英文:
Is "%#x" equivalent to "0x%x" in the format string of printf family functions?
问题
这两个函数调用等效吗？（假设变量 x 的类型为 unsigned int）
printf ("%#x\n", x);
printf ("0x%x\n", x);
以及以下两个函数调用：
printf ("%#.8x\n", x);
printf ("0x%08x\n", x);
为什么一些嵌入式程序偏向于使用后者而不是前者？
英文:
Are these two function calls equivalent? ( Assuming that the type of variable x is unsigned int)
printf ("%#x\n", x);
printf ("0x%x\n", x);
And the following two function calls:
printf ("%#.8x\n", x);
printf ("0x%08x\n", x);
Why do some embedded programs prefers the latter one to the former one?
答案1
得分: 2
从C标准7.9.16.1第6段中关于fprintf中的#：
结果被转换为“替代形式”...对于x（或X）转换，非零结果前缀为0x（或0X）。
所以，%#x 将在非零结果前缀为 0x。所以它与 0x%x 相同，除非结果为零。
英文:
From the C standard 7.9.16.1 para 6, on # in fprintf:
> The result is converted to an ‘‘alternative form’’... For x (or X) conversion, a nonzero
> result has 0x (or 0X) prefixed to it.
So there you have it: %#x prefixes 0x to any non-zero result. So it's the same as 0x%x except when the result is zero.
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