C++ 布尔类型 True 和 False
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Bool type True and False
提问by Jake2k13
I was playing around with the Bool type (Boolean Variable) and typed this:
我在玩 Bool 类型（布尔变量）并输入以下内容：
#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+1would cause it to evaluate to true. Can someone explain why this:
我知道使用按位运算符&和逻辑运算符之间的区别&&，但我看不出这是如何产生假 (0) 值的。我知道如果我交换按位运算符并使用 a+表达式0+1会导致它计算为true. 有人可以解释为什么会这样：
bool $ok = false & true;
evaluates to false?
评估为假？
回答by Santosh Sahu
false=0(0x00000000) true=1(0x00000001)
假=0（0x00000000）真=1（0x00000001）
Now when we do bitwise and operator of (false & true)---(0&1=0).
现在，当我们执行 (false & true) ---(0&1=0) 的按位和运算符时。
0x00000000
& 0x00000001
-------------
0x00000000
Hence the result is 0(0x00000000)
因此结果是 0(0x00000000)
回答by Cornstalks
It's because falseis 0 (when converted from boolean-land to integer-land), while trueis 1 (when converted from boolean-land to integer-land).
这是因为false是 0（从 boolean-land 转换为 integer-land 时），而true1（从 boolean-land 转换为 integer-land 时）。
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.
如果 的魔法对您来说&是个谜，那么按位与 上有很多很棒的资源。
回答by Corbin
Why would this be true? falseconverts to a 0-valued integer. trueconverts to a non-zero valued integer (normally 1, but this is not guaranteed). 0 & xfor any xis always 0. 0 == falseby definition of the integer/boolean interactions, thus the false branch is entered.
为什么这是真的？false转换为 0 值的整数。true转换为非零值整数（通常为 1，但不能保证）。0 & x因为任何x总是0。0 == false根据整数/布尔交互的定义，因此进入了错误分支。
For what it's worth, over a domain of 0 and 1, with 0 as false and 1 as true, *maps to ANDwhereas +maps to OR. Given this, I'm not quite sure why you'd expect +and &to give the sameresults.
对于它的价值，在 0 和 1 的域上，0 为假，1 为真，*映射到AND而+映射到OR. 鉴于此，我不太确定为什么您会期望+并&给出相同的结果。
x * y != 0 iff x != 0 and y != 0
x + y != 0 iff x != 0 or y != 0
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.
还值得一提的是，对有符号类型进行按位操作往往是一个坏主意。如果您要将整数视为位域，请使用无符号整数类型，其中运算规则更加自然和直观。
