这种大小调整是如何工作的 [英] How is this size alignment working

问题描述

我不能够理解下面的code相对于所提供的评论。这是什么code呢，而且这将是相当于code为 8对齐？

I am not able to understand the below code with respect to the comment provided. What does this code does, and what would be the equivalent code for 8-aligned?

/* segment size must be 4-aligned */
attr->options.ssize &= ~3;

在这里， ssize 是 unsigned int型的键入

推荐答案

由于4为100，对准以4字节边界的任何值（即4的倍数）将具有最后两个比特设置为零。

Since 4 in binary is 100, any value aligned to 4-byte boundaries (i.e. a multiple of 4) will have the last two bits set to zero.

二进制3为11，和〜3是这些位的位反，即，... 1111100。执行位与该值将保持每一点相同的，除了最后两个将被清除（比特及1 ==位，和比特及0 == 0）。这为我们提供了一个下一个较低或相等值，该值是4的倍数。

3 in binary is 11, and ~3 is the bitwise negation of those bits, i.e., ...1111100. Performing a bitwise AND with that value will keep every bit the same, except the last two which will be cleared (bit & 1 == bit, and bit & 0 == 0). This gives us a the next lower or equal value that is a multiple of 4.

要8（二进制1000）做同样的操作，我们需要清除最低的三个位。我们可以做到这一点与二进制111的按位求反，即7〜

To do the same operation for 8 (1000 in binary), we need to clear out the lowest three bits. We can do that with the bitwise negation of the binary 111, i.e., ~7.

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