Re: boost::lambda in std::map

From:

porparek@gmail.com

Newsgroups:

comp.lang.c++

Date:

Fri, 19 Dec 2014 03:58:57 -0800 (PST)

Message-ID:

<22c7b0fe-1f36-4a59-8059-a4fba49f9ca6@googlegroups.com>

On Monday, December 15, 2014 5:54:41 PM UTC+1, =D6=F6 Tiib wrote:

On Monday, 15 December 2014 16:51:15 UTC+2, porp...@gmail.com wrote:

I would like to implement my own comparator for std::map using
boost::lambda (no C++11 capable compiler).
I wrote my very first version of such comparator.
Unfortunately I cannot compile it. I get the following compilation erro=

'boost::lambda::{anonymous}::_1' cannot appear in a constant-expression
'boost::lambda::{anonymous}::_2' cannot appear in a constant-expression

Please help me to overcome this error.

thanks in advance

#include <map>
#include <string>
#include <boost/lambda/lambda.hpp>

using namespace std;
using namespace boost::lambda;

int main(void)
{
map<string, string, (_1 < _2)> myMap;
return 0;
}

That is because you are mixing the values and the types of the
values up. As result the compiler does not understand what you
want to do. Perhaps you should try it step-by-step?
DISCLAIMER: I have no compiler nearby so I can't guarantee it does
not have typos or something but it must illustrate the idea:

// 1) Should such code work?:
typedef boost::function<bool(std::string const&, std::string const&)>=

Comparator;

    // what happened on above line?

    // 2) After it should that work?:
    Comparator c = _1 < _2;
    // what happened on above line?

    // 3) Now ... does that work?:
    typedef std::map<std::string, std::string, Comparator> StrToStr;
    // what happened on above line?

    // 4) So ... does that work?:
    StrToStr myMap(c);
    // what happened on above line?

When you understand these elementary steps then you can perhaps start
to write more complex lines of C++:

    int main()
    {
        map<string, string, function<bool (string const&, string const&)>=

> myMap(_1 < _2);

return 0;
}

I myself avoid writing such complex lines. It looks too lot like
cryptic garbage.

Thanks for reply.
I thought that lambda expression like (_1<_2) generates functor. It looks l=
ike it rather generates an object of a functor, right ? This is the source =
of my confusion.
I also wonder why the following does not compile
map<string, string> myMap(_1<_2)

I suspect that most probably the 3-rd template parameter is difficult to de=
duce and a compiler just gives up. Am I right ?