[Enable-if] Selectively instantiating methods.
Hi All This doesn't work. template<typename T> struct S { typename disable_if<is_const<T>, int>::type mutating_method(); }; I understand why not, although only after the compiler threw it out, but obviously (in hindsight), the method itself is not a template, so can't be selectively instantiated. The question is whether and how I can achieve the effect I'm after, ie., that the method is not present if the struct is instantiated with a const type? Thx, Rob.
On Fri, Sep 23, 2011 at 11:48 PM, Igor R <boost.lists@gmail.com> wrote:
I didn't test, but I think something like this should do the job:
template<typename T> struct S { template <class T1 = T> typename disable_if<is_const<T1>, int>::type mutating_method(); };
Default template arguments are disallowed for function templates.
True, but they are allowed in C++0x/11 (and maybe the restriction applies only to free-standing functions, not methods, didn't have time to dig up more). http://stackoverflow.com/questions/2447458/default-template-arguments-for-fu... Julien
2011/9/23 Robert Jones <robertgbjones@gmail.com>
Hi All
This doesn't work.
template<typename T> struct S { typename disable_if<is_const<T>, int>::type mutating_method(); };
I understand why not, although only after the compiler threw it out, but obviously (in hindsight), the method itself is not a template, so can't be selectively instantiated.
The question is whether and how I can achieve the effect I'm after, ie., that the method is not present if the struct is instantiated with a const type?
You'll need to specialize S or do something to the similar effect (e.g., have two base classes and inherit from one of them depending on the template argument). Roman Perepelitsa.
On Fri, 23 Sep 2011 00:51:40 -0700, Robert Jones <robertgbjones@gmail.com> wrote:
Hi All
This doesn't work.
template<typename T> struct S { typename disable_if<is_const<T>, int>::type mutating_method(); };
I understand why not, although only after the compiler threw it out, but obviously (in hindsight), the method itself is not a template, so can't be selectively instantiated.
The question is whether and how I can achieve the effect I'm after, ie., that the method is not present if the struct is instantiated with a const type?
Thx, Rob.
See http://thread.gmane.org/gmane.comp.lib.boost.devel/222181/focus=222402 Short answer (from the cited post):
template <class U> struct T { void f() { f_impl(<static_cast<void *>(0)); }
private: template <class T> typename enable_if< mpl::and_< is_convertible<T, void *>, some_condition_on<U>
::type f_impl(T) { .... } };
Basically introduce a dummy parameter with a dummy template type that's not used in f_impl. Mostafa
participants (5)
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Igor R -
Julien Nitard -
Mostafa -
Robert Jones -
Roman Perepelitsa