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值的布尔值。
语法:
template < class T > struct is_compound;
参数:此模板包含单个参数T(Trait类),以检查T是否为复合类型。
返回值:该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。
下面的程序说明了C++ STL中的is_compound模板:
程序1 :
// C++ program to illustrate
// is_compound template
#include
#include
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::value
<< endl;
cout << "GFG2: "
<< is_compound::value
<< endl;
cout << "int: "
<< is_compound::value
<< endl;
cout << "int*: "
<< is_compound::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
#include
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::value
<< endl;
cout << "GFG2: "
<< is_compound::value
<< endl;
cout << "int[10]: "
<< is_compound::value
<< endl;
cout << "int &: "
<< is_compound::value
<< endl;
cout << "char: "
<< is_compound::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
#include
using namespace std;
// driver code
int main()
{
class gfg {
};
cout << boolalpha;
cout << "is_compound:"
<< endl;
cout << "int(gfg::*): "
<< is_compound::value
<< endl;
cout << "float: "
<< is_compound::value
<< endl;
cout << "double: "
<< is_compound::value
<< endl;
cout << "int(int): "
<< is_compound::value
<< endl;
return 0;
}
输出:
is_compound:
int(gfg::*): true
float: false
double: false
int(int): true
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