问题
Dan's answer to this question: Is There Anything Like a Templatized Case-Statement takes a DefaultType
template parameter.
Is it possible to pass something there which will force a compile time fail?
For example, given this templatized function:
template <typename T, typename R = enable_if_t<is_integral<T>::value, int>>
R foo(T bar) {return static_cast<R>(bar);}
This code will compile fine: foo(13)
But this code will fail: foo(13.0)
.
The reason that foo(13.0)
will fail at compile time is that enable_if_t
is undefined. Is there a name for "undefined" that I can pass to the afore mentioned DefaultType
?
If so I should be able to test this by calling foo<int, undefined>(13)
and having it fail, if in fact undefined
was the type I'm looking for.
EDIT:
Apparently more explanation is called for.
Dan's `static_case can be called like this:
template<class T>
typename static_switch<sizeof(T),
int, // default case
static_case<sizeof(char),char>,
static_case<sizeof(short),short>,
static_case<sizeof(long),long>,
static_case<sizeof(long long),long long>>::type foo(T bar){ return reinterpret_cast<decltype(foo(bar))&>(bar);}
I want to pass "undefined" or whatever to that second parameter to make it fail to compile only when that parameter is evaluated. (The parameter marked "default case".)
In my simple test passing any two parameters to foo
will succeed (for example foo<int, int>(13);
) I want a parameter that would mimic the effect of enable_if_t
causing it to fail (like when we call foo<double>(13.0);
) In the example I said call foo<in, undefined>(13)
. I just want to know what "undefined" is.
回答1:
The simplest way to have the undefined
type you are talking about is:
std::enable_if_t<false>
So you can call your foo
template like this:
foo<int, std::enable_if_t<false>>(13)
And have it fail as you asked.
来源:https://stackoverflow.com/questions/28724017/pass-a-failing-template-argument