C++中的std :: has_virtual_destructor以及示例

C++ STL的std :: has_virtual_destructor用于检查给定的类型T是否具有虚拟的析构函数。它返回布尔值true或false。以下是相同的语法：

头文件：

#include

句法：

template
  
  struct has_virtual_destructor;

参数：模板std :: has_virtual_destructor接受单个参数T(Trait类)以检查T是否具有虚拟析构函数。

返回值：

• 正确：如果存在虚拟析构函数。
• False：如果虚拟析构函数不存在。

下面的程序说明了C++ STL中的std :: has_virtual_destructor模板：

程序1：

// C++ program to illustrate
// has_virtual_destructor example
  
#include 
#include 
using namespace std;
  
struct gfg1 {
};
struct gfg2 {
    virtual ~gfg2() {}
};
struct gfg3 : gfg2 {
};
  
// Driver Code
int main()
{
    cout << boolalpha;
    cout << "has_virtual_destructor:"
         << endl;
    cout << "int: "
         << has_virtual_destructor::value
         << endl;
    cout << "gfg1: "
         << has_virtual_destructor::value
         << endl;
    cout << "gfg2: "
         << has_virtual_destructor::value
         << endl;
    cout << "gfg3: "
         << has_virtual_destructor::value
         << endl;
  
    return 0;
}

输出：

has_virtual_destructor:
int: false
gfg1: false
gfg2: true
gfg3: true

程式2：

// C++ program to illustrate
// has_virtual_destructor example
  
#include 
#include 
using namespace std;
  
struct gfg1 {
    virtual ~gfg1() {}
};
struct gfg2 {
};
struct gfg3 : gfg1 {
};
  
// Driver Code
int main()
{
    cout << boolalpha;
    cout << "has_virtual_destructor:"
         << endl;
    cout << "int: "
         << has_virtual_destructor::value
         << endl;
    cout << "gfg1: "
         << has_virtual_destructor::value
         << endl;
    cout << "gfg2: "
         << has_virtual_destructor::value
         << endl;
    cout << "gfg3: "
         << has_virtual_destructor::value
         << endl;
  
    return 0;
}

输出：

has_virtual_destructor:
int: false
gfg1: true
gfg2: false
gfg3: true

参考： http://www.cplusplus.com/reference/type_traits/has_virtual_destructor/

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