I have a string:
s = "\t\n"
I intend to output:
\t\n
When I do
puts s
I see nothing. How should I do it?
That's because \n and \t are escape sequences
A single-quoted strings don’t process ASCII escape codes, and they don’t do string interpolation while double-quoted does both.
So if you want that behaviour, you should do
s = '\t\n'
Just use inspect to see the "raw" escape sequences.
s = "\t\n"
puts s.inspect
# >> "\t\n"
p s # >> "\t\n"
or
puts s.inspect[1...-1] # >> \t\n
Related
I don't understand this Ruby code:
>> puts '\\ <- single backslash'
# \ <- single backslash
>> puts '\\ <- 2x a, because 2 backslashes get replaced'.sub(/\\/, 'aa')
# aa <- 2x a, because two backslashes get replaced
so far, all as expected. but if we search for 1 with /\\/, and replace with 2, encoded by '\\\\', why do we get this:
>> puts '\\ <- only 1 ... replace 1 with 2'.sub(/\\/, '\\\\')
# \ <- only 1 backslash, even though we replace 1 with 2
and then, when we encode 3 with '\\\\\\', we only get 2:
>> puts '\\ <- only 2 ... 1 with 3'.sub(/\\/, '\\\\\\')
# \\ <- 2 backslashes, even though we replace 1 with 3
anyone able to understand why a backslash gets swallowed in the replacement string? this happens on 1.8 and 1.9.
Quick Answer
If you want to sidestep all this confusion, use the much less confusing block syntax. Here is an example that replaces each backslash with 2 backslashes:
"some\\path".gsub('\\') { '\\\\' }
Gruesome Details
The problem is that when using sub (and gsub), without a block, ruby interprets special character sequences in the replacement parameter. Unfortunately, sub uses the backslash as the escape character for these:
\& (the entire regex)
\+ (the last group)
\` (pre-match string)
\' (post-match string)
\0 (same as \&)
\1 (first captured group)
\2 (second captured group)
\\ (a backslash)
Like any escaping, this creates an obvious problem. If you want include the literal value of one of the above sequences (e.g. \1) in the output string you have to escape it. So, to get Hello \1, you need the replacement string to be Hello \\1. And to represent this as a string literal in Ruby, you have to escape those backslashes again like this: "Hello \\\\1"
So, there are two different escaping passes. The first one takes the string literal and creates the internal string value. The second takes that internal string value and replaces the sequences above with the matching data.
If a backslash is not followed by a character that matches one of the above sequences, then the backslash (and character that follows) will pass through unaltered. This is also affects a backslash at the end of the string -- it will pass through unaltered. It's easiest to see this logic in the rubinius code; just look for the to_sub_replacement method in the String class.
Here are some examples of how String#sub is parsing the replacement string:
1 backslash \ (which has a string literal of "\\")
Passes through unaltered because the backslash is at the end of the string and has no characters after it.
Result: \
2 backslashes \\ (which have a string literal of "\\\\")
The pair of backslashes match the escaped backslash sequence (see \\ above) and gets converted into a single backslash.
Result: \
3 backslashes \\\ (which have a string literal of "\\\\\\")
The first two backslashes match the \\ sequence and get converted to a single backslash. Then the final backslash is at the end of the string so it passes through unaltered.
Result: \\
4 backslashes \\\\ (which have a string literal of "\\\\\\\\")
Two pairs of backslashes each match the \\ sequence and get converted to a single backslash.
Result: \\
2 backslashes with character in the middle \a\ (which have a string literal of "\\a\\")
The \a does not match any of the escape sequences so it is allowed to pass through unaltered. The trailing backslash is also allowed through.
Result: \a\
Note: The same result could be obtained from: \\a\\ (with the literal string: "\\\\a\\\\")
In hindsight, this could have been less confusing if String#sub had used a different escape character. Then there wouldn't be the need to double escape all the backslashes.
This is an issue because backslash (\) serves as an escape character for Regexps and Strings. You could do use the special variable \& to reduce the number backslashes in the gsub replacement string.
foo.gsub(/\\/,'\&\&\&') #for some string foo replace each \ with \\\
EDIT: I should mention that the value of \& is from a Regexp match, in this case a single backslash.
