isspace

Verifica se un elemento in impostazioni locali è uno spazio vuoto.

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

Parametri

• _Ch
  l'elemento da testare.

• _Loc
  Le impostazioni locali che contengono l'elemento da testare.

Valore restituito

true se l'elemento testato da uno spazio vuoto; false caso contrario.

Note

La funzione di modello restituisce use_facet<tipo C<char> > (_Loc).viene(ctype< >_double_colon_space, _Chdichar).

Esempio

// 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;
}

Output

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.

Requisiti

intestazione: <locale>

Spazio dei nomi: deviazione standard