I want to split the following Scala code line like this:
ConditionParser.parseSingleCondition("field=*value1*").description
must equalTo("field should contain value1")
But which is the line continuation character?
Wrap it in parentheses:
(ConditionParser.parseSingleCondition("field=*value1*").description
must equalTo("field should contain value1"))
Scala does not have a "line continuation character" - it infers a semicolon always when:
An expression can end
The following (not whitespace) line begins not with a token that can start a statement
There are no unclosed ( or [ found before
Thus, to "delay" semicolon inference one can place a method call or the dot at the end of the line or place the dot at the beginning of the following line:
ConditionParser.
parseSingleCondition("field=*value1*").
description must equalTo("field should contain value1")
a +
b +
c
List(1,2,3)
.map(_+1)
Related
My code has a ton of occurrences of something like:
idof(some_object)
I want to replace them with:
some_object["id"]
It sounds simple:
sed -i 's/idof(\([^)]\+\))/\1["id"]/g' source.py
The problem is that some_object might be something like idof(get_some_object()), or idof(my_class().get_some_object()), in which case, instead of getting what I want (get_some_object()["id"] or my_class().get_some_object()["id"]), I get get_some_object(["id"]) or my_class(["id"].get_some_object()).
Is there a way to have sed match closing bracket, so that it internally keeps track of any opening/closing brackets inside my (), and ignores those?
It needs to keep everything that's between those brackets: idof(ANYTHING) becomes ANYTHING["id"].
Using sed
$ sed -E 's/idof\(([[:alpha:][:punct:]]*)\)/\1["id"]/g' input_file
Using ERE, exclude idof and the first opening parenthesis.
As a literal closing parenthesis is also excluded, everything in-between the capture parenthesis including additional parenthesis will be captured.
[[:alpha:]] will match all alphabetic characters including upper and lower case while [[:punct:]] will capture punctuation characters including ().-{} and more.
The g option will make the substitution as many times as the pattern is found.
Theoretically, you can write a regex that will handle all combinations of idof(....) up to some limit of nested () calls inside ..... Such regex would have to list with all possible combinations of calls, like idof(one(two(three))) or idof(one(two(three)four(five)) you can match with an appropriate regex like idof([^()]*([^()]*([^()]*)[^()]*)[^()]*) or idof([^()]*([^()]*([^()]*)[^()]*([^()]*)[^()]*) respectively.
The following regex handles only some cases, but shows the complexity and general path. Writing a regex to handle all possible cases to "eat" everything in front of the trailing ) is left to OP as an exercise why it's better to use something else. Note that handling string literals ")" becomes increasingly complex.
The following Bash code:
sed '
: begin
# No idof? Just print the line!
/^\(.*\)idof(\([^)]*)\)/!n
# Note: regex is greedy - we start from the back!
# Note: using newline as a stack separator.
s//\1\n\2/
# hold the front
{ h ; x ; s/\n.*// ; x ; s/[^\n]*\n// ; }
: handle_brackets
# Eat everything before final ) up to some number of nested ((())) calls.
# Insert more jokes here.
: eat_brackets
/^[^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*\(([^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)\?[^()]*)\)/{
s//&\n/
# Hold the front.
{ H ; x ; s/\n\([^\n]*\)\n.*/\1/ ; x ; s/[^\n]*\n// ; }
b eat_brackets
}
/^\([^()]*\))/!{
s/^/ERROR: eating brackets did not work: /
q1
}
# Add the id after trailing ) and remove it.
s//\1["id"]/
# Join with hold space and clear the hold space for next round
{ H ; s/.*// ; x ; s/\n//g ; }
# Restart for another idof if in input.
b begin
' <<EOF
before idof(some_object) after
before idof(get_some_object()) after
before idof(my_class().get_some_object()) after
before idof(one(two(three)four)five) after
before idof(one(two(three)four)five) between idof(one(two(three)four)five) after
before idof( one(two(three)four)five one(two(three)four)five ) after
before idof(one(two(three(four)five)six(seven(eight)nine)ten) between idof(one(two(three(four)five)six(seven(eight)nine)ten) after
EOF
Will output:
before some_object["id"] after
before get_some_object()["id"] after
before my_class().get_some_object()["id"] after
before one(two(three)four)five["id"] after
before one(two(three)four)five["id"] between one(two(three)four)five["id"] after
before one(two(three)four)five one(two(three)four)five ["id"] after
ERROR: eating brackets did not work: one(two(three(four)five)six(seven(eight)nine)ten) after
The last line is not handled correctly, because (()()) case is not correctly handled. One would have to write a regex to match it.
I am trying to parse all the files and verify if any of the file content has strings TESTDIR or TEST_DIR
Files contents might look something like:-
TESTDIR = foo
include $(TESTDIR)/chop.mk
...