Also, I thought that there was a special way to create a string that disabled the escape character, but apparently not. None of these will produce two slashes:
puts "\\"
puts '\\'
puts %q{\\}
puts %Q{\\}
puts """\\"""
puts '''\\'''
puts <<EOF
\\
EOF
argh, right after I typed all this out, I realised that \ is used to refer to groups in the replacement string. I guess this means that you need a literal \\ in the replacement string to get one replaced \. To get a literal \\ you need four \s, so to replace one with two you actually need eight(!).
# Double every occurrence of \. There's eight backslashes on the right there!
>> puts '\\'.sub(/\\/, '\\\\\\\\')
anything I'm missing? any more efficient ways?
Clearing up a little confusion on the author's second line of code.
You said:
>> puts '\\ <- 2x a, because 2 backslashes get replaced'.sub(/\\/, 'aa')
# aa <- 2x a, because two backslashes get replaced
2 backslashes aren't getting replaced here. You're replacing 1 escaped backslash with two a's ('aa'). That is, if you used .sub(/\\/, 'a'), you would only see one 'a'
'\\'.sub(/\\/, 'anything') #=> anything
the pickaxe book mentions this exact problem, actually. here's another alternative (from page 130 of the latest edition)
str = 'a\b\c' # => "a\b\c"
str.gsub(/\\/) { '\\\\' } # => "a\\b\\c"
I don't understand this Ruby code:
>> puts '\\ <- single backslash'
# \ <- single backslash
>> puts '\\ <- 2x a, because 2 backslashes get replaced'.sub(/\\/, 'aa')
# aa <- 2x a, because two backslashes get replaced
so far, all as expected. but if we search for 1 with /\\/, and replace with 2, encoded by '\\\\', why do we get this:
>> puts '\\ <- only 1 ... replace 1 with 2'.sub(/\\/, '\\\\')
# \ <- only 1 backslash, even though we replace 1 with 2
and then, when we encode 3 with '\\\\\\', we only get 2:
>> puts '\\ <- only 2 ... 1 with 3'.sub(/\\/, '\\\\\\')
# \\ <- 2 backslashes, even though we replace 1 with 3
anyone able to understand why a backslash gets swallowed in the replacement string? this happens on 1.8 and 1.9.
Quick Answer
If you want to sidestep all this confusion, use the much less confusing block syntax. Here is an example that replaces each backslash with 2 backslashes:
"some\\path".gsub('\\') { '\\\\' }
Gruesome Details
The problem is that when using sub (and gsub), without a block, ruby interprets special character sequences in the replacement parameter. Unfortunately, sub uses the backslash as the escape character for these:
\& (the entire regex)
\+ (the last group)
\` (pre-match string)
\' (post-match string)
\0 (same as \&)
\1 (first captured group)
\2 (second captured group)
\\ (a backslash)
Like any escaping, this creates an obvious problem. If you want include the literal value of one of the above sequences (e.g. \1) in the output string you have to escape it. So, to get Hello \1, you need the replacement string to be Hello \\1. And to represent this as a string literal in Ruby, you have to escape those backslashes again like this: "Hello \\\\1"
So, there are two different escaping passes. The first one takes the string literal and creates the internal string value. The second takes that internal string value and replaces the sequences above with the matching data.
If a backslash is not followed by a character that matches one of the above sequences, then the backslash (and character that follows) will pass through unaltered. This is also affects a backslash at the end of the string -- it will pass through unaltered. It's easiest to see this logic in the rubinius code; just look for the to_sub_replacement method in the String class.
Here are some examples of how String#sub is parsing the replacement string:
1 backslash \ (which has a string literal of "\\")
Passes through unaltered because the backslash is at the end of the string and has no characters after it.
Result: \
2 backslashes \\ (which have a string literal of "\\\\")
The pair of backslashes match the escaped backslash sequence (see \\ above) and gets converted into a single backslash.
Result: \
3 backslashes \\\ (which have a string literal of "\\\\\\")
The first two backslashes match the \\ sequence and get converted to a single backslash. Then the final backslash is at the end of the string so it passes through unaltered.
Result: \\
4 backslashes \\\\ (which have a string literal of "\\\\\\\\")
Two pairs of backslashes each match the \\ sequence and get converted to a single backslash.
Result: \\
2 backslashes with character in the middle \a\ (which have a string literal of "\\a\\")
The \a does not match any of the escape sequences so it is allowed to pass through unaltered. The trailing backslash is also allowed through.
