是否有可能编写一个模板，根据某个成员函数是否定义在类上而改变行为?
下面是我想写的一个简单的例子:
template<class T>
std::string optionalToString(T* obj)
{
if (FUNCTION_EXISTS(T->toString))
return obj->toString();
else
return "toString not defined";
}
因此，如果类T定义了toString()，那么它就使用它;否则，它就不会。我不知道如何做的神奇部分是“FUNCTION_EXISTS”部分。
当前回答
我修改了https://stackoverflow.com/a/264088/2712152中提供的解决方案，使其更加通用。此外，由于它不使用任何新的c++ 11特性，我们可以将它与旧的编译器一起使用，并且应该也可以与msvc一起使用。但是编译器应该允许C99使用这个，因为它使用可变宏。
下面的宏可用于检查特定类是否具有特定类型定义。
/**
* @class : HAS_TYPEDEF
* @brief : This macro will be used to check if a class has a particular
* typedef or not.
* @param typedef_name : Name of Typedef
* @param name : Name of struct which is going to be run the test for
* the given particular typedef specified in typedef_name
*/
#define HAS_TYPEDEF(typedef_name, name) \
template <typename T> \
struct name { \
typedef char yes[1]; \
typedef char no[2]; \
template <typename U> \
struct type_check; \
template <typename _1> \
static yes& chk(type_check<typename _1::typedef_name>*); \
template <typename> \
static no& chk(...); \
static bool const value = sizeof(chk<T>(0)) == sizeof(yes); \
}
下面的宏可以用来检查一个特定的类是否有一个特定的成员函数，是否有给定数量的参数。
/**
* @class : HAS_MEM_FUNC
* @brief : This macro will be used to check if a class has a particular
* member function implemented in the public section or not.
* @param func : Name of Member Function
* @param name : Name of struct which is going to be run the test for
* the given particular member function name specified in func
* @param return_type: Return type of the member function
* @param ellipsis(...) : Since this is macro should provide test case for every
* possible member function we use variadic macros to cover all possibilities
*/
#define HAS_MEM_FUNC(func, name, return_type, ...) \
template <typename T> \
struct name { \
typedef return_type (T::*Sign)(__VA_ARGS__); \
typedef char yes[1]; \
typedef char no[2]; \
template <typename U, U> \
struct type_check; \
template <typename _1> \
static yes& chk(type_check<Sign, &_1::func>*); \
template <typename> \
static no& chk(...); \
static bool const value = sizeof(chk<T>(0)) == sizeof(yes); \
}
我们可以使用上面的两个宏来检查has_typedef和has_mem_func:
class A {
public:
typedef int check;
void check_function() {}
};
class B {
public:
void hello(int a, double b) {}
void hello() {}
};
HAS_MEM_FUNC(check_function, has_check_function, void, void);
HAS_MEM_FUNC(hello, hello_check, void, int, double);
HAS_MEM_FUNC(hello, hello_void_check, void, void);
HAS_TYPEDEF(check, has_typedef_check);
int main() {
std::cout << "Check Function A:" << has_check_function<A>::value << std::endl;
std::cout << "Check Function B:" << has_check_function<B>::value << std::endl;
std::cout << "Hello Function A:" << hello_check<A>::value << std::endl;
std::cout << "Hello Function B:" << hello_check<B>::value << std::endl;
std::cout << "Hello void Function A:" << hello_void_check<A>::value << std::endl;
std::cout << "Hello void Function B:" << hello_void_check<B>::value << std::endl;
std::cout << "Check Typedef A:" << has_typedef_check<A>::value << std::endl;
std::cout << "Check Typedef B:" << has_typedef_check<B>::value << std::endl;
}
其他回答
我在另一个线程中对此写了一个答案(与上面的解决方案不同)，也检查继承的成员函数:
SFINAE检查继承的成员函数
以下是该解决方案的一些例子:
例二:
我们正在检查一个具有以下签名的成员: T::const_iterator begin(
template<class T> struct has_const_begin
{
typedef char (&Yes)[1];
typedef char (&No)[2];
template<class U>
static Yes test(U const * data,
typename std::enable_if<std::is_same<
typename U::const_iterator,
decltype(data->begin())
>::value>::type * = 0);
