例如:
int a = 12;
cout << typeof(a) << endl;
预期的输出:
int
例如:
int a = 12;
cout << typeof(a) << endl;
预期的输出:
int
当前回答
如前所述,typeid().name()可能返回一个错误的名称。在GCC(和其他一些编译器)中,你可以使用以下代码来解决它:
#include <cxxabi.h>
#include <iostream>
#include <typeinfo>
#include <cstdlib>
namespace some_namespace { namespace another_namespace {
class my_class { };
} }
int main() {
typedef some_namespace::another_namespace::my_class my_type;
// mangled
std::cout << typeid(my_type).name() << std::endl;
// unmangled
int status = 0;
char* demangled = abi::__cxa_demangle(typeid(my_type).name(), 0, 0, &status);
switch (status) {
case -1: {
// could not allocate memory
std::cout << "Could not allocate memory" << std::endl;
return -1;
} break;
case -2: {
// invalid name under the C++ ABI mangling rules
std::cout << "Invalid name" << std::endl;
return -1;
} break;
case -3: {
// invalid argument
std::cout << "Invalid argument to demangle()" << std::endl;
return -1;
} break;
}
std::cout << demangled << std::endl;
free(demangled);
return 0;
}
其他回答
正如Scott Meyers在《Effective Modern c++》中所解释的那样,
对std::type_info::name的调用不能保证返回任何有意义的东西。
最好的解决方案是让编译器在类型推断期间生成错误消息,例如:
template<typename T>
class TD;
int main(){
const int theAnswer = 32;
auto x = theAnswer;
auto y = &theAnswer;
TD<decltype(x)> xType;
TD<decltype(y)> yType;
return 0;
}
根据不同的编译器,结果会是这样的:
test4.cpp:10:21: error: aggregate ‘TD<int> xType’ has incomplete type and cannot be defined TD<decltype(x)> xType;
test4.cpp:11:21: error: aggregate ‘TD<const int *> yType’ has incomplete type and cannot be defined TD<decltype(y)> yType;
因此,我们知道x的类型是int, y的类型是const int*
复制这个答案:https://stackoverflow.com/a/56766138/11502722
我能够在c++ static_assert()中获得这一点。这里的问题是static_assert()只接受字符串字面量;Constexpr string_view将不起作用。你需要接受typename周围的额外文本,但它可以工作:
template<typename T>
constexpr void assertIfTestFailed()
{
#ifdef __clang__
static_assert(testFn<T>(), "Test failed on this used type: " __PRETTY_FUNCTION__);
#elif defined(__GNUC__)
static_assert(testFn<T>(), "Test failed on this used type: " __PRETTY_FUNCTION__);
#elif defined(_MSC_VER)
static_assert(testFn<T>(), "Test failed on this used type: " __FUNCSIG__);
#else
static_assert(testFn<T>(), "Test failed on this used type (see surrounding logged error for details).");
#endif
}
}
MSVC输出:
error C2338: Test failed on this used type: void __cdecl assertIfTestFailed<class BadType>(void)
... continued trace of where the erroring code came from ...
涉及RTTI (typeid)的其他答案可能是您想要的,只要:
您可以承担内存开销(对于某些编译器,这可能相当大) 编译器返回的类名很有用
另一种选择(类似于Greg Hewgill的答案)是建立一个特征的编译时表。
template <typename T> struct type_as_string;
// declare your Wibble type (probably with definition of Wibble)
template <>
struct type_as_string<Wibble>
{
static const char* const value = "Wibble";
};
注意,如果你将声明包装在宏中,你将在声明带有多个参数的模板类型时遇到麻烦(例如std::map),这是由于逗号的原因。
要访问变量类型的名称,您所需要的是
template <typename T>
const char* get_type_as_string(const T&)
{
return type_as_string<T>::value;
}
Howard Hinnant使用魔法数字提取类型名称。康桓瑋建议字符串前缀和后缀。但是前缀/后缀一直在变化。 使用" probe_type " type_name自动计算" probe_type "的前缀和后缀大小,以提取类型名称:
#include <string_view>
using namespace std;
namespace typeName {
template <typename T>
constexpr string_view wrapped_type_name () {
#ifdef __clang__
return __PRETTY_FUNCTION__;
#elif defined(__GNUC__)
return __PRETTY_FUNCTION__;
#elif defined(_MSC_VER)
return __FUNCSIG__;
#endif
}
class probe_type;
constexpr string_view probe_type_name ("typeName::probe_type");
constexpr string_view probe_type_name_elaborated ("class typeName::probe_type");
constexpr string_view probe_type_name_used (wrapped_type_name<probe_type> ().find (probe_type_name_elaborated) != -1 ? probe_type_name_elaborated : probe_type_name);
constexpr size_t prefix_size () {
return wrapped_type_name<probe_type> ().find (probe_type_name_used);
}
constexpr size_t suffix_size () {
return wrapped_type_name<probe_type> ().length () - prefix_size () - probe_type_name_used.length ();
}
template <typename T>
string_view type_name () {
constexpr auto type_name = wrapped_type_name<T> ();
return type_name.substr (prefix_size (), type_name.length () - prefix_size () - suffix_size ());
}
}
#include <iostream>
using typeName::type_name;
using typeName::probe_type;
class test;
int main () {
cout << type_name<class test> () << endl;
cout << type_name<const int*&> () << endl;
cout << type_name<unsigned int> () << endl;
const int ic = 42;
const int* pic = ⁣
const int*& rpic = pic;
cout << type_name<decltype(ic)> () << endl;
cout << type_name<decltype(pic)> () << endl;
cout << type_name<decltype(rpic)> () << endl;
cout << type_name<probe_type> () << endl;
}
输出
gcc 10.2:
test
const int *&
unsigned int
const int
const int *
const int *&
typeName::probe_type
铿锵声11.0.0:
test
const int *&
unsigned int
const int
const int *
const int *&
typeName::probe_type
VS 2019版本16.7.6:
class test
const int*&
unsigned int
const int
const int*
const int*&
class typeName::probe_type
对于那些还在访问的人,我最近也遇到了同样的问题,并决定根据这篇文章的答案编写一个小型库。它提供了constexpr类型名称和类型索引,并且在Mac, Windows和Ubuntu上进行了测试。
库代码在这里:https://github.com/TheLartians/StaticTypeInfo