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reverting [80445] which is still under discussion on the mailing list

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[SVN r80452]
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Daniel Walker 2012-09-08 15:32:35 +00:00
parent e8440e8855
commit e9bbb50eb4
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@ -230,51 +230,6 @@ typedef boost::result_of<
&gt;::type type; // type is int</pre> &gt;::type type; // type is int</pre>
</blockquote> </blockquote>
<p>The <code>result</code> template must be specialized for every valid calling signature of the function object.
If the <code>operator()</code> accepts arguments by (possibly <code>const</code>) reference and/or is <code>const</code>
qualified, the <code>result</code> specialization must take this into account. <a href="../type_traits/doc/html/index.html">Type traits</a>
and more generic specializations may help to reduce the number of <code>result</code> specializations. This way <code>result_of</code> users
will be able to specify argument types exactly according to the function object call expression. For example:</p>
<blockquote>
<pre>struct functor {
template&lt;class&gt; struct result;
// Use template parameter F to match the function object. This will allow result deduction for both const and non-const functor.
template&lt;class F, class T&gt;
struct result&lt;F(T)&gt; {
// When argument type is matched like above, remember that the type may be a (const-qualified) reference.
// Use type traits to transform the argument type.
typedef typename remove_cv&lt;typename remove_reference&lt;T&gt;::type&gt;::type argument_type;
typedef argument_type type;
};
// The operator can be called on both const and non-const functor. The argument can be lvalue or rvalue.
template&lt;class T&gt;
T operator()(T const&amp; x) const
{
return x;
}
};
// All following result_of uses are valid and result in int
typedef boost::result_of&lt; functor(int) &gt;::type type1; // the argument is rvalue
functor f;
type1 r1 = f(10);
typedef boost::result_of&lt; const functor(int) &gt;::type type2; // the function object is const
const functor cf;
type2 r2 = cf(10);
typedef boost::result_of&lt; functor(int&amp;) &gt;::type type3; // the argument is lvalue
int a = 10;
type3 r3 = f(a);
typedef boost::result_of&lt; functor(int const&amp;) &gt;::type type4; // the argument is const lvalue
const int ca = 10;
type4 r4 = f(ca);</pre>
</blockquote>
<p>Since <code>decltype</code> is a new language <p>Since <code>decltype</code> is a new language
feature recently standardized in C++11, feature recently standardized in C++11,
if you are writing a function object if you are writing a function object