static No test(...);
static const bool value = sizeof(Yes) == sizeof(has_const_begin::test((typename std::remove_reference<T>::type*)0));
};
请注意，它甚至检查方法的常量，并且也适用于基本类型。(我的意思是has_const_begin<int>::value为false，不会导致编译时错误。)
示例2
现在我们正在寻找签名:void foo(MyClass&， unsigned)
template<class T> struct has_foo
{
typedef char (&Yes)[1];
typedef char (&No)[2];
template<class U>
static Yes test(U * data, MyClass* arg1 = 0,
typename std::enable_if<std::is_void<
decltype(data->foo(*arg1, 1u))
>::value>::type * = 0);
static No test(...);
static const bool value = sizeof(Yes) == sizeof(has_foo::test((typename std::remove_reference<T>::type*)0));
};
请注意，MyClass不一定是默认可构造的或满足任何特殊的概念。该技术也适用于模板成员。
我急切地等待有关这方面的意见。
下面是我的版本，它可以任意处理所有可能的成员函数重载，包括模板成员函数，可能带有默认实参。当成员函数调用某个类类型时，它区分了3种互斥的情况，给定的arg类型:(1)有效，或(2)模糊，或(3)不可用。使用示例:
#include <string>
#include <vector>
HAS_MEM(bar)
HAS_MEM_FUN_CALL(bar)
struct test
{
void bar(int);
void bar(double);
void bar(int,double);
template < typename T >
typename std::enable_if< not std::is_integral<T>::value >::type
bar(const T&, int=0){}
template < typename T >
typename std::enable_if< std::is_integral<T>::value >::type
bar(const std::vector<T>&, T*){}
template < typename T >
int bar(const std::string&, int){}
};
现在你可以这样使用它:
int main(int argc, const char * argv[])
{
static_assert( has_mem_bar<test>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(char const*,long)>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(std::string&,long)>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(std::vector<int>, int*)>::value , "");
static_assert( has_no_viable_mem_fun_call_bar<test(std::vector<double>, double*)>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(int)>::value , "");
static_assert( std::is_same<void,result_of_mem_fun_call_bar<test(int)>::type>::value , "");
static_assert( has_valid_mem_fun_call_bar<test(int,double)>::value , "");
static_assert( not has_valid_mem_fun_call_bar<test(int,double,int)>::value , "");
static_assert( not has_ambiguous_mem_fun_call_bar<test(double)>::value , "");
static_assert( has_ambiguous_mem_fun_call_bar<test(unsigned)>::value , "");
static_assert( has_viable_mem_fun_call_bar<test(unsigned)>::value , "");
static_assert( has_viable_mem_fun_call_bar<test(int)>::value , "");
static_assert( has_no_viable_mem_fun_call_bar<test(void)>::value , "");
return 0;
}
下面是用c++11编写的代码，但是，你可以很容易地将它移植到具有typeof扩展的非c++11(例如gcc)。你可以用你自己的宏替换HAS_MEM宏。
#pragma once
#if __cplusplus >= 201103
#include <utility>
#include <type_traits>
#define HAS_MEM(mem) \
\
template < typename T > \
struct has_mem_##mem \
{ \
struct yes {}; \
struct no {}; \
\
struct ambiguate_seed { char mem; }; \
template < typename U > struct ambiguate : U, ambiguate_seed {}; \
\
template < typename U, typename = decltype(&U::mem) > static constexpr no test(int); \
template < typename > static constexpr yes test(...); \
\
static bool constexpr value = std::is_same<decltype(test< ambiguate<T> >(0)),yes>::value ; \
typedef std::integral_constant<bool,value> type; \
};
#define HAS_MEM_FUN_CALL(memfun) \
\
template < typename Signature > \
struct has_valid_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct has_valid_mem_fun_call_##memfun< T(Args...) > \
{ \
struct yes {}; \
struct no {}; \
\
template < typename U, bool = has_mem_##memfun<U>::value > \
struct impl \
{ \
template < typename V, typename = decltype(std::declval<V>().memfun(std::declval<Args>()...)) > \
struct test_result { using type = yes; }; \
\
template < typename V > static constexpr typename test_result<V>::type test(int); \
