#include <iostream>
#include <typeinfo>
using namespace std;
#define show_type_name(_t) \
    system(("echo " + string(typeid(_t).name()) + " | c++filt -t").c_str())

int main() {
    auto a = {"one", "two", "three"};
    cout << "Type of a: " << typeid(a).name() << endl;
    cout << "Real type of a:\n";
    show_type_name(a);
    for (auto s : a) {
        if (string(s) == "one") {
            cout << "Type of s: " << typeid(s).name() << endl;
            cout << "Real type of s:\n";
            show_type_name(s);
        }
        cout << s << endl;
    }

    int i = 5;
    cout << "Type of i: " << typeid(i).name() << endl;
    cout << "Real type of i:\n";
    show_type_name(i);
    return 0;
}

输出:

Type of a: St16initializer_listIPKcE
Real type of a:
std::initializer_list<char const*>
Type of s: PKc
Real type of s:
char const*
one
two
three
Type of i: i
Real type of i:
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 ...

一个没有函数重载的更通用的解决方案:

template<typename T>
std::string TypeOf(T){
    std::string Type="unknown";
    if(std::is_same<T,int>::value) Type="int";
    if(std::is_same<T,std::string>::value) Type="String";
    if(std::is_same<T,MyClass>::value) Type="MyClass";

    return Type;}

这里的MyClass是用户定义的类。这里还可以添加更多的条件。

例子:

#include <iostream>



class MyClass{};


template<typename T>
std::string TypeOf(T){
    std::string Type="unknown";
    if(std::is_same<T,int>::value) Type="int";
    if(std::is_same<T,std::string>::value) Type="String";
    if(std::is_same<T,MyClass>::value) Type="MyClass";
    return Type;}


int main(){;
    int a=0;
    std::string s="";
    MyClass my;
    std::cout<<TypeOf(a)<<std::endl;
    std::cout<<TypeOf(s)<<std::endl;
    std::cout<<TypeOf(my)<<std::endl;

    return 0;}

输出:

int
String
MyClass

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*

Try:

#include <typeinfo>

// …
std::cout << typeid(a).name() << '\n';

您可能必须在编译器选项中激活RTTI才能使其工作。此外，它的输出取决于编译器。它可能是一个原始类型名称或名称混乱符号或介于两者之间的任何东西。

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。