TEST_DIR := goldimage
MAKE_TESTDIR = var_make
NEW_TEST_DIR = tesing_var
Actually I am only interested in TESTDIR ,$(TESTDIR),TEST_DIR but in my case last two lines should be ignored. I am new to perl , Can anyone help me out with re-rex.
/\bTEST_?DIR\b/
\b means a "word boundary", i.e. the place between a word character and a non-word character. "Word" here has the Perl meaning: it contains characters, numbers, and underscores.
_? means "nothing or an underscore"
Look at "characterset".
Only (space) surrounding allowed:
/^(.* )?TEST_?DIR /
^ beginning of the line
(.* )? There may be some content .* but if, its must be followed by a space
at the and says that a whitespace must be there. Otherwise use ( .*)?$ at the end.
One of a given characterset is allowed:
Should the be other characters then a space be possible you can use a character class []:
/^(.*[ \t(])?TEST_?DIR[) :=]/
(.*[ \t(])? in front of TEST_?DIR may be a (space) or a \t (tab) or ( or nothing if the line starts with itself.
afterwards there must be one of (space) or : or = or ). Followd by anything (to "anything" belongs the "=" of ":=" ...).
One of a given group is allowed:
So you need groups within () each possible group in there devided by a |:
/^(.*( |\t))?TEST_?DIR( | := | = )/
In this case, at the beginning is no change to [ \t] because each group holds only one character and \t.
At the end, there must be (single space) or := (':=' surrounded by spaces) or = ('=' surrounded by spaces), following by anything...
You can use any combination...
/^(.*[ \t(])?TEST_?DIR([) =:]| :=| =|)/
Test it on Debuggex.com. (Use 'PCRE')
I am working on the exercises from https://www.scala-exercises.org/std_lib/formatting
For the following question, m answer seems incorrect but I do not know why.
val c = 'a' //unicode for a
val d = '\141' //octal for a
val e = '\"'
val f = '\\'
"%c".format(c) should be("a") //my answers
"%c".format(d) should be("a")
"%c".format(e) should be(")
"%c".format(f) should be(\)
your answer should be enclosed in quotes
"%c".format(e) should be("\"")
"%c".format(f) should be("\\")
because it wouldn't recognize string unless it's enclosed in quotes
Your last two lines are invalid Scala code and cannot be compiled:
// These are wrong
"%c".format(e) should be(")
"%c".format(f) should be(\)
The be() function needs to be passed a String, and neither of those calls are being passed a String. A String needs to start and end with a double-quote (there are some exceptions).
// In this case you started a String with a double-quote, but you are never
// closing the string with a second double-quote
"%c".format(e) should be(")
// In this case you are missing both double-quotes
"%c".format(f) should be(\)
In this case the code should be:
"%c".format(e) should be("\"")
"%c".format(f) should be("\\")
If you want a character to be treated literally in a String, you need to "escape" it with a backslash. So if you want to literally show a double-quote, you need to prefix it with a backslash:
\"
And as a String:
"\""
Similarily for a backslash:
\\
As a String:
"\\"
Using an IDE makes this easier to see. Using IntelliJ the String is green but the special non-literal characters are highlighted in orange.
Check quote signs.
https://www.tutorialspoint.com/scala/scala_strings.htm
https://docs.scala-lang.org/overviews/core/string-interpolation.html
https://learnxinyminutes.com/docs/scala/
You can run Scala code online and check yourself here:
https://scastie.scala-lang.org
https://ideone.com/
I can't fix this error...
#temp=split(/(/)/,$headerLine);
this error appears
Unmatched ( in regex; marked by <-- HERE in m/( <-- HERE
Use
#temp=split(/(\/)/,$headerLine);
or
#temp=split(m&(/)&,$headerLine);
The slash in the parentheses prematurely terminates your regular expression.
Your second / character is terminating the regular expression, so Perl is interpreting your code as:
#temp=split /(/
followed by garbage.
Simply escape the literal /:
#temp=split(/(\/)/, $headerLine)
I've modified the variable paragraph-start to count lines starting with .*: as a paragraph start:
(setq paragraph-start "\f\\|[ \t]*$\\|[ \t]*[0-9.]\.\\|.*:$\\|" )
However, if I have a buffer:
foo:
bar:
baz: some stuff
more
_
(Where _ indicates point location)
Then the first backward-paragraph skips to the beginning of the line 'bar:', not the line starting with 'baz:' as expected. How do I change this behaviour/why is it behaving this way?
It's because you have $ after ::
"\f\\|[ \t]*$\\|[ \t]*[0-9.]\.\\|.*:$\\|"
$ matches at the end of a line. So the part of your regexp that matches something followed by : also requires that nothing follow the :.
The first line (going backward from point) that ends in a : is the bar: line.
(Note too that you might not want .*:, if you want to exclude the possibility that what precedes the : not include a :, e.g., if you want to exclude a:b:c foo. To exclude :, use [^:]* instead of .*. And to exclude a lone :, use [^:]+.)