<type_traits>頭文件中提供了C++ STL的std::is_nothrow_destructible模板。 C++ STL的std::is_nothrow_denstructible模板用於檢查T是否為可破壞類型,並且眾所周知它不會引發任何異常。如果T是可銷毀類型,則返回布爾值true,否則返回false。
頭文件:
#include<type_traits>
模板類別:
template <class T> struct is_nothrow_destructible;
用法:
std::is_nothrow_destructible<T>::value
參數:模板std::is_nothrow_destructible接受單個參數T(特質類)以檢查T是否為可破壞類型。
返回值:模板std::is_nothrow_destructible返回一個布爾值:
- True:如果類型T是可破壞的類型。
- False:如果類型T是不可破壞的類型。
以下是在C++中演示std::is_nothrow_destructible的程序:
程序1:
// C++ program to illustrate
// std::is_nothrow_destructible
#include <bits/stdc++.h>
#include <type_traits>
using namespace std;
// Declare a structures
struct X {
};
struct Y {
// Destructors
~Y() = delete;
};
struct Z {
~Z() = default;
};
struct A:Y {
};
// Driver Code
int main()
{
cout << boolalpha;
// Check if int is nothrow
// destructible or not
cout << "int is nothrow destructible? "
<< is_nothrow_destructible<int>::value
<< endl;
// Check if float is nothrow
// destructible or not
cout << "float is nothrow destructible? "
<< is_nothrow_destructible<float>::value
<< endl;
// Check if struct X is
// nothrow destructible or not
cout << "struct X is nothrow destructible? "
<< is_nothrow_destructible<X>::value
<< endl;
// Check if struct Y is
// nothrow destructible or not
cout << "struct Y is nothrow destructible? "
<< is_nothrow_destructible<Y>::value
<< endl;
// Check if struct Z is
// nothrow destructible or not
cout << "struct Z is nothrow destructible? "
<< is_nothrow_destructible<Z>::value
<< endl;
// Check if struct A is
// nothrow destructible or not
cout << "struct A is nothrow destructible? "
<< is_nothrow_destructible<A>::value
<< endl;
return 0;
}
輸出:
int is nothrow destructible? true float is nothrow destructible? true struct X is nothrow destructible? true struct Y is nothrow destructible? false struct Z is nothrow destructible? true struct A is nothrow destructible? false
程序2:
// C++ program to illustrate
// std::is_nothrow_destructible
#include <bits/stdc++.h>
#include <type_traits>
using namespace std;
// Class GfG
class GfG {
int v1;
float v2;
public:
GfG(int n)
:v1(n), v2()
{
}
GfG(int n, double f) noexcept
:v1(n),
v2(f) {}
};
// Declare Structure
struct X {
int n;
X() = default;
};
// Driver Code
int main()
{
cout << boolalpha;
cout << "GfG is Nothrow-destructible for int? "
<< is_nothrow_destructible<int>::value
<< '\n';
cout << "GfG is Nothrow-destructible for GfG? "
<< is_nothrow_destructible<GfG>::value
<< '\n';
cout << "GfG is Nothrow-destructible for struct X? "
<< is_nothrow_destructible<X>::value
<< '\n';
}
輸出:
GfG is Nothrow-destructible for int? true GfG is Nothrow-destructible for GfG? true GfG is Nothrow-destructible for struct X? true
參考: http://www.cplusplus.com/reference/type_traits/is_nothrow_destructible/
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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::is_nothrow_destructible 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。