1> e:\my examples\с++\language tests\maptest\maptest\maptest.cpp(8) : see reference to class template instantiation 'std::pair<_Ty1,_Ty2>' being compiled
1> with
1> [
1> _Ty1=const int,
1> _Ty2=std::wstring
1> ]
I can't understand why opertor = is called though I pass val in lambda-function by reference.
Could you explain what I am doing wrong?

Answer 1

You cannot use std::remove_if with an associative container, because that algorithm works by overwriting removed elements with the subsequent ones: the problem here is that keys of a map are constant, in order to prevent you (or the std::remove_if algorithm) from messing up with the internal ordering of the container.
To remove elements from a map conditionally, rather do this:
for (auto iter = testMap.begin(); iter != testMap.end();)
{
if (!iter->second.compare(L"one")) // Or whatever your condition is...
{
testMap.erase(iter++);
}
else
{
++iter;
}
}
Here is a live example.