logical_and Struct

Artikkeli
08/03/2021

A predefined function object that performs the logical conjunction operation (operator&&) on its arguments.

Syntax

template <class Type = void>
struct logical_and : public binary_function<Type, Type, bool>
{
    bool operator()(const Type& Left, const Type& Right) const;
};

// specialized transparent functor for operator&&
template <>
struct logical_and<void>
{
  template <class T, class U>
  auto operator()(T&& Left, U&& Right) const
    -> decltype(std::forward<T>(Left) && std::forward<U>(Right));
};

Parameters

Type, T, U
Any type that supports an operator&& that takes operands of the specified or inferred types.

Left
The left operand of the logical conjunction operation. The unspecialized template takes an lvalue reference argument of type Type. The specialized template does perfect forwarding of lvalue and rvalue reference arguments of inferred type T.

Right
The right operand of the logical conjunction operation. The unspecialized template takes an lvalue reference argument of type Type. The specialized template does perfect forwarding of lvalue and rvalue reference arguments of inferred type U.

Return Value

The result of Left && Right. The specialized template does perfect forwarding of the result, which has the type that's returned by operator&&.

Remarks

For user-defined types, there is no short-circuiting of operand evaluation. Both arguments are evaluated by operator&&.

Example

// functional_logical_and.cpp
// compile with: /EHsc

#define _CRT_RAND_S
#include <stdlib.h>
#include <deque>
#include <algorithm>
#include <functional>
#include <iostream>

int main( )
{
   using namespace std;
   deque<bool> d1, d2, d3( 7 );
   deque<bool>::iterator iter1, iter2, iter3;

   unsigned int randomValue;

   int i;
   for ( i = 0 ; i < 7 ; i++ )
   {
      if ( rand_s( &randomValue ) == 0 )
      {
         d1.push_back((bool)(( randomValue % 2 ) != 0));
      }

   }

   int j;
   for ( j = 0 ; j < 7 ; j++ )
   {
      if ( rand_s( &randomValue ) == 0 )
      {
         d2.push_back((bool)(( randomValue % 2 ) != 0));
      }
   }

   cout << boolalpha;    // boolalpha I/O flag on

   cout << "Original deque:\n d1 = ( " ;
   for ( iter1 = d1.begin( ) ; iter1 != d1.end( ) ; iter1++ )
      cout << *iter1 << " ";
   cout << ")" << endl;

   cout << "Original deque:\n d2 = ( " ;
   for ( iter2 = d2.begin( ) ; iter2 != d2.end( ) ; iter2++ )
      cout << *iter2 << " ";
   cout << ")" << endl;

   // To find element-wise conjunction of the truth values
   // of d1 & d2, use the logical_and function object
   transform( d1.begin( ), d1.end( ), d2.begin( ),
      d3.begin( ), logical_and<bool>( ) );
   cout << "The deque which is the conjunction of d1 & d2 is:\n d3 = ( " ;
   for ( iter3 = d3.begin( ) ; iter3 != d3.end( ) ; iter3++ )
      cout << *iter3 << " ";
   cout << ")" << endl;
}

Original deque:
d1 = ( true true true true true false false )
Original deque:
d2 = ( true false true true false true false )
The deque which is the conjunction of d1 & d2 is:
d3 = ( true false true true false false false )

Jaa