Result: \a\
Note: The same result could be obtained from: \\a\\ (with the literal string: "\\\\a\\\\")
In hindsight, this could have been less confusing if String#sub had used a different escape character. Then there wouldn't be the need to double escape all the backslashes.
This is an issue because backslash (\) serves as an escape character for Regexps and Strings. You could do use the special variable \& to reduce the number backslashes in the gsub replacement string.
foo.gsub(/\\/,'\&\&\&') #for some string foo replace each \ with \\\
EDIT: I should mention that the value of \& is from a Regexp match, in this case a single backslash.
Also, I thought that there was a special way to create a string that disabled the escape character, but apparently not. None of these will produce two slashes:
puts "\\"
puts '\\'
puts %q{\\}
puts %Q{\\}
puts """\\"""
puts '''\\'''
puts <<EOF
\\
EOF
argh, right after I typed all this out, I realised that \ is used to refer to groups in the replacement string. I guess this means that you need a literal \\ in the replacement string to get one replaced \. To get a literal \\ you need four \s, so to replace one with two you actually need eight(!).
# Double every occurrence of \. There's eight backslashes on the right there!
>> puts '\\'.sub(/\\/, '\\\\\\\\')
anything I'm missing? any more efficient ways?
Clearing up a little confusion on the author's second line of code.
You said:
>> puts '\\ <- 2x a, because 2 backslashes get replaced'.sub(/\\/, 'aa')
# aa <- 2x a, because two backslashes get replaced
2 backslashes aren't getting replaced here. You're replacing 1 escaped backslash with two a's ('aa'). That is, if you used .sub(/\\/, 'a'), you would only see one 'a'
'\\'.sub(/\\/, 'anything') #=> anything
the pickaxe book mentions this exact problem, actually. here's another alternative (from page 130 of the latest edition)
str = 'a\b\c' # => "a\b\c"
str.gsub(/\\/) { '\\\\' } # => "a\\b\\c"
I'm trying to do a regex with lookbehind that changes \n to but not if it's a \\n.
My closest attempt has no effect:
text.gsub /(?<!\\)\n/, ''
Unfortunately, no number of backslashes in the lookbehind seem to fix the problem. How can I address this?
You need to double the backslash before the n in the regex, otherwise it's looking for a newline instead of a literal backslash followed by n:
irb(main):001:0> puts "hello\\nthere\\\\n".gsub(/(?<!\\)\\n/, ' ')
hello there\\n
You don't need anything special. "\n" is a single character. It does not include a "\" or "n" character.
text.gsub(/\n/, "")
But instead of that, you should do:
text.gsub("\n", "")
or
text.tr("\n", "")
But I would do:
text.tr($/, "")
How can i say "all symbols except backslash" in Ruby character class?
/'[^\]*'/.match("'some string \ hello'") => should be nil
Variant with two backslashed doesn't work
/'[^\\]*'/.match("'some string \ hello'") => 'some string \ hello' BUT should be nil
Your problem is not with your regex; you got that right. Your problem is that your test string does not have a backslash in it. It has an escaped space, instead. Try this:
str = "'some string \\ hello'"
puts str #=> 'some string \ hello'
p /'[^\\]*'/.match(str) #=> nil
You need to escape the backslash:
[^\\]*
because backslash is the escape character in regular expressions, thus escaping the closing bracket here.
If you want to verify that the whole string contains non-backslash characters, then you need anchors:
^[^\\]*$
There's actually not a backslash in your string to match against. Try taking a look at just your input:
"'some string \ hello'".length # => 20
"a\ b".length # => 3
The "\ " in double quotes is being escaped into just a space:
irb(main):014:0> " "[0].to_i # => 32
irb(main):015:0> "\ "[0].to_i # => 32
irb(main):016:0> "\ ".size #=> 1
If you want to match no slash, you'll need two, like in your second example, which looks good to me:
/'[^\\]*'/.match("'some string \\ hello'") # => nil
This is relevant to Csharp.. but idk it might help you.
[^\\\\]*
four slashes instead of two.
There is a string:
str = "foo\nbar"
How can I escape it to:
'foo\nbar'
?
I noticed "#{str}" doesn't work.
str.inspect Should do it for you
Add another \:
"foo\\nbar" == 'foo\nbar' #=> true
Single-quoted strings do not have interpolation, so #{str} does nothing special in them.