C语言 C：以十六进制打印“unsigned long”的正确方法
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C: Correct way to print "unsigned long" in hex
提问by SomethingSomething
I have a function that gets an unsigned longvariable as parameter and I want to print it in Hex.
我有一个函数，它获取一个unsigned long变量作为参数，我想以十六进制打印它。
What is the correct way to do it?
正确的做法是什么？
Currently, I use printfwith "%lx"
目前，我将printf与“%lx”一起使用
void printAddress(unsigned long address) {
printf("%lx\n", address);
}
Should I look for a printf pattern for unsignedlong hex? (and not just "long hex" as mentioned above)
我应该为无符号长十六进制寻找 printf 模式吗？（而不仅仅是上面提到的“长十六进制”）
Or does printfconvert numbers to hex only using the bits? - so I should not care about the sign anyway?
还是printf仅使用位将数字转换为十六进制？- 所以我不应该关心这个标志吗？
回答by unwind
You're doing it right.
你做得对。
From the manual page:
从手册页：
o, u, x, X
The unsigned int argument is converted to unsigned octal (o), unsigned decimal (u), or unsigned hexadecimal (x and X) notation.
o, u, x, x
unsigned int 参数被转换为无符号八进制 (o)、无符号十进制 (u) 或无符号十六进制（x 和 X）表示法。
So the value for xshould always be unsigned. To make it longin size, use:
所以它的值x应该总是unsigned. 要使其long尺寸变大，请使用：
l
(ell) A following integer conversion corresponds to a long int or unsigned long int argument [...]
升
(ell) 以下整数转换对应于 long int 或 unsigned long int 参数 [...]
So %lxis unsigned long. An address (pointer value), however, should be printed with %pand cast to void *.
所以%lx是unsigned long。但是，地址（指针值）应该打印%p并转换为void *.
回答by Himanshu Sourav
I think the following format specifier should work give it a try
我认为以下格式说明符应该可以尝试一下
printf("%#lx\n",address);
printf("%#lx\n",address);
