TL;DR我想写一个模板函数Process(T value)
根据非成员函数的存在,对于不同的值表现不同CreateProcessor<T>()
。我能为此做些什么呢?
我对 SFINAE 有疑问。假设我们需要支持功能CreateProcessor
返回接口的实现IProcessor<T>
对于某些类型类型T
.
在 C++ 中,我们不能创建仅返回类型不同的函数的多个重载,因此我们必须使函数CreateProcessor
也可以是参数化的模板函数T
.
现在假设我们要写一个模板函数Process<T>(T value)
根据是否存在而工作不同CreateProcessor<T>()
,即它应该处理value
使用处理器以防万一CreateProcessor<T>()
已执行,否则会导致错误。
我尝试编写以下代码:
#include <cstdio>
#include <type_traits>
// A workaround for void_t as described here: http://en.cppreference.com/w/cpp/types/void_t.
template<typename... Ts> struct make_void { typedef void type;};
template<typename... Ts> using void_t = typename make_void<Ts...>::type;
// An interface for a processor that receives a value of specific type.
template<class T>
class IProcessor {
public:
virtual void process(T value) = 0;
};
// A processor for int.
class IntProcessor : public IProcessor<int> {
public:
virtual void process(int value) override {
printf("IntProcessor::process is called for value = %d\n", value);
}
};
// Template prototype.
template<class T>
IProcessor<T>* CreateProcessor();
// Template specialization for int.
template<>
IProcessor<int>* CreateProcessor() {
return new IntProcessor();
}
// Detector of CreateProcessor.
template<class, class=void>
struct CreateProcessorImplemented : std::false_type { };
template<class T>
struct CreateProcessorImplemented<T, void_t<decltype(CreateProcessor<T>())>> : std::true_type { };
// Specializations depending on existence of CreateProcessor.
template <typename T>
typename std::enable_if<CreateProcessorImplemented<T>::value, void>::type Process(T value) {
IProcessor<T>* processor = CreateProcessor<T>();
processor->process(value);
}
template <typename T>
typename std::enable_if<!CreateProcessorImplemented<T>::value, void>::type Process(T value) {
printf("Processor for requested typename is unavailable\n");
}
int main() {
Process(42);
Process("abc");
// static_assert(!CreateProcessorImplemented<char const*>::value, ":(");
/* This static_assert fails with an error:
* code.cpp:56:5: error: static assertion failed: :(
* static_assert(!CreateProcessorImplemented<char const*>::value, ":(");
*/
}
虽然这会导致链接错误:
/tmp/ccTQRc9N.o:code.cpp:function std::enable_if<CreateProcessorImplemented<char const*, void>::value, void>::type Process<char const*>(char const*): error: undefined reference to 'IProcessor<char const*>* CreateProcessor<char const*>()'
collect2: error: ld returned 1 exit status
我的想法是,当我们解决CreateProcessorImplemented<char const*>
, decltype(CreateProcessor<const char*>())
不会失败,因为有模板原型IProcessor<T> CreateProcessor()
并且编译器认为 decltype 等于IProcessor<T>
这在某种程度上是合乎逻辑的,但不是我需要的。