MESSAGE
| DATE | 2011-07-07 |
| FROM | Ruben Safir
|
| SUBJECT | Subject: [NYLXS - HANGOUT] (fwd) Re: iostream manipulator construction and design
|
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Path: reader1.panix.com!panix!usenet.stanford.edu!postnews.google.com!j28g2000vbp.googlegroups.com!not-for-mail
From: Michael DOUBEZ
Newsgroups: comp.lang.c++
Subject: Re: iostream manipulator construction and design
Date: Wed, 6 Jul 2011 01:00:41 -0700 (PDT)
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On Jul 6, 4:55=A0am, Ruben Safir wrote:
> I've covered the iostream operations in Stroustrup but there is a section
> that I'm failing to understand in section 21.4.6.1 (of the third edition
> of C++ - The Programming Language) which I hope that someone can help
> me decipher. =A0The section discusses Manipultors that take arguments and
> Stroustrup states
>
>
> "Manipulators that take arguments can also be useful. =A0For example, we
> might want to write
>
> =A0 =A0cout << setprecision(4) << angle;
>
> to print the value of the floating-point variable angle with four
> digits.
>
> To do this, setprecision must return an object that is initialized by 4
> and that calls cout::setprcision when involked. =A0Such a manipulator is =
a
> function object thatr is involed by << rather than by (). =A0The exact
> tyoe of function object is implementation-defined, but it might be
> defined like this:
>
> =A0 =A0struct smanip{
> =A0 =A0 =A0 ios_base& (*f)(ios_base&, int); //function to be called
> =A0 =A0 =A0 int i;
>
> =A0 =A0 =A0 smanip(ios_base& (*ff) (ios_base&, int), int ii) : f(ff, i(ii=
){}
> =A0 =A0};
>
> =A0 =A0template
> =A0 =A0ostream& operator<<(ostream& os, smanip& m)
> =A0 =A0{
> =A0 =A0 =A0 return m.f(os, m.i);
> =A0 =A0}
>
> The smanip constructor stores its arguments in f and i and the
> operator<< calls f(i). =A0We can now define setprecision() like this:
>
> =A0 =A0ios_base& set_precision(ios)base& s, int n) =A0//helper
> =A0 =A0{
> =A0 =A0 =A0 return s.setprecision(n); //call member function
> =A0 =A0}
>
> =A0 =A0inline smap setprecision(int n)
> =A0 =A0{
> =A0 =A0 =A0 =A0 =A0return smanip(set_precision, n);
> =A0 =A0}
>
> We can now write:
> =A0 =A0cout << setprecision(4) << angle;
> " =A0
>
>
> What don't understand a couple of things here, and this might be an
> accumulation of increasing confusion on my part as I've continued
> through this chapter.
>
> First, I just don't understand what the problem is that is being solved.
The problem being solved is being able to use a parametrized modifier
with the same syntax.
By example, if you want to uppercase your output, you write:
out<
This works because ostream (in fact basic_ostream<>) defines an
overload like
basic_ostream<>& operator<<(ios_base& (*pf)(ios_base&))
{
*pt(*this);
return *this;
}
And std::uppercase simply set the relevant flag
ios_base& std::uppercase(ios_base& os)
{
os.setf(ios_base::uppercase);
return os;
}
> Why can't a manipulator just take an argument of an int and return a
> base_stream object, like the manipulators that don't take an argument?
But now, if you want a parameter to uppercase:
ios_base& set_uppercase(ios_base& os, int isUppercase)
{
if(isUppercase)os.setf(ios_base::uppercase);
else os.unsetf(ios_base::uppercase);
return os;
}
You cannot directly use set_uppercase because, there is no matching
overload in ostream. C++ provides a magic implementation specific
class called smanip which provide a mechanism for calling your
function.
smanip setUppercase(bool b)
{
return smanip(set_uppercase, b);
}
> Secondly, the exqmple completely befuddles me. =A0I'm guessing that don't
> understand the order of operations of
>
> =A0 =A0cout << setprecision(4) << angle;
>
> =A0 =A0is ostream<< called first and then the function setprecision?
No, it can help if you rewrite it with function call:
operator<<( operator<<(cout ,setprecision(4)) , angle );
--
Michael
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