C++ STL的std::has_virtual_destructor用於檢查給定的類型T是否具有虛構的析構函數。它返回布爾值true或false。以下是相同的語法:
頭文件:
#include<type_traits>
用法:
template <class T> struct has_virtual_destructor;
參數:模板std::has_virtual_destructor接受單個參數T(Trait類)以檢查T是否具有虛擬析構函數。
返回值:
- 正確:如果存在虛擬析構函數。
- False:如果虛擬析構函數不存在。
以下示例程序旨在說明C++ STL中的std::has_virtual_destructor模板:
程序1:
// C++ program to illustrate
// has_virtual_destructor example
#include <bits/stdc++.h>
#include <type_traits>
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<int>::value
<< endl;
cout << "gfg1:"
<< has_virtual_destructor<gfg1>::value
<< endl;
cout << "gfg2:"
<< has_virtual_destructor<gfg2>::value
<< endl;
cout << "gfg3:"
<< has_virtual_destructor<gfg3>::value
<< endl;
return 0;
}
輸出:
has_virtual_destructor: int:false gfg1:false gfg2:true gfg3:true
程序2:
// C++ program to illustrate
// has_virtual_destructor example
#include <bits/stdc++.h>
#include <type_traits>
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<int>::value
<< endl;
cout << "gfg1:"
<< has_virtual_destructor<gfg1>::value
<< endl;
cout << "gfg2:"
<< has_virtual_destructor<gfg2>::value
<< endl;
cout << "gfg3:"
<< has_virtual_destructor<gfg3>::value
<< endl;
return 0;
}
輸出:
has_virtual_destructor: int:false gfg1:true gfg2:false gfg3:true
參考: http://www.cplusplus.com/reference/type_traits/has_virtual_destructor/
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注:本文由純淨天空篩選整理自bansal_rtk_大神的英文原創作品 std::has_virtual_destructor in C++ with Examples。非經特殊聲明,原始代碼版權歸原作者所有,本譯文未經允許或授權,請勿轉載或複製。