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)
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'm trying to loop through a string and count its characters in Swift. This code successfully outputs the character count, but I receive this warning:
warning: immutable value 'character' was never used; consider
replacing with '_' or removing it for character in quote {
^~~~~~~~~
_
This is my code:
var quote = "hello there"
var count = 0
for character in quote {
count = count + 1
}
print("\(count)")
Does anyone know why I have this warning? Also, is this the best way to approach this task? Thanks.
Please read the error message carefully, it tells you precisely what's wrong and what you can do.
immutable value 'character' was never used
That's indeed true, the variable character is unused. The compiler provides two fixes:
consider replacing with '_' or removing it
The latter is not an option in a loop, so use the first, replace character with an underscore
for _ in quote {
I have to write a lex program that has these rules:
Identifiers: String of alphanumeric (and _), starting with an alphabetic character
Literals: Integers and strings
Comments: Start with ! character, go to until the end of the line
Here is what I came up with
[a-zA-Z][a-zA-Z0-9]+ return(ID);
[+-]?[0-9]+ return(INTEGER);
[a-zA-Z]+ return ( STRING);
!.*\n return ( COMMENT );
However, I still get a lot of errors when I compile this lex file.
What do you think the error is?
It would have helped if you'd shown more clearly what the problem was with your code. For example, did you get an error message or did it not function as desired?
There are a couple of problems with your code, but it is mainly correct. The first issue I see is that you have not divided your lex program into the necessary parts with the %% divider. The first part of a lex program is the declarations section, where regular expression patterns are specified. The second part is where the action that match patterns are specified. The (optional) third section is where any code (for the compiler) is placed. Code for the compiler can also be placed in the declaration section when delineated by %{ and %} at the start of a line.
If we put your code through lex we would get this error:
"SoNov16.l", line 1: bad character: [
"SoNov16.l", line 1: unknown error processing section 1
"SoNov16.l", line 1: unknown error processing section 1
"SoNov16.l", line 1: bad character: ]
"SoNov16.l", line 1: bad character: +
"SoNov16.l", line 1: unknown error processing section 1
"SoNov16.l", line 1: bad character: (
"SoNov16.l", line 1: unknown error processing section 1
"SoNov16.l", line 1: bad character: )
"SoNov16.l", line 1: bad character: ;
Did you get something like that? In your example code you are specifying actions (the return(ID); is an example of an action) and thus your code is for the second section. You therefore need to put a %% line ahead of it. It will then be a valid lex program.
You code is dependant on (probably) a parser, which consumes (and declares) the tokens. For testing purposes it is often easier to just print the tokens first. I solved this problem by making a C macro which will do the print and can be redefined to do the return at a later stage. Something like this:
%{
#define TOKEN(t) printf("String: %s Matched: " #t "\n",yytext)
%}
%%
[a-zA-Z][a-zA-Z0-9]+ TOKEN(ID);
[+-]?[0-9]+ TOKEN(INTEGER);
[a-zA-Z]+ TOKEN (STRING);
!.*\n TOKEN (COMMENT);
If we build and test this, we get the following:
abc
String: abc Matched: ID
abc123
String: abc123 Matched: ID
! comment text
String: ! comment text
Matched: COMMENT
Not quite correct. We can see that the ID rule is matching what should be a string. This is due to the ordering of the rules. We have to put the String rule first to ensure it matches first - unless of course you were supposed to match strings inside some quotes? You also missed the underline from the ID pattern. Its also a good idea to match and discard any whitespace characters:
%{
#define TOKEN(t) printf("String: %s Matched: " #t "\n",yytext)
%}
%%
[a-zA-Z]+ TOKEN (STRING);
[a-zA-Z][a-zA-Z0-9_]+ TOKEN(ID);
[+-]?[0-9]+ TOKEN(INTEGER);
!.*\n TOKEN (COMMENT);
[ \t\r\n]+ ;
Which when tested shows:
abc
String: abc Matched: STRING
abc123_
String: abc123_ Matched: ID
-1234
String: -1234 Matched: INTEGER
abc abc123 ! comment text
String: abc Matched: STRING
String: abc123 Matched: ID
String: ! comment text
Matched: COMMENT
Just in case you wanted strings in quotes, that is easy too:
%{
#define TOKEN(t) printf("String: %s Matched: " #t "\n",yytext)
%}
%%
\"[^"]+\" TOKEN (STRING);
[a-zA-Z][a-zA-Z0-9_]+ TOKEN(ID);
[+-]?[0-9]+ TOKEN(INTEGER);
!.*\n TOKEN (COMMENT );
[ \t\r\n] ;
"abc"
String: "abc" Matched: STRING
I am having difficulties replacing a string containing special characters using sed. My old and new string are shown below
oldStr = "# td=(nstates=20) cam-b3lyp/6-31g geom=connectivity"
newStr = "# opt b3lyp/6-31g geom=connectivity"
My sed command is the following
sed -i 's/\# td\=\(nstates\=20\) cam\-b3lyp\/6\-31g geom\=connectivity/\# opt b3lyp\/6\-31g geom\=connectivity/g' myfile.txt
I dont get any errors, however there is no match. Any ideas on how to fix my patterns.
Thanks
try s|# td=(nstates=20) cam-b3lyp/6-31g geom=connectivity|# opt b3lyp/6-31g geom=connectivity|g'
you can use next to anything after s instead of /, as your expression contains slashes I used | instead. -, = and # don't have to be escaped (minus only in character sets [...]), escaped parens indicate a group, nonescaped parens are literals.
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)