按位运算符 & 比较每一对位。仅当左右操作数至少有一个匹配位设置为1时，结果才为非空。

The bitwise operator & compares each individual pair of bits. The result will be non-null only if the left and right operands have at least one matching bit set to 1.

示例：0100 AND 0010→0000但0110 AND 0010 →0010。

Example : 0100 AND 0010 → 0000 but 0110 AND 0010 → 0010.

此运算符允许您使用单个整数值在不同位上存储多个布尔值，然后使用第二个值（称为掩码）来过滤比特。

This operator allows you to use a single integer value to store several booleans on different bits, then use a second value (known as a mask) to filter the bits.

kSCNetworkFlagsReachable 等于 1<< 1 （ 2 ）。因此， flag&仅当 flag 的第二个最低有效位设置为1时，kSCNetworkFlagsReachable 才为真。

kSCNetworkFlagsReachable is equal to 1<<1 (2). Thus, flag & kSCNetworkFlagsReachable is true only if the second least significant bit of flag is set to 1.

使用&& 代替& 是一个常见的错误。编译器将尝试检测该错误。在您的示例中， kSCNetworkFlagsReachable 是一个常量值。由于 kSCNetworkFlagsReachable 是常量且始终为真，因此测试 flag&& kSCNetworkFlagsReachable 为true与测试 flag 是否为真相同。因此，您不太可能真的想在逻辑运算中使用常量值。这就是编译器发出警告的原因。

Using && instead of & is a common mistake. The compiler will try to detect that mistake. In your example, kSCNetworkFlagsReachable is a constant value. As kSCNetworkFlagsReachable is constant and always true, testing whether flag && kSCNetworkFlagsReachable is true is the same as testing whether flag is true. Thus it is very unlikely that you really wanted to use a constant value in a logical operation. That's why the compiler emits the warning.