template < typename > static constexpr no test(...); \
\
static constexpr bool value = std::is_same<decltype(test<U>(0)),yes>::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename U > \
struct impl<U,false> : std::false_type {}; \
\
static constexpr bool value = impl<T>::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename Signature > \
struct has_ambiguous_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct has_ambiguous_mem_fun_call_##memfun< T(Args...) > \
{ \
struct ambiguate_seed { void memfun(...); }; \
\
template < class U, bool = has_mem_##memfun<U>::value > \
struct ambiguate : U, ambiguate_seed \
{ \
using ambiguate_seed::memfun; \
using U::memfun; \
}; \
\
template < class U > \
struct ambiguate<U,false> : ambiguate_seed {}; \
\
static constexpr bool value = not has_valid_mem_fun_call_##memfun< ambiguate<T>(Args...) >::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename Signature > \
struct has_viable_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct has_viable_mem_fun_call_##memfun< T(Args...) > \
{ \
static constexpr bool value = has_valid_mem_fun_call_##memfun<T(Args...)>::value \
or has_ambiguous_mem_fun_call_##memfun<T(Args...)>::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename Signature > \
struct has_no_viable_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct has_no_viable_mem_fun_call_##memfun < T(Args...) > \
{ \
static constexpr bool value = not has_viable_mem_fun_call_##memfun<T(Args...)>::value; \
using type = std::integral_constant<bool, value>; \
}; \
\
template < typename Signature > \
struct result_of_mem_fun_call_##memfun; \
\
template < typename T, typename... Args > \
struct result_of_mem_fun_call_##memfun< T(Args...) > \
{ \
using type = decltype(std::declval<T>().memfun(std::declval<Args>()...)); \
};
#endif
奇怪的是，竟然没有人建议我在这个网站上看到的下面这个漂亮的把戏:
template <class T>
struct has_foo
{
struct S { void foo(...); };
struct derived : S, T {};
template <typename V, V> struct W {};
template <typename X>
char (&test(W<void (X::*)(), &X::foo> *))[1];
template <typename>
char (&test(...))[2];
static const bool value = sizeof(test<derived>(0)) == 1;
};
你必须确保T是一个类。查找foo时的模糊性似乎是替换失败。我让它在gcc上工作，但不确定它是否是标准的。
这是个不错的小难题——好问题!
这里有一个替代Nicola Bonelli的解决方案，它不依赖于非标准typeof运算符。
不幸的是，它不能在GCC (MinGW) 3.4.5或Digital Mars 8.42n上工作，但它可以在所有版本的MSVC(包括VC6)和Comeau c++上工作。
较长的注释块有关于它如何工作(或应该如何工作)的详细信息。正如它所说，我不确定哪些行为符合标准-我欢迎对此发表评论。
更新- 2008年11月7日:
看起来，虽然这段代码在语法上是正确的，但MSVC和Comeau c++所显示的行为并不符合标准(感谢Leon Timmermans和litb为我指明了正确的方向)。c++ 03标准说:
14.6.2依赖名称[temp.dep] 段3 在类模板定义中 或类模板的成员，如果 类模板的基类 类型取决于模板参数 基类范围不检查 在非限定名称查找期间 在定义的时候 类的模板或成员 类模板的实例化或 成员。
因此，当MSVC或Comeau考虑T的toString()成员函数在模板实例化时在doToString()中的调用站点执行名称查找时，这看起来是不正确的(尽管它实际上是我在本例中寻找的行为)。
GCC和Digital Mars的行为看起来是正确的——在这两种情况下，非成员toString()函数都绑定到调用。
老鼠-我以为我可能找到了一个聪明的解决方案，但我发现了几个编译器错误…
#include <iostream>
#include <string>
struct Hello
{
std::string toString() {
return "Hello";
}
};
struct Generic {};
// the following namespace keeps the toString() method out of
// most everything - except the other stuff in this
// compilation unit
namespace {
std::string toString()
{
return "toString not defined";
}
template <typename T>
class optionalToStringImpl : public T
{
public:
std::string doToString() {
// in theory, the name lookup for this call to
// toString() should find the toString() in
// the base class T if one exists, but if one
// doesn't exist in the base class, it'll
// find the free toString() function in
// the private namespace.
