C++ STL的std::is_compound模板用于检查类型是否为复合类型。它返回一个显示相同的布尔值。
用法:
template < class T > struct is_compound;
参数:此模板包含单个参数T(特质类),以检查T是否为复合类型。
返回值:此模板返回一个布尔值,如下所示:
- True:如果类型是复合类型。
- False:如果类型是非复合类型。
以下示例程序旨在说明C++ STL中的is_compound模板:
程序1::
// C++ program to illustrate
// is_compound template
#include <iostream>
#include <type_traits>
using namespace std;
// main program
struct GFG1 {
};
union GFG2 {
int var1;
float var2;
};
int main()
{
cout << boolalpha;
cout << "is_compound:"
<< endl;
cout << "GFG1:"
<< is_compound<GFG1>::value
<< endl;
cout << "GFG2:"
<< is_compound<GFG2>::value
<< endl;
cout << "int:"
<< is_compound<int>::value
<< endl;
cout << "int*:"
<< is_compound<int*>::value
<< endl;
return 0;
}
输出:
is_compound: GFG1:true GFG2:true int:false int*:true
程序2::
// C++ program to illustrate
// is_compound template
#include <iostream>
#include <type_traits>
using namespace std;
class GFG1 {
};
enum class GFG2 { var1,
var2,
var3,
var4
};
// main program
int main()
{
cout << boolalpha;
cout << "is_compound:"
<< endl;
cout << "GFG1:"
<< is_compound<GFG1>::value
<< endl;
cout << "GFG2:"
<< is_compound<GFG2>::value
<< endl;
cout << "int[10]:"
<< is_compound<int[10]>::value
<< endl;
cout << "int &:"
<< is_compound<int&>::value
<< endl;
cout << "char:"
<< is_compound<char>::value
<< endl;
return 0;
}
输出:
is_compound: GFG1:true GFG2:true int[10]:true int &:true char:false
程序3::
// C++ program to illustrate
// is_compound template
#include <iostream>
#include <type_traits>
using namespace std;
// driver code
int main()
{
class gfg {
};
cout << boolalpha;
cout << "is_compound:"
<< endl;
cout << "int(gfg::*):"
<< is_compound<int(gfg::*)>::value
<< endl;
cout << "float:"
<< is_compound<float>::value
<< endl;
cout << "double:"
<< is_compound<double>::value
<< endl;
cout << "int(int):"
<< is_compound<int(int)>::value
<< endl;
return 0;
}
输出:
is_compound: int(gfg::*):true float:false double:false int(int):true
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注:本文由纯净天空筛选整理自rajasethupathi大神的英文原创作品 std::is_compound Template in C++。非经特殊声明,原始代码版权归原作者所有,本译文未经允许或授权,请勿转载或复制。