blocks|key|180782|text|因为%25d是一个有符号的int说明符。使用%25u。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|180783|在我的机器上打印4294967295。|180784|正如其他人所提到的，如果您将最高的无符号值解释为签名，则得到-1，请参阅二补的维基百科条目。|180785|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Signed_number_representations#Two.27s_complement^0|2|2|K|2|0|8|A|0|10|2|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|Z|8|@]|D|@$9|10|A|11|1|12]]|E|$]]|$1|J|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

Because <code>%d</code> is a signed int specifier. Use <code>%u</code>.

which prints <code>4294967295</code> on my machine.

As others mentioned, if you interpret the highest unsigned value as signed, you get -1, see the wikipedia entry for <a href="http://en.wikipedia.org/wiki/Signed_number_representations#Two.27s_complement" rel="noreferrer">two's complement</a>.

blocks|key|1206891|text|您的系统使用负数的http://en.wikipedia.org/wiki/Two%2527s_complement表示形式。在这种表示法中，由所有二进制数组成的二进制数表示最大负数-1。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1206892|因为倒置零的所有位会给出一个由所有数字组成的数字，所以当你用一个期望有符号数字而不是无符号数字的%25d来重新解释这个数字为一个有符号的数字时，你会得到它。|1206893|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Two%2527s_complement^0|2H|2|9|1B|0|0|1C|2|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

Your system uses <a href="http://en.wikipedia.org/wiki/Two%27s_complement" rel="nofollow">two's complement</a> representation of negative numbers. In this representation a binary number composed of all ones represent the biggest negative number <code>-1</code>.

Since inverting all bits of a zero gives you a number composed of all ones, you get <code>-1</code> when you re-interpret the number as a signed number by printing it with a <code>%d</code> which expects a signed number, not an unsigned one.

blocks|key|180810|text|printf()只知道通过格式字符串中使用的格式说明符传递它的变量类型。所以这里发生的事情是，您将x和y打印为有符号整数，因为您在格式字符串中使用了%25d。尝试%25u，您将得到一个更符合您可能期望的结果。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|180811|entityMap^0|0|8|1D|1|1F|1|22|2|27|2|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|G|$]]

<code>printf()</code> only knows what type of variable you passed it by what format specifiers you used in your format string. So what's happening here is that you're printing <code>x</code> and <code>y</code> as signed integers, because you used <code>%d</code> in your format string. Try <code>%u</code> instead, and you'll get a result more in line with what you're probably expecting.

blocks|key|180829|text|首先，在使用printf时，您告诉它将数字打印为已签名("%25d")，而不是无符号("%25u")。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|180830|其次，您是对的，因为它“与数字存储在内存中的方式有关”。int+(签名或无符号)不是计算机上的一个位，而是k位的集合。K的确切值取决于计算机体系结构的具体情况，但很可能您有k=32。|180831|为了简洁起见，让我们假设您的ints有8位长，所以k=8+(当然不是这样，除非您在一个非常有限的嵌入式系统上工作)。在这种情况下，(int)0实际上是00000000，(int)~0+(等于所有位)是11111111。|180832|最后，在两个补码(这是最常见的数字二进制表示)中，11111111实际上是-1。有关两个补语的描述，请参见http://en.wikipedia.org/wiki/Two's_complement。|offset|length|180833|如果您将打印更改为使用"%25u"，则它将打印一个表示(2%5Ek-1)的正整数，其中k是整数中的位数(因此很可能它将打印4294967295)。|180834|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Two^0|0|0|0|1H|W|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@$H|X|I|Y|1|Z]]|A|$]]|$1|J|3|K|5|6|7|10|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

First, in your use of printf you are telling it to print the number as signed ("%d") instead of unsigned ("%u").

Second, you are right in that it has "something to do with the way numbers are stored in memory". An int (signed or unsigned) is not a single bit on your computer, but a collection of k bits. The exact value of k depends on the specifics of your computer architecture, but most likely you have k=32.

For the sake of succinctness, lets assume your ints are 8 bits long, so k=8 (this is most certainly not the case, unless you are working on a very limited embedded system,). In that case (int)0 is actually 00000000, and (int)~0 (which negates all the bits) is 11111111.

Finally, in two's complement (which is the most common binary representation of signed numbers), 11111111 is actually -1. See <a href="http://en.wikipedia.org/wiki/Two" rel="nofollow">http://en.wikipedia.org/wiki/Two</a>'s_complement for a description of two's complement.

If you changed your print to use "%u", then it will print a positive integer that represents (2^k-1) where k is the number of bits in an integer (so probably it will print 4294967295).

What is the difference between <code>(unsigned)~0</code> and <code>(unsigned)1</code>. Why is <code>unsigned</code> of <code>~0</code> is <code>-1</code> and <code>unsigned</code> of <code>1</code> is <code>1</code>? Does it have something to do with the way unsigned numbers are stored in the memory. Why does an unsigned number give a signed result. It didn't give any overflow error either. I am using <code>GCC</code> compiler: 

<pre><code>#include&lt;sdio.h&gt;
main()
{
 unsigned int x=(unsigned)~0; 
 unsigned int y=(unsigned)1; 
 printf("%d\n",x); //prints -1
 printf("%d\n",y); //prints 1
}
</code></pre>

Difference between '(unsigned)1' and '(unsigned)~0'

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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(unsigned)~0和(unsigned)1的区别是什么？为什么unsigned of ~0是-1而unsigned of 1是1？它与无符号数字存储在内存中的方式有关吗？为什么没有符号的数字会给出一个有符号的结果。它也没有产生任何溢出错误。我正在使用GCC编译器：#include<sdio.h>main(){ u...

问“(无符号)1”与“(无符号)~0”之间的差异
EN

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问“(无符号)1”与“(无符号)~0”之间的差异EN