isspace
Comprueba si un elemento en una configuración regional es un carácter de espacio en blanco.
template<Class CharType>
bool isspace(
CharType _Ch,
const locale& _Loc
)
Parámetros
_Ch
El elemento que se va a probar._Loc
Configuración regional que contiene el elemento que se va a probar.
Valor devuelto
true si el elemento probado es un carácter de espacio en blanco; false si no es.
Comentarios
La función de la plantilla devuelve use_facet<C<CharType> >(_Loc).es(::<space>, _ChdectypeCharType).
Ejemplo
// 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;
}
Resultados
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.
Requisitos
configuración regional <deEncabezado: >
Espacio de nombres: std