尝试打印UINT_MAX时得到-1 [英] When trying to print UINT_MAX I get -1
问题描述
当我尝试全部打印'UINT_MAX'时，我得到-1，这是为什么?这是我的"main()"中唯一的东西，它是一个打印"UINT_MAX"的printf语句
When trying to print 'UINT_MAX' all I get it -1, why is this? It's the only thing I have in my 'main()', a printf statement that prints 'UINT_MAX'
推荐答案
您可以使用printf()打印它
You can print it with printf()
printf("%u\n", UINT_MAX);
您需要使用u说明符， man printf .
为什么用d说明符得到-1?这是因为内存中有符号整数的工作方式.
Why you get -1 with d specifier? It's because of how work a signed integer in memory.
unsigned integer从0开始，在内存中是一个简单的二进制系统.
unsigned integer start at 0, in memory it a simple binary system.
十进制=>二进制
- 1 => 1
- 2 => 10
- 8 => 1000
- 255 => 11111111
带符号整数使用一位被解释为负数，这令人困惑，因为将无符号整数像整数一样解释时的最大值是-1.
signed integer use a bit to be interpreted as a negative number, it's confusing because the max value of an unsigned int when it is interpreted like an int is -1.
具有8位整数的示例:
- -1 => 11111111
- -128 => 10000000
- -9 => 11110111
就像您看到的那样，第一位曾经是符号位.
Like you see, the first bit is used to be the sign bit.
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