C 编程 - 打印最高有效位为 1 的十六进制字符
代码片段
char c1;
char c2;
c1 = 0X7F;
c2 = 0X80;
printf("c1 = %X\nc2 = %X\n", c1, c2);
输出
c1 = 7F
c2 = FFFFFF80
为什么c2不打印为80？ （我猜测这与 0X80 的最高有效位为 1 而 0x7F 的 MSB 为 0 有关）。我怎样才能让c2打印为简单的80？
The code snippet
char c1;
char c2;
c1 = 0X7F;
c2 = 0X80;
printf("c1 = %X\nc2 = %X\n", c1, c2);
outputs
c1 = 7F
c2 = FFFFFF80
why is it that c2 does not print as 80? (i'm guessing it's something to do with the fact that the most significant bit of 0X80 is 1 while the MSB of 0x7F is 0). how can i get c2 to print as simply 80?
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因为
printf是一个可变参数函数。可变参数函数的参数经历默认提升；这样做的效果之一是char参数被转换为int。在您的平台上，char是signed，因此c2实际上是负数。因此，printf打印int中出现的所有前导（这些是由于二进制补码表示）。有（至少）3 种方法可以解决此问题：
使用%02X而不是%X作为printf格式说明符。unsigned char而不是char（这会提升为unsigned int，因此没有前导）int，然后掩码：(int)c2 & 0xFF。[注：其推论是
char c2 = 0x80的行为实际上是实现定义的；如果char是signed那么这将会溢出。]Because
printfis a variadic function. Arguments to variadic functions undergo the default promotions; one effect of this is thatchararguments are converted toint. On your platform,charissigned, soc2is actually negative.printfis therefore printing all the leading ones that occur in theint(these are due to two's-complement representation).There are (at least) 3 methods to work around this:
Use%02Xrather than%Xas yourprintfformat specifier.unsigned charrather thanchar(this gets promoted tounsigned int, so no leading ones)chars toint, and then mask:(int)c2 & 0xFF.[Note: The corollary of this is that the behaviour of
char c2 = 0x80is actually implementation-defined; ifcharissignedthen this will overflow.]c2得到（符号）扩展为int。以下将满足您的要求：c2gets (sign-)extended toint. The following will do what you require:尝试使用
unsigned char根据您的需要进行打印。这是一个 ideone 示例正如已经提到的，您看到的输出是由于符号扩展造成的。
希望这有帮助！
Try
unsigned charto print as per your need. Here is a ideone sampleAs mentioned already the output you are seeing is due to sign extension.
Hope this helps!
为了强制 printf 将其视为字符，您需要给出“hh”修饰符。尝试：
in order to force printf to treat that as a character, you need to give the "hh" modifier. try:
您正在传递
char并将它们读取为unsigned int，您很幸运它能够正常工作。也就是说，char已签名。 0x80 实际上使它成为负数。它们显然被转换为整数，然后无符号。这扩展了符号位，从f开始。You're passing
chars and having them read asunsigned ints, you're lucky it works at all. That said,charis signed. 0x80 actually makes it negative. They are apparently being converted to ints and then unsigned. That extends the sign bit, thence thefs.