布尔类型True和False

Question

I was playing around with the Bool type (Boolean Variable) and typed this:

#include <iostream>

int main(void)
{
using std::cout;
using std::cin;
using std::endl;

bool $ok = false & true;

if($ok == true)
{
    cout << "The value is True." << endl;
}
else if($ok == false)
{
    cout << "The value is false." << endl;
}

cin.get();
cin.get();
return 0;
}

I know the differences between using the bitwise operator & and the logical operator && , but I do not see how this produces a false (0) value. I know if I swapped the bitwise operator and used a + the expression 0+1 would cause it to evaluate to true . Can someone explain why this:

bool $ok = false & true;

evaluates to false?

Answer 1

false=0(0x00000000) true=1(0x00000001)

Now when we do bitwise and operator of (false & true)---(0&1=0).

            0x00000000
         &  0x00000001
          -------------
           0x00000000

Hence the result is 0(0x00000000)

Answer 2

Why would this be true? false converts to a 0-valued integer. true converts to a non-zero valued integer (normally 1, but this is not guaranteed). 0 & x for any x is always 0 . 0 == false by definition of the integer/boolean interactions, thus the false branch is entered.

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For what it's worth, over a domain of 0 and 1, with 0 as false and 1 as true, * maps to AND whereas + maps to OR . Given this, I'm not quite sure why you'd expect + and & to give the sameresults.

x * y != 0 iff x != 0 and y != 0
x + y != 0 iff x != 0 or y != 0

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It's also worth mentioning that bit-wise operations on signed types tend to be a bad idea. If you're going to treat integers as bitfields, use unsigned integral types where the rules around the operations are much more natural and intuitive.

Answer 3

It's because false is 0 (when converted from boolean-land to integer-land), while true is 1 (when converted from boolean-land to integer-land).

false & true == 0 & 1 == 0 == false
false + true == 0 + 1 == 1 == true

If the magic of & is a mystery to you, there are lots of great resources on bitwise-and .