根据Howard的解决方案，如果你不喜欢神奇的数字，我认为这是一种很好的表示方式，看起来很直观:

#include <string_view>

template <typename T>
constexpr auto type_name() {
  std::string_view name, prefix, suffix;
#ifdef __clang__
  name = __PRETTY_FUNCTION__;
  prefix = "auto type_name() [T = ";
  suffix = "]";
#elif defined(__GNUC__)
  name = __PRETTY_FUNCTION__;
  prefix = "constexpr auto type_name() [with T = ";
  suffix = "]";
#elif defined(_MSC_VER)
  name = __FUNCSIG__;
  prefix = "auto __cdecl type_name<";
  suffix = ">(void)";
#endif
  name.remove_prefix(prefix.size());
  name.remove_suffix(suffix.size());
  return name;
}

演示。

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 = &ic;
  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

你也可以使用c++filt选项-t (type)来要求类型名:

#include <iostream>
#include <typeinfo>
#include <string>

using namespace std;

int main() {
  auto x = 1;
  string my_type = typeid(x).name();
  system(("echo " + my_type + " | c++filt -t").c_str());
  return 0;
}

仅在linux上测试。

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;

}

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 ...

在c++ 11中，我们有decltype。在标准c++中，没有办法显示使用decltype声明的变量的确切类型。我们可以使用boost typeindex，即type_id_with_cvr (cvr代表const, volatile, reference)来打印如下所示的类型。

#include <iostream>
#include <boost/type_index.hpp>

using namespace std;
using boost::typeindex::type_id_with_cvr;

int main() {
  int i = 0;
  const int ci = 0;
  cout << "decltype(i) is " << type_id_with_cvr<decltype(i)>().pretty_name() << '\n';
  cout << "decltype((i)) is " << type_id_with_cvr<decltype((i))>().pretty_name() << '\n';
  cout << "decltype(ci) is " << type_id_with_cvr<decltype(ci)>().pretty_name() << '\n';
  cout << "decltype((ci)) is " << type_id_with_cvr<decltype((ci))>().pretty_name() << '\n';
  cout << "decltype(std::move(i)) is " << type_id_with_cvr<decltype(std::move(i))>().pretty_name() << '\n';
  cout << "decltype(std::static_cast<int&&>(i)) is " << type_id_with_cvr<decltype(static_cast<int&&>(i))>().pretty_name() << '\n';
  return 0;
}