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