Does anyone know how to provide a single backslash as the replacement value in Ruby's gsub method? I thought using double backslashes for the replacement value would result in a single backslash but it results in two backslashes.
Example: "a\b".gsub("\", "\\")
Result: a\\b
I also get the same result using a block:
Example: "a\b".gsub("\"){"\\"}
Result: a\\b
Obviously I can't use a single backslash for the replacement value since that would just serve to escape the quote that follows it. I've also tried using single (as opposed to double) quotes around the replacement value but still get two backslashes in the result.
EDIT: Thanks to the commenters I now realize my confusion was with how the Rails console reports the result of the operation (i.e. a\\b). Although the strings 'a\b' and 'a\\b' appear to be different, they both have the same length:
'a\b'.length (3)
'a\\b'.length (3)
You can represent a single backslash by either "\\" or '\\'. Try this in irb, where
"\\".size
correctly outputs 1, showing that you indeed have only one character in this string, not 2 as you think. You can also do a
puts "\\"
Similarily, your example
puts("a\b".gsub("\", "\\"))
correctly prints
a\b
Related
I've found interesting thing in ruby. Do anybody know why is behavior?
tried '+'.gsub!('+', '\+') and expected "\\+" but got ""(empty string)
gsub is implemented, after some indirection, as rb_sub_str_bang in C, which calls rb_reg_regsub.
Now, gsub is supposed to allow the replacement string to contain backreferences. That is, if you pass a regular expression as the first argument and that regex defines a capture group, then your replacement string can include \1 to indicate that that capture group should be placed at that position.
That behavior evidently still happens if you pass an ordinary, non-regex string as the pattern. Your verbatim string obviously won't have any capture groups, so it's a bit silly in this case. But trying to replace, for instance, + with \1 in the string + will give the empty string, since \1 says to go get the first capture group, which doesn't exist and hence is vacuously "".
Now, you might be thinking: + isn't a number. And you'd be right. You're replacing + with \+. There are several other backreferences allowed in your replacement string. I couldn't find any official documentation where these are written down, but the source code does quite fine. To summarize the code:
Digits \1 through \9 refer to numbered capture groups.
\k<...> refers to a named capture group, with the name in angled brackets.
\0 or \& refer to the whole substring that was matched, so (\0) as a replacement string would enclose the match in parentheses.
A backslash followed by a backtick (I have no idea how to write that using StackOverflow's markdown) refers to the entire string up to the match.
\' refers to the entire string following the match.
\+ refers to the final capture group, i.e. the one with the highest number.
\\ is a literal backslash.
(Most of these are based on Perl variables of a similar name)
So, in your examples,
\+ as the replacement string says "take the last capture group". There is no capture group, so you get the empty string.
\- is not a valid backreference, so it's replaced verbatim.
\ok is, likewise, not a backreference, so it's replaced verbatim.
In \\+, Ruby eats the first backslash sequence, so the actual string at runtime is \+, equivalent to the first example.
For \\\+, Ruby processes the first backslash sequence, so we get \\+ by the time the replacement function sees it. \\ is a literal backslash, and + is no longer part of an escape sequence, so we get \+.
I am using Tcl_StringCaseMatch function in C++ code for string pattern matching. Everything works fine until input pattern or string has [] bracket. For example, like:
str1 = pq[0]
pattern = pq[*]
Tcl_StringCaseMatch is not working i.e returning false for above inputs.
How to avoid [] in pattern matching?
The problem is [] are special characters in the pattern matching. You need to escape them using a backslash to have them treated like plain characters
pattern= "pq\\[*\\]"
I don't think this should affect the string as well. The reason for double slashing is you want to pass the backslash itself to the TCL engine.
For the casual reader:
[] have a special meaning in TCL in general, beyond the pattern matching role they take here - "run command" (like `` or $() in shells), but [number] will have no effect, and the brackets are treated normally - thus the string str1 does not need escaping here.
For extra confusion:
TCL will interpret ] with no preceding [ as a normal character by default. I feel that's getting too confusing, and would rather that TCL complains on unbalanced brackets. As OP mentions though, this allows you to forgo the final two backslashes and use "pq\\[*]". I dislike this, and rather make it obvious both are treated normally and not the usual TCL way, but to each her/is own.
I want to match the path "/". I've tried the following alternatives, and the first two do match, but I don't know why the third doesn't:
/\A\/\z/.match("/") # <MatchData "/">
"/\A\/\z/".match("/") # <MatchData "/">
Regexp.new("/\A\/\z/").match("/") # nil
What's going on here? Why are they different?
