c++ - 对于带移位的循环

Question

谁能解释这个for循环是如何工作的？ for (bitMask = 0x01; bitMask; bitMask <<= 1)这是我第一次在for循环中遇到这样的语法，并且很想知道循环将如何结束。

Answer 1

如果您获得了无符号 int32变量bitMask。在第32个周期，其位表示为

10000000 00000000 00000000 00000000

然后向左移一位，它溢出，仅保留低32位，因此该值变为0，循环条件变为假。

1 00000000 00000000 00000000 00000000
↑
this bit is discarded

如果 bitMask是 签名的整数，该怎么办？这是未定义的行为。

C标准(N2716，6.5.7位移位运算符)说:

The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined

C++标准(N4713，8.5.7 Shift运算符)说:

The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are zero-filled. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. Otherwise, if E1 has a signed type and non-negative value, and E1 × 2^E2 is representable in the corresponding unsigned type of the result type, then that value, converted to the result type, is the resulting value; otherwise, the behavior is undefined.

我的观点是永远不要使用这种循环，因为我们很容易忘记这仅适用于无符号整数。相反，您应该使用类似以下的方法为每个位生成掩码。

for (int i = 0; i < 32; i++) {
    int bitMask = 1 << i;
}