>>>之间的差异和>>运营商 [英] Difference between >>> and >> operators

问题描述

如果移位的数字为正>> >>且>>工作相同。

If the shifted number is positive >>> and >> work the same.

如果移位的数字为负>>>用1s填充最高有效位而>>操作将MSB填充为0。

If the shifted number is negative >>> fills the most significant bits with 1s whereas >> operation shifts filling the MSBs with 0.

我的理解是否正确？

如果负数存储时MSB设置为1，而不是Java使用的2s补码方式，运算符的行为完全不同，正确吗？

If the negative numbers are stored with the MSB set to 1 and not the 2s complement way that Java uses the the operators would behave entirely differently, correct?

推荐答案

负数表示的方式称为2的补码。为了演示这是如何工作的，以-12为例。 12，二进制，是00001100（假设整数是8位，但实际上它们要大得多）。通过简单地反转每一位得到2的补码，你得到11110011.然后，只需加1得到11110100.注意，如果你再次应用相同的步骤，你会得到正12回。

The way negative numbers are represented is called 2's complement. To demonstrate how this works, take -12 as an example. 12, in binary, is 00001100 (assume integers are 8 bits though in reality they are much bigger). Take the 2's complement by simply inverting every bit, and you get 11110011. Then, simply add 1 to get 11110100. Notice that if you apply the same steps again, you get positive 12 back.

>>>无论如何都会变为零，所以12 >>> 1应该给你00000110，这是6，而（-12）>>> 1应该给你01111010，即122。你实际上在Java中尝试这个，你会得到一个更大的数字，因为Java整数实际上远大于8位。

The >>> shifts in zero no matter what, so 12 >>> 1 should give you 00000110, which is 6, and (-12) >>> 1 should give you 01111010, which is 122. If you actually try this in Java, you'll get a much bigger number since Java ints are actually much bigger than 8 bits.

>>转换有点相同最高位，正数保持正数，负数保持负数。 12 >> 1是00000110（仍然是6）和（ - 12）>> 1将是11111010，这是负6。

The >> shifts in a bit identical to the highest bit, so that positive numbers stay positive and negative numbers stay negative. 12 >> 1 is 00000110 (still 6) and (-12) >> 1 would be 11111010 which is negative 6.

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