英文:
Why "-1 & postive integer" alway return the postive integer?
问题
例如：
printf("%d", -1 & 3);
输出结果为3。我尝试了其他数字，结果都一样。
抱歉问题很简短，但我认为没有必要无谓地延长它。
我搜索了一下，没有找到答案。
英文:
For example:
printf("%d", -1 & 3);
The output is 3. I tried with other number and it's the same.
sorry for short question, but I don't think I should make it long for no reason.
I searched around and did not find answer to it.
答案1
得分: 3
Some1) (not all) of the bitwise operators don't care about signedness of the operands but instead look at the raw binary representation of the variable.
In case of the negative value -1, the raw binary in the common two's complement format looks as 1111....1111 (all ones).
As an example, lets assume 16 bit int, 2's complement:
Dec Bin
-1 1111 1111 1111 1111
3 0000 0000 0000 0011
-------------------------
& 0000 0000 0000 0011 = 3 dec
1) Specifically, & | ^ ~ and their compound assignment equivalents only look at the raw binary representation.
英文:
Some<sup>1)</sup> (not all) of the bitwise operators don't care about signedness of the operands but instead look at the raw binary representation of the variable.
In case of the negative value -1, the raw binary in the common two's complement format looks as 1111....1111 (all ones).
As an example, lets assume 16 bit int, 2's complement:
Dec Bin
-1 1111 1111 1111 1111
3 0000 0000 0000 0011
-------------------------
& 0000 0000 0000 0011 = 3 dec
<sup>1)</sup> Specifically, & | ^ ~ and their compound assignment equivalents only look at the raw binary representation.
答案2
得分: 0
因为这是按位运算。两个补码的负数最高位设置为1，正数将这一位设置为零。
如果你执行 1 & 0 操作，结果将始终为零。因此，在这个操作之后，最高位将变为零，整数将变为正数。
示例（x - 不重要）
1xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx - 负数
0xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx - 正数
1xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx（负数）& 0xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx（正数）= 0xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx（正数）
英文:
Because it is bitwise operation. Two complement's negative numbers have the most significant bit set, positive have this bit set to zero.
If you 1 & 0 the result will be always zero. So after this operation the MSB will be zero and the integer will be positive
Example (x - does not matter)
1xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx - negative number
0xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx - positive number
1xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx (negative) & 0xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx (positive) = 0xxx xxxx xxxx xxxx xxxx xxxx xxxx xxxx (positive)
答案3
得分: 0
为什么"-1 & 正整数"总是返回正整数？
C语言支持各种int编码。所有都有一个"符号"位。"必须恰好有一个符号位"。只有设置了该位，值才能为负数。
使用"&"永远不会设置任何位，包括"符号"位。"&"只会减少或保持已设置的位数。
"任何内容 & 正整数"总是会得到一个正数（非负数）。
英文:
> Why "-1 & positive integer" always return the positive integer?
C allows various int encodings. All have a sign bit. "There shall be exactly one sign bit." Only with that bit set may the value be negative.
Using & will never set any bit including the sign bit. & will only reduce - or keep the same - the bits that are set.
"anything & positive integer" will always result in a positive (non-negative) number.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论