为什么 uint8_t 和 uint16_t 的格式说明符相同 (%u)? [英] Why is the format specifier for uint8_t and uint16_t the same (%u)?

问题描述

由于搜索 printf() 的大量结果，我只发现了非常不相关的问题.

I only found pretty unrelated questions due to the tons of results searching for printf().

为什么 uint8_t 不指定自己的格式字符串，而其他任何类型都指定?

Why does uint8_t not specify its own format string but any other type does?

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.

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.

由于 uint8_t 和 uint16_t 使用相同的格式说明符 %u，printf() 怎么知道"要处理多少字节?或者在提供 uint8_t 时是否涉及到 uint16_t 的隐式转换?

Since uint8_t and uint16_t use the same format specifier %u, how does printf() "know" how many bytes to process? Or is there somehow an implicit cast to uint16_t involved when supplying uint8_t?

也许我遗漏了一些明显的东西.

Maybe I am missing something obvious.

推荐答案

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.

printf() is a variadic function. Its optional arguments( and only those ) get promoted according to default argument promotions( 6.5.2.2. p6 ).

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 )

Since you are asking for integers, integer promotions are applied in this case, and types you mention get promoted to int. ( and not unsigned int because C )

如果您在 printf() 中使用 "%u"，并将其传递给 uint16_t 变量，则该函数会将其转换为 int，然后到一个 unsigned int(因为你用 %u 要求它)然后打印它.

If you use "%u" in printf(), and pass it an uint16_t variable, then the function converts that to an int, then to an unsigned int( because you asked for it with %u ) and then prints it.

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