非常丑陋，但如果你只想要编译时信息(例如调试):

auto testVar = std::make_tuple(1, 1.0, "abc");
decltype(testVar)::foo= 1;

返回:

Compilation finished with errors:
source.cpp: In function 'int main()':
source.cpp:5:19: error: 'foo' is not a member of 'std::tuple<int, double, const char*>'

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;
}

你可以上特质课。喜欢的东西:

#include <iostream>
using namespace std;

template <typename T> class type_name {
public:
    static const char *name;
};

#define DECLARE_TYPE_NAME(x) template<> const char *type_name<x>::name = #x;
#define GET_TYPE_NAME(x) (type_name<typeof(x)>::name)

DECLARE_TYPE_NAME(int);

int main()
{
    int a = 12;
    cout << GET_TYPE_NAME(a) << endl;
}

DECLARE_TYPE_NAME定义的存在是为了让您更容易地为所有需要的类型声明这个trait类。

这可能比涉及typeid的解决方案更有用，因为您可以控制输出。例如，在我的编译器上使用typeid For long long会给出“x”。

非常丑陋，但如果你只想要编译时信息(例如调试):

auto testVar = std::make_tuple(1, 1.0, "abc");
decltype(testVar)::foo= 1;

返回:

Compilation finished with errors:
source.cpp: In function 'int main()':
source.cpp:5:19: error: 'foo' is not a member of 'std::tuple<int, double, const char*>'

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。