C++獲取完整類型名

typeid 運算符

• 查詢類型的信息。
• 用於必須知曉多態對象的動態類型的場合以及靜態類型鑒別。
• 語法
  (1) typeid( 類型 )
  (2) typeid( 表達式 )
• typeid 表達式為左值表達式，指代一個具有靜態存儲期的，多態類型 或某個其派生類型的const std::type_info 對象。
• 類 type_info 指定一個類型的信息，包括類型的名稱和比較二個類型相等的方法。這是 typeid 運算符所返回的類。
• 獲取完整類型名方法：typeid(類型名/表達式).name()

#include <iostream>
#include <string>
#include <typeinfo>
 
struct Base {}; // 非多態
struct Derived : Base {};
 
struct Base2 { virtual void foo() {} }; // 多態
struct Derived2 : Base2 {};
 
int main() {
    int myint = 50;
    std::string mystr = "string";
    double *mydoubleptr = nullptr;
 
    std::cout << "myint has type: " << typeid(myint).name() << '\n'
              << "mystr has type: " << typeid(mystr).name() << '\n'
              << "mydoubleptr has type: " << typeid(mydoubleptr).name() << '\n';
 
    // std::cout << myint 為多態類型的泛左值表達式；求值
    const std::type_info& r1 = typeid(std::cout << myint);
    std::cout << '\n' << "std::cout<<myint has type : " << r1.name() << '\n';
 
    // std::printf() 不是多態類型的泛左值表達式；不求值
    const std::type_info& r2 = typeid(std::printf("%d\n", myint));
    std::cout << "printf(\"%d\\n\",myint) has type : " << r2.name() << '\n';
 
    // 非多態左值時為靜態類型
    Derived d1;
    Base& b1 = d1;
    std::cout << "reference to non-polymorphic base: " << typeid(b1).name() << '\n';
 
    Derived2 d2;
    Base2& b2 = d2;
    std::cout << "reference to polymorphic base: " << typeid(b2).name() << '\n';
 
    try {
        // 解引用空指針：對於非多態表達式 OK
        std::cout << "mydoubleptr points to " << typeid(*mydoubleptr).name() << '\n'; 
        // 解引用空指針：對多態左值則不行
        Derived2* bad_ptr = nullptr;
        std::cout << "bad_ptr points to... ";
        std::cout << typeid(*bad_ptr).name() << '\n';
    } catch (const std::bad_typeid& e) {
         std::cout << " caught " << e.what() << '\n';
    }
}

可能的輸出：

myint has type: int
mystr has type: std::basic_string<char, std::char_traits<char>, std::allocator<char> >
mydoubleptr has type: double*
50
std::cout<<myint has type : std::basic_ostream<char, std::char_traits<char> >
printf("%d\n",myint) has type : int
reference to non-polymorphic base: Base
reference to polymorphic base: Derived2
mydoubleptr points to double
bad_ptr points to...  caught std::bad_typeid