iscntrl

Testet, ob ein Element in einem Gebietsschema ein Steuerzeichen ist.

template<Class CharType>
   bool iscntrl(
      CharType _Ch, 
      const locale& _Loc
   )

Parameter

• _Ch
  Das zu testenden Element.

• _Loc
  Das Gebietsschema, das das zu testenden Element enthält.

Rückgabewert

true, wenn das getestete Element ein Steuerzeichen ist; false, wenn nicht ist.

Hinweise

Die Vorlagenfunktion gibt use_facet<ctype<CharType> >(_Loc).is(ctype<CharType>::cntrl, _Ch) zurück.

Beispiel

// locale_iscntrl.cpp
// compile with: /EHsc
#include <locale>
#include <iostream>

using namespace std;

int main( )   
{
   locale loc ( "German_Germany" );
   bool result1 = iscntrl ( 'L', loc );
   bool result2 = iscntrl ( '\n', loc );
   bool result3 = iscntrl ( '\t', loc );

   if ( result1 )
      cout << "The character 'L' in the locale is "
           << "a control character." << endl;
   else
      cout << "The character 'L' in the locale is "
           << " not a control character." << endl;

   if ( result2 )
      cout << "The character-set 'backslash-n' in the locale\n is "
           << "a control character." << endl;
   else
      cout << "The character-set 'backslash-n' in the locale\n is "
           << " not a control character." << endl;

   if ( result3 )
      cout << "The character-set 'backslash-t' in the locale\n is "
           << "a control character." << endl;
   else
      cout << "The character-set 'backslash-n' in the locale \n is "
           << " not a control character." << endl;
}

Output

The character 'L' in the locale is  not a control character.
The character-set 'backslash-n' in the locale
 is a control character.
The character-set 'backslash-t' in the locale
 is a control character.

Anforderungen

Header: <locale>

Namespace: std