按位移位运算符。已签名和未签名
我正在利用互联网上的临时笔记练习 SCJP 考试。
根据我的笔记， >> 运算符应该是有符号右移，符号位从左侧引入。而左移运算符 << 应该保留符号位。
然而，我可以使用 << 运算符来移动符号（fe Integer.MAX_VALUE << 1 计算结果为 -2，虽然我永远无法使用 >> 运算符来移动符号，但
我一定是误解了一些东西，但是什么呢？
I'm practising for the SCJP exam using cram notes from the Internet.
According to my notes the >> operator is supposed to be signed right shift, with the sign bit being brought in from the left. While the left shift operator << is supposed to preserve the sign bit.
Playing around however, I'm able to shift the sign with the << operator (f.e. Integer.MAX_VALUE << 1 evaluates to -2, while I'm never able to shift the sign with the >> operator.
I must be misunderstanding something here, but what?
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“>>”已签名，因为它保留了该标志。它使用数字的二进制表示形式中最左边的数字作为填充符。例如：
“>>”是该运算符的无符号版本。它总是使用零作为填充符：
在二进制表示中，最左边的数字决定数字的符号。因此，如果它是“1”，那么我们的值为负值，如果它是“0”，那么我们的数字就是正值。这就是为什么使用最左边的数字作为填充符可以保持符号永久。
">>" is signed because it keeps the sign. It uses the most left digit in binary representation of a number as a filler. For example:
">>>" is unsigned version of this operator. It always use zero as a filler:
In binary representation the most left digit determines sign of the number. So, if it's '1' then we have negative value and if it's '0' - then our number is positive. That's why using the most left digit as a filler allows to keep sign permanent.
这些移位背后的想法是，它们可以充当 2 的幂的乘法和除法（<< 1 相当于 *= 2，>> 2 相当于 /= 4），这就是带符号版本的原因的转变存在。但无符号移位不一定保留符号。 <<正如您所建议的，运算符实际上并不保留符号；它恰好发生在你的例子中。尝试对 2,147,483,647 进行左移；它不会保持积极的状态。他们不费心尝试进行“有符号”左移的原因是，如果数字从正数变为负数（反之亦然），那么无论如何您都会超出变量类型的范围。
The idea behind the shifts is that they can act as multiplying and dividing by powers of 2 ( << 1 is equivalent to *= 2, >> 2 is equivalent to /= 4), which is why the signed version of shifting exists. Unsigned shifting doesn't preserve the sign, necessarily, though. The << operator doesn't actually preserve the sign, as you suggest; it simply happens to in your example. Try doing a left shift on 2,147,483,647; it doesn't stay positive. The reason that they don't bother trying to make a 'signed' left shift is because, if the number shifts from positive to negative (or viceversa), then you went outside the bounds of the variable type anyway.