I'm using Visual Studio 2008. I have this class:
template <bool T1>
class Foo {
public:
void doSomething() {}
Foo<T1>& operator=(int a) {
doSomething();
return *this;
}
};
But I want that the method operator=
be hidden (by simply doing: return *this
) if the template parameter T1
is false.
I need that for instances of Foo, the lines:
Foo<false> foo;
foo = 20; //this should give a compilation error
So I tried specializing the class definition:
template<>
class Foo<false> {
private:
Foo<false>& operator=(int a) {
return *this;
}
};
However, by doing this I lose the method doSomething()
on instances that are Foo<false>
, which is not what I need.
I've tried removing the operator=
with boost::enable_if, like this:
typename boost::enable_if<
boost::mpl::bool_<T1>
, Foo<T1>
>::type&
operator=(int a) {
callProxy();
return *this;
}
But that makes me unable to have a class like the following:
class Bar {
public:
Foo<true> assignable;
Foo<false> unassignable;
};
I've also tried putting both methods in Foo and removing them with boost::enable_if
and boost::disable_if
, like this:
template <bool T1>
class Foo {
public:
void doSomething() {}
typename boost::enable_if<
boost::mpl::bool_<T1>
, Foo<T1>
>::type&
operator=(int a) {
doSomethi开发者_运维问答ng();
return *this;
}
private:
typename boost::disable_if<
boost::mpl::bool_<T1>
, Foo<T1>
>::type&
operator=(int a) {
return *this;
}
};
Which didn't work too (I expected that, but it was worth trying).
So, is it possible to get the behaviour I need, and if it is, how could I do it?
why not just use a regular if()?
if(T1) doSomething();
You can statically assert the condition:
Foo<T1>& operator=(int a) {
BOOST_STATIC_ASSERT(T1);
doSomething();
return *this;
}
Instead of special-casing the false
case, you could special-case the true
case, and only include the operator=
in that case.
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