//
// This theory works for MSVC (all versions
// from VC6 to VC9) and Comeau C++, but
// does not work with MinGW 3.4.5 or
// Digital Mars 8.42n
//
// I'm honestly not sure what the standard says
// is the correct behavior here - it's sort
// of like ADL (Argument Dependent Lookup -
// also known as Koenig Lookup) but without
// arguments (except the implied "this" pointer)
return toString();
}
};
}
template <typename T>
std::string optionalToString(T & obj)
{
// ugly, hacky cast...
optionalToStringImpl<T>* temp = reinterpret_cast<optionalToStringImpl<T>*>( &obj);
return temp->doToString();
}
int
main(int argc, char *argv[])
{
Hello helloObj;
Generic genericObj;
std::cout << optionalToString( helloObj) << std::endl;
std::cout << optionalToString( genericObj) << std::endl;
return 0;
}
c++ 20 -需要表达式
c++ 20带来了一些概念和各种工具，比如require表达式，这是一种检查函数是否存在的内置方式。有了它们，你可以重写optionalToString函数如下:
template<class T>
std::string optionalToString(T* obj)
{
constexpr bool has_toString = requires(const T& t) {
t.toString();
};
if constexpr (has_toString)
return obj->toString();
else
return "toString not defined";
}
pre - c++ 20 -检测工具包
N4502 proposes a detection toolkit for inclusion into the C++17 standard library that eventually made it into the library fundamentals TS v2. It most likely won't ever get into the standard because it has been subsumed by requires expressions since, but it still solves the problem in a somewhat elegant manner. The toolkit introduces some metafunctions, including std::is_detected which can be used to easily write type or function detection metafunctions on the top of it. Here is how you could use it:
template<typename T>
using toString_t = decltype( std::declval<T&>().toString() );
template<typename T>
constexpr bool has_toString = std::is_detected_v<toString_t, T>;
注意，上面的例子是未经测试的。标准库中还没有检测工具包，但建议包含了一个完整的实现，如果您确实需要它，可以很容易地复制它。它可以很好地使用c++ 17的特性，如果constexpr:
template<class T>
std::string optionalToString(T* obj)
{
if constexpr (has_toString<T>)
return obj->toString();
else
return "toString not defined";
}
C++14 - 助推哈娜
提振。Hana显然建立在这个特定的例子之上，并在其文档中提供了c++ 14的解决方案，所以我将直接引用它:
[...] Hana provides a is_valid function that can be combined with C++14 generic lambdas to obtain a much cleaner implementation of the same thing: auto has_toString = hana::is_valid([](auto&& obj) -> decltype(obj.toString()) { }); This leaves us with a function object has_toString which returns whether the given expression is valid on the argument we pass to it. The result is returned as an IntegralConstant, so constexpr-ness is not an issue here because the result of the function is represented as a type anyway. Now, in addition to being less verbose (that's a one liner!), the intent is much clearer. Other benefits are the fact that has_toString can be passed to higher order algorithms and it can also be defined at function scope, so there is no need to pollute the namespace scope with implementation details.
提振。创科实业
执行这种检查的另一个惯用工具包是Boost，尽管它没有那么优雅。TTI，在Boost 1.54 4.0中引入。对于您的示例，您必须使用宏BOOST_TTI_HAS_MEMBER_FUNCTION。下面是你如何使用它:
#include <boost/tti/has_member_function.hpp>
// Generate the metafunction
BOOST_TTI_HAS_MEMBER_FUNCTION(toString)
// Check whether T has a member function toString
// which takes no parameter and returns a std::string
constexpr bool foo = has_member_function_toString<T, std::string>::value;
然后，您可以使用bool来创建SFINAE检查。
解释
宏BOOST_TTI_HAS_MEMBER_FUNCTION生成元函数has_member_function_toString，该函数将选中的类型作为其第一个模板参数。第二个模板形参对应于成员函数的返回类型，下面的形参对应于函数形参的类型。如果类T有成员函数std::string toString()，则成员值为true。
或者，has_member_function_toString可以接受成员函数指针作为模板形参。因此，可以将has_member_function_toString<T, std::string>::value替换为has_member_function_toString<std::string T::* ()>::value。