The first snippet is the only correct one.
The second example is... misleading. That string literal "/\A\/\z/" is, obviously, not a regex. It's a string. Strings have #match method which converts its argument to a regexp (if not already one) and match against it. So, in this example, it's '/' that is the regular expression, and it matches a forward slash found in the other string.
The third line is completely broken: don't need the surrounding slashes there, they are part of regex literal, which you didn't use. Also use single quoted strings, not double quoted (which try to interpret escape sequences like \A)
Regexp.new('\A/\z').match("/") # => #<MatchData "/">
And, of course, none of the above is needed if you just want to check if a string consists of only one forward slash. Just use the equality check in this case.
s == '/'
I'm trying to replace single quotes (') with escaped single quotes (\') in a string in ruby 1.9.3 and 1.8.7.
The exact problem string is "Are you sure you want to delete '%#'". This string should become "Are you sure you want to delete \'%#\'"
Using .gsub!(/\'/,"\'") leads to the following string "Are you sure you want to %#'%#".
Any ideas on what's going on?
String#gsub in the form gsub(exp,replacement) has odd quirks affecting the replacement string which sometimes require lots of escaping slashes. Ruby users are frequently directed to use the block form instead:
str.gsub(/'/){ "\\'" }
If you want to do away with escaping altogether, consider using an alternate string literal form:
str.gsub(/'/){ %q(\') }
Once you get used to seeing these types of literals, using them to avoid escape sequences can make your code much more readable.
\' in a substitution replacement string means "The portion of the original string after the match". So str.gsub!(/\'/,"\\'") replaces the ' character with everything after it - which is what you've noticed.
You need to further escape the backslash in the replacement. .gsub(/'/,"\\\\'") works in my irb console:
irb(main):059:0> puts a.gsub(/'/,"\\\\'")
Are you sure you want to delete \'%#\'
You need to escape the backslash. What about this?
"Are you sure you want to delete '%#'".gsub(/(?=')/, "\\")
# => "Are you sure you want to delete \\'%#\\'"
The above should be what you want. Your expected result is wrong. There is no way to literally see a single backslash when it means literally a backslash.
How do I excape a backslash before a captured group?
Example:
"foo+bar".gsub(/(\+)/, '\\\1')
What I expect (and want):
foo\+bar
what I unfortunately get:
foo\\1bar
How do I escape here correctly?
As others have said, you need to escape everything in that string twice. So in your case the solution is to use '\\\\\1' or '\\\\\\1'. But since you asked why, I'll try to explain that part.
The reason is that replacement sequence is being parsed twice--once by Ruby and once by the underlying regular expression engine, for whom \1 is its own escape sequence. (It's probably easier to understand with double-quoted strings, since single quotes introduce an ambiguity where '\\1' and '\1' are equivalent but '\' and '\\' are not.)
So for example, a simple replacement here with a captured group and a double quoted string would be:
"foo+bar".gsub(/(\+)/, "\\1") #=> "foo+bar"
This passes the string \1 to the regexp engine, which it understands as a reference to a capture group. In Ruby string literals, "\1" means something else entirely (ASCII character 1).
What we actually want in this case is for the regexp engine to receive \\\1. It also understands \ as an escape character, so \\1 is not sufficient and will simply evaluate to the literal output \1. So, we need \\\1 in the regexp engine, but to get to that point we need to also make it past Ruby's string literal parser.
To do that, we take our desired regexp input and double every backslash again to get through Ruby's string literal parser. \\\1 therefore requires "\\\\\\1". In the case of single quotes one slash can be omitted as \1 is not a valid escape sequence in single quotes and is treated literally.
Addendum
One of the reasons this problem is usually hidden is thanks to the use of /.+/ style regexp quotes, which Ruby treats in a special way to avoid the need to double escape everything. (Of course, this doesn't apply to gsub replacement strings.) But you can still see it in action if you use a string literal instead of a regexp literal in Regexp.new:
Regexp.new("\.").match("a") #=> #<MatchData "a">
Regexp.new("\\.").match("a") #=> nil
As you can see, we had to double-escape the . for it to be understood as a literal . by the regexp engine, since "." and "\." both evaluate to . in double-quoted strings, but we need the engine itself to receive \..
This happens due to a double string escaping. You should use 5 slashes in this case.
"foo+bar".gsub(/([+])/, '\\\\\1')
Adding \ two more times escapes this properly.
irb(main):011:0> puts "foo+bar".gsub(/(\+)/, '\\\\\1')
foo\+bar
=> nil