问题
Consider this simple check for whether a (global) function is defined:
template <typename T>
concept has_f = requires ( const T& t ) { Function( t ); };
// later use in MyClass<T>:
if constexpr ( has_f<T> ) Function( value );
unfortunately this allows for implicit conversions. This is obviously a big risk for mess-ups.
Question: How to check if Function( const T& t ) 'explicitly' exists?
Something like
if constexpr ( std::is_same_v<decltype( Function( t ) ), void> )
should be free of implict conversions, but I can't get it working.
Note: The point of the concept approach was to get rid of old 'detection patterns' and simplify.
回答1:
Before explaining how to do this, I will explain why you shouldn't want to do any of this.
You mentioned "old 'detection patterns'" without adding any specifics as to what you are referring to. There are a fair number of idioms C++ users sometimes employ that can do something like detecting if a function takes a particular parameter. Which ones of these count as "detection patterns" by your reckoning is not known.
However, the vast majority of these idioms exist to serve a specific, singular purpose: to see if a particular function call with a given set of arguments is valid, legal C++ code. They don't really care if a function exactly takes T; testing for T specifically is just how a few of those idioms work to produce the important information. Namely whether you can pass a T to said function.
Looking for a specific function signature was almost always a means to an end, not the final goal.
Concepts, particularly requires expressions, is the end itself. It allows you to ask the question directly. Because really, you don't care if Function has a parameter that takes a T; you care whether Function(t) is legitimate code or not. Exactly how that happens is an implementation detail.
The only reason I can think of that someone might want to constrain a template on an exact signature (rather than an argument match) is to defeat implicit conversion. But you really shouldn't try to break basic language features like that. If someone writes a type that is implicitly convertible to another, they have the right to the benefits of that conversion, as defined by the language. Namely, the ability to use it in many ways as if it were that other type.
That is, if Function(t) is what your constrained template code is actually going to do, then the user of that template has every right to provide code that makes that compiler within the limits of the C++ language. Not within the limits of your personal ideas of what features are good or bad in that language.
Concepts are not like base classes, where you decide the exact signature for each method and the user must strictly abide by that. Concepts are patterns that constrain template definitions. Expressions in concept constraints are expressions that you expect to use in your template. You only put an expression in a concept if you plan on using it in your templates constrained by that concept.
You don't use a function signature; you call functions. So you constrain a concept on what functions can be called with which arguments. You're saying "you must let me do this", not "provide this signature".
That having been said... what you want is not generally possible ;)
There are several mechanisms that you might employ to achieve it, but none of them do exactly what you want in all cases.
The name of a function resolves to an overload set consisting of all of the functions that could be called. This name can be converted into a pointer to a specific function signature if and only if that signature is one of the functions in the overload set. So in theory, you might do this:
template <typename T>
concept has_f = requires () { static_cast<void (*)(T const&)>(&Function); };
However, because the name Function is not dependent on T (as far as C++ is concerned), it must be resolved during the first pass of two-phase name lookup for templates. That means any and all Function overloads you intend to care about have to be declared before has_f is defined, not merely instantiated with an appropriate T.
I think this is sufficient to declare that this is non-functional as a solution. Even if it worked though, it would only "work" given 3 circumstances:
Functionis known/required to be an actual function, rather than a global object with anoperator()overload. So if a provider ofTwants to provide a global functor instead of a regular function (for any number of reasons) this method will not work, even thoughFunction(t)is 100% perfectly valid, legitimate, and does none of those terrible implicit conversions that for some reason must be stopped.The expression
Function(t)is not expected to use ADL to find the actualFunctionto call.Functionis not a template function.
And not one of these possibilities has anything to do with implicit conversions. If you're going to call Function(t), then it's 100% OK for ADL to find it, template argument deduction to instantiate it, or for the user to fulfill this with some global lambda.
Your second-best bet is to rely on how overload resolution works. C++ only permits a single user-defined conversion in operator overloading. As such, you can create a type which will consume that one user-defined conversion in the function call expression in lieu of T. And that conversion should be a conversion to T itself.
You would use it like this:
template<typename T>
class udc_killer
{
public:
//Will never be called.
operator T const&();
};
template <typename T>
concept has_f = requires () { Function(udc_killer<T>{}); };
This of course still leaves the standard conversions, so you can't differentiate between a function taking a float if T is int, or derived classes from bases. You also can't detect if Function has any default parameters after the first one.
Overall, you're still not detecting the signature, merely call-ability. Because that's all you should care about to begin with.
来源：https://stackoverflow.com/questions/63385719/check-existence-of-global-function-but-disallow-implicit-conversion