[lambda] std::sort problem

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Re: [Boost-users] boost geometry...

Michał Nowotka

25 Jan 2009 25 Jan '09

12:17 a.m.

Following code does not compile: sort(m_results->begin(), m_results->end(), lambda::_1->second > lambda::_2->second); where: m_result is type of std::vector* > vs2008 compiler says: error C2780: 'void std::sort(_RanIt,_RanIt)' : expects 2 arguments - 3 provided, see declaration of 'std::sort' but in lamda documentation there is similar example: sort(vp.begin(), vp.end(), *_1 > *_2); so what's wrong with my code? -- Michał Nowotka

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Steven Watanabe

25 Jan 25 Jan

12:30 a.m.

AMDG Michał Nowotka wrote:

...

Following code does not compile:

sort(m_results->begin(), m_results->end(), lambda::_1->second > lambda::_2->second);

where:

m_result is type of std::vector* >

vs2008 compiler says:

error C2780: 'void std::sort(_RanIt,_RanIt)' : expects 2 arguments - 3 provided, see declaration of 'std::sort'

but in lamda documentation there is similar example:

sort(vp.begin(), vp.end(), *_1 > *_2);

so what's wrong with my code

Boost.Lambda doesn't overload operator-> because C++ doesn't allow it to work in general. You have to do it like this (lambda::_1->*&std::pair::second) using member pointers. In Christ, Steven Watanabe

Michał Nowotka

9:46 a.m.

Ok, but there is another problem. Suppose i defined: typedef MyType std::pair; So if I want to keep DRY i should write: lambda::_1 ->*&MyType::value_type::second > lambda::_2 ->*&MyType::value_type::second); I also tried: bind(&MyType::value_type::second, lambda::_1) > bind(&MyType::value_type::second, lambda::_2) but none of this expression works for me.

Stuart Dootson

10:42 a.m.

On Sun, Jan 25, 2009 at 9:46 AM, Michał Nowotka wrote:

...

Ok, but there is another problem.

Suppose i defined: typedef MyType std::pair;

So if I want to keep DRY i should write:

lambda::_1 ->*&MyType::value_type::second > lambda::_2 ->*&MyType::value_type::second);

I also tried: bind(&MyType::value_type::second, lambda::_1) > bind(&MyType::value_type::second, lambda::_2)

but none of this expression works for me.

Shouldn't that be &MyType::second - std::pair has no value_type? Stuart Dootson

Michał Nowotka

11:58 a.m.

Right, there is a mistake. So the problem is: Suppose i defined: typedef std::vector* > MyType; So if I want to keep DRY i should write: lambda::_1 ->*&MyType::value_type::second > lambda::_2 ->*&MyType::value_type::second); I also tried: bind(&MyType::value_type::second, lambda::_1) > bind(&MyType::value_type::second, lambda::_2) but none of this expression works for me. -- Michał Nowotka

Stuart Dootson

1:48 p.m.

2009/1/25 Michał Nowotka :

...

Right, there is a mistake.

So the problem is:

Suppose i defined: typedef std::vector* > MyType;

So if I want to keep DRY i should write:

lambda::_1 ->*&MyType::value_type::second > lambda::_2 ->*&MyType::value_type::second);

I also tried: bind(&MyType::value_type::second, lambda::_1) > bind(&MyType::value_type::second, lambda::_2)

but none of this expression works for me.

Well, it's hardly surprising, as the value_type of the vector is a pointer, so won't have a member called 'second'. Let's refactor the DRY-ness a little: typedef std::pair ElemType; typedef std::vector MyType; MyType v; Then you can use this: std::sort(v.begin(), v.end(), bind(&ElemType::second, _1) > bind(&ElemType::second, _2)); ? Stuart Dootson

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Michał Nowotka
Steven Watanabe
Stuart Dootson