[英]How is é bitwise shifted to a E in C++?
Hello I having trouble understanding this and how for example it take an accent char like é
and converts it E9
.您好,我在理解这一点时遇到了麻烦,例如它如何采用像
é
这样的重音字符并将其转换为E9
。 I could be missing something i get that it bit-shifts right 4. é = 11101000
and E = 01000101
shifting 4 doesn't make E
right?我可能会遗漏一些我知道它向右移位 4 的
E = 01000101
é = 11101000
和E = 01000101
移位 4 不会使E
正确吗?
static const char *digits = "0123456789ABCDEF";
unsigned char ch;
*dest++ = digits[(ch >> 4) & 0x0F];//this returns E
*dest++ = digits[ch & 0x0F];//this returns 9
The code doesn't convert é
to E9
- it converts an 8-bit number to its hexadecimal representation, in four-bit pieces ("nybbles").代码不会将
é
转换为E9
- 它将 8 位数字转换为其十六进制表示,以四位部分(“nybbles”)表示。
digits[(ch >> 4) & 0x0F]
is the digit that represents the high nybble, and digits[ch & 0x0F]
is the digit that represents the low nybble. digits[(ch >> 4) & 0x0F]
是代表高 nybble 的digits[ch & 0x0F]
, digits[ch & 0x0F]
是代表低 nybble 的数字。
If you see é
become E9
, it's because é
has the value 233 in your character encoding.如果您看到
é
变为E9
,那是因为é
在您的字符编码中具有值 233。
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