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looking for explanation of bind2nd() source code

Hi,
Digging through the STL source code, i found this gem..

template <class _Operation, class _Tp>
inline binder2nd<_Oper ation>
bind2nd(const _Operation& __fn, const _Tp& __x)
{
typedef typename _Operation::sec ond_argument_ty pe _Arg2_type;
return binder2nd<_Oper ation>(__fn, _Arg2_type(__x) );
}

I am trying to understand what this does. Specifically, i don't get
what the first two lines signify.

Why is it that bind2nd can be called like this... bind2nd (b, t) and
NOT bind2nd<binary_ function, int(b, t)??

Thanks a lot in advance
Aug 5 '08 #1
4 1837
responsible wrote:
Hi,
Digging through the STL source code, i found this gem..

template <class _Operation, class _Tp>
inline binder2nd<_Oper ation>
bind2nd(const _Operation& __fn, const _Tp& __x)
{
typedef typename _Operation::sec ond_argument_ty pe _Arg2_type;
return binder2nd<_Oper ation>(__fn, _Arg2_type(__x) );
}

I am trying to understand what this does. Specifically, i don't get
what the first two lines signify.

Why is it that bind2nd can be called like this... bind2nd (b, t) and
NOT bind2nd<binary_ function, int(b, t)??
Because it's a function, and the type arguments can be deduced by the
compiler. In fact, that's the entire purpose of bind2nd, so that the
programmer doesn't have to provide template arguments to binder2nd.

The first two lines:

1. template<class _Operation, class _Tp>, declares that this is a
function template.

2. inline binder2nd<_Oper ationstates that this is an inline function,
returning a binder2nd<_Oper ation>

Aug 5 '08 #2
On Aug 5, 7:53 am, responsible <msd...@gmail.c omwrote:
template <class _Operation, class _Tp>
inline binder2nd<_Oper ation>
bind2nd(const _Operation& __fn, const _Tp& __x)
{
typedef typename _Operation::sec ond_argument_ty pe _Arg2_type;
return binder2nd<_Oper ation>(__fn, _Arg2_type(__x) );
}
I am trying to understand what this does. Specifically, i don't get
what the first two lines signify.

Why is it that bind2nd can be called like this... bind2nd (b, t) and
NOT bind2nd<binary_ function, int(b, t)??
bind2nd is a helper function that returns binder2nd function object
adapter. (The binder2nd transforms a binary function object to
unary ).

The fact that we are able to call the function as bind2nd(b, t) is
because of "Template Argument Deduction". ie. the compiler is able to
determine the intended arguments without the programmer having to
explicitly specify them.
eg:
find_if(v.begin (), v.end(), bind2nd(greater <int>(), 10));
could also be rewritten as

find_if(v.begin (), v.end(), bind2nd<greater <int>, int>(greater<in t>(),
10));
This works just fine on VS2008 & gcc 3.4.4 but the code becomes
unreadable.

regards,
Aman Angrish.
Aug 5 '08 #3
On Aug 5, 2:04*am, "aman.c++" <aman....@gmail .comwrote:
On Aug 5, 7:53 am, responsible <msd...@gmail.c omwrote:
template <class _Operation, class _Tp>
inline binder2nd<_Oper ation>
bind2nd(const _Operation& __fn, const _Tp& __x)
{
* typedef typename _Operation::sec ond_argument_ty pe _Arg2_type;
* return binder2nd<_Oper ation>(__fn, _Arg2_type(__x) );
}
I am trying to understand what this does. Specifically, i don't get
what the first two lines signify.
Why is it that bind2nd can be called like this... bind2nd (b, t) and
NOT bind2nd<binary_ function, int(b, t)??

bind2nd is a helper function that returns binder2nd function object
adapter. (The binder2nd transforms a binary function object to
unary ).

The fact that we are able to call the function as bind2nd(b, t) is
because of "Template Argument Deduction". ie. the compiler is able to
determine the intended arguments without the programmer having to
explicitly specify them.
eg:
find_if(v.begin (), v.end(), bind2nd(greater <int>(), 10));
* *could also be rewritten as

find_if(v.begin (), v.end(), bind2nd<greater <int>, int>(greater<in t>(),
10));
* This works just fine on VS2008 & gcc 3.4.4 but the code becomes
unreadable.

regards,
Aman Angrish.
How is this simalar : find_if(v.begin (), v.end(),
bind2nd<greater <int>, int>(greater<in t>(),
10)); ?
Aug 5 '08 #4
On Aug 5, 6:27*am, puzzlecracker <ironsel2...@gm ail.comwrote:
On Aug 5, 2:04*am, "aman.c++" <aman....@gmail .comwrote:
On Aug 5, 7:53 am, responsible <msd...@gmail.c omwrote:
template <class _Operation, class _Tp>
inline binder2nd<_Oper ation>
bind2nd(const _Operation& __fn, const _Tp& __x)
{
* typedef typename _Operation::sec ond_argument_ty pe _Arg2_type;
* return binder2nd<_Oper ation>(__fn, _Arg2_type(__x) );
}
I am trying to understand what this does. Specifically, i don't get
what the first two lines signify.
Why is it that bind2nd can be called like this... bind2nd (b, t) and
NOT bind2nd<binary_ function, int(b, t)??
bind2nd is a helper function that returns binder2nd function object
adapter. (The binder2nd transforms a binary function object to
unary ).
The fact that we are able to call the function as bind2nd(b, t) is
because of "Template Argument Deduction". ie. the compiler is able to
determine the intended arguments without the programmer having to
explicitly specify them.
eg:
find_if(v.begin (), v.end(), bind2nd(greater <int>(), 10));
* *could also be rewritten as
find_if(v.begin (), v.end(), bind2nd<greater <int>, int>(greater<in t>(),
10));
* This works just fine on VS2008 & gcc 3.4.4 but the code becomes
unreadable.
regards,
Aman Angrish.

How is this simalar : find_if(v.begin (), v.end(),
bind2nd<greater <int>, int>(greater<in t>(),
10)); ?
It's a typo. He meant binder2nd in the second example.
Aug 5 '08 #5

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