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'的右十六进制值为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!
由于
答案 0 :(得分:13)
String#hex
没有给你一个字符的ASCII索引,它用于将一个基数为16的数字( hex adecimal)从一个字符串转换为一个整数:
% 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
使用基数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
答案 1 :(得分:0)
比hex方法更有优势。 ' 10-0'至256.
考虑一下你要比较' 100' > ' 20&#39 ;.应该返回true但返回false。使用' 100' .hex>' 20' .hex。返回true。哪个更准确。