我发现
Ruby中的String#hex没有为给定的char返回正确的十六进制值,这很奇怪.我可能误解了该方法,但请采用以下示例：
'a'.hex => 10
而’a’的正确十六进制值为61：
'a'.unpack('H*')
=> 61
我错过了什么吗？什么是十六进制？任何提示赞赏！
谢谢
解决方法
String#hex不提供字符的ASCII索引,它用于将字符串的16位数字(十六进制)转换为整数：
% ri String\#hex String#hex (from ruby site) ------------------------------------------------------------------------------ str.hex -> integer ------------------------------------------------------------------------------ Treats leading characters from str as a string of hexadecimal digits (with an optional sign and an optional 0x) and returns the corresponding number. Zero is returned on error. "0x0a".hex #=> 10 "-1234".hex #=> -4660 "0".hex #=> 0 "wombat".hex #=> 0
所以它使用法线贴图：
'0'.hex #=> 0 '1'.hex #=> 1 ... '9'.hex #=> 9 'a'.hex #=> 10 == 0xA 'b'.hex #=> 11 ... 'f'.hex #=> 15 == 0xF == 0x0F '10'.hex #=> 16 == 0x10 '11'.hex #=> 17 == 0x11 ... 'ff'.hex #=> 255 == 0xFF
当使用base 16时,它与String#to_i非常相似：
'0xff'.to_i(16) #=> 255 'FF'.to_i(16) #=> 255 '-FF'.to_i(16) #=> -255
来自文档：
% ri String\#to_i String#to_i (from ruby site) ------------------------------------------------------------------------------ str.to_i(base=10) -> integer ------------------------------------------------------------------------------ Returns the result of interpreting leading characters in str as an integer base base (between 2 and 36). Extraneous characters past the end of a valid number are ignored. If there is not a valid number at the start of str,0 is returned. This method never raises an exception when base is valid. "12345".to_i #=> 12345 "99 red balloons".to_i #=> 99 "0a".to_i #=> 0 "0a".to_i(16) #=> 10 "hello".to_i #=> 0 "1100101".to_i(2) #=> 101 "1100101".to_i(8) #=> 294977 "1100101".to_i(10) #=> 1100101 "1100101".to_i(16) #=> 17826049