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>() .我能为此做什么？

我对 SFINAE 有疑问。假设我们需要支持函数 CreateProcessor返回接口(interface) IProcessor<T> 的实现对于某些类型 T .

在 C++ 中，我们不能为一个函数创建多个仅在返回类型上不同的重载，因此我们必须创建函数 CreateProcessor也是由 T 参数化的模板函数.

现在假设我们要写一个模板函数Process<T>(T value)根据 CreateProcessor<T>() 的存在而不同，即它应该处理 value在情况下使用处理器 CreateProcessor<T>()已实现，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(prototype) 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。

最佳答案

使其工作的一种方法是使用包装器结构来运行 CreateProcessor，如下所示:

#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>
struct ProcessorCreator: std::false_type { 
   static IProcessor<T>* CreateProcessor();
};

// Template specialization for int.
template<>
struct ProcessorCreator<int>: std::true_type {
static IProcessor<int>* CreateProcessor() {
    return new IntProcessor();
}
};

// Detector of CreateProcessor.
template<class, class=void>
struct CreateProcessorImplemented : std::false_type { };

template<class T>
struct CreateProcessorImplemented<T, typename std::enable_if<ProcessorCreator<T>::value>::type > : 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 = ProcessorCreator<T>::CreateProcessor();
    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, ":(");
 */
}

或者，您可以删除模板声明并使用函数重载传递 IProcessor 模板参数类型——通过创建虚拟参数:

#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);
    }
};


IProcessor<int>* CreateProcessor(const int&) {
    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(std::declval<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(value);
    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, ":(");
 */
}

关于c++ - SFINAE 用于检测非成员模板函数的存在，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35939855/