isspace

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

template<Class CharType> 
   bool isspace( 
      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 Leerraumzeichen ist; false, wenn nicht.

Hinweise

Die Vorlagenfunktion gibt use_facet<C<CharType zurück > >(_Loc).ist(::<ctype>CharTypespace, _Ch).

Beispiel

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

using namespace std;

int main( )   
{
   locale loc ( "German_Germany" );
   bool result1 = isspace ( 'L', loc );
   bool result2 = isspace ( '\n', loc );
   bool result3 = isspace ( ' ', loc );

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

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

   if ( result3 )
      cout << "The character ' ' in the locale is "
           << "a whitespace character." << endl;
   else
      cout << "The character ' ' in the locale is "
           << " not a whitespace character." << endl;
}

Ausgabe

The character 'L' in the locale is  not a whitespace character.
The character 'backslash-n' in the locale is a whitespace character.
The character ' ' in the locale is a whitespace character.

Anforderungen

Gebietsschema Header: <>

Namespace: std