C语言 为什么 uint8_t 和 uint16_t 的格式说明符相同 (%u)？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/26362386/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Why is the format specifier for uint8_t and uint16_t the same (%u)?
提问by Rev1.0
I only found pretty unrelated questions due to the tons of results searching for printf().
由于搜索了大量结果，我只发现了非常不相关的问题printf()。
Why does uint8_tnot specify its own format string but any other type does?
为什么不uint8_t指定自己的格式字符串，而其他类型呢？
As far as I understand printf(), it has to know the length of the supplied parameters to be able to parse the variable argument list.
据我了解printf()，它必须知道提供的参数的长度才能解析变量参数列表。
Since uint8_tand uint16_tuse the same format specifier %u, how does printf()"know" how many bytes to process? Or is there somehow an implicit cast to uint16_tinvolved when supplying uint8_t?
由于uint8_t并uint16_t使用相同的格式说明符%u，如何printf()“知道”要处理多少字节？或者uint16_t在提供时是否以某种方式隐式转换uint8_t？
Maybe I am missing something obvious.
也许我错过了一些明显的东西。
采纳答案by 2501
printf()is a variadic function. Its optional arguments( and only those ) get promoted according to default argument promotions( 6.5.2.2. p6 ).
printf()是一个可变参数函数。它的可选参数（并且只有那些）根据默认参数提升（6.5.2.2.p6）得到提升。
Since you are asking for integers, integer promotions are applied in this case, and types you mention get promoted to int. ( and not unsigned intbecause C )
由于您要求使用整数，因此在这种情况下会应用整数提升，并且您提到的类型将提升为int。（而不是unsigned int因为 C ）
If you use "%u"in printf(), and pass it an uint16_tvariable, then the function converts that to an int, then to an unsigned int( because you asked for it with %u ) and then prints it.
如果您"%u"在 printf() 中使用，并向其传递一个uint16_t变量，则该函数会将其转换为 an int，然后转换为 an unsigned int（因为您使用 %u 要求它），然后打印它。
回答by wick
Because %u stands for "unsigned", it well may be uint64_t and is architecture dependent. According to man 3 printf, you may want to use length modifier to get sought behaviour, i.e. %hu (uint16_t) and %hhu (uint8_t).
因为 %u 代表“无符号”，它很可能是 uint64_t 并且依赖于体系结构。根据man 3 printf，您可能希望使用长度修饰符来获得所寻求的行为，即 %hu (uint16_t) 和 %hhu (uint8_t)。
