I get an "Expected Identifier" message against the if line. Any ideas why?
if ([inputA.text isEqualToString:#""]) && ([inputB.text <> isEqualToString:#""]) {
c = 1;
}
I'm trying to say it both inputs are empty...
I presume there isn't an easier way to say if the text is null in Obj C++?
An if statement requires that its condition expression be enclosed in parentheses. You have a compound expression. You've used parentheses around the subexpressions of the logical AND operation (&&), but you haven't surrounded the entire expression in parentheses. (The subexpressions don't actually require parentheses in this case, but they don't hurt.)
Next, you have a random <> in the second subexpression. What is that doing in there? In some languages that is the "not equal" operator, but a) it's not an operator in C or Objective-C, b) it wouldn't go inside a message-send expression like that, and c) you claim you were trying to check that both inputs are empty, so I wouldn't expect you to try to negate the test for equality with the empty string.
So, fixing just those problems yields:
if ([inputA.text isEqualToString:#""] && [inputB.text isEqualToString:#""]) {
c = 1;
}
That said, pie's answer is good, too. It also works if either of the inputs has a nil text property, too.
if ([inputA.text length]==0 && [inputB.text length]==0)
{
c = 1;
}
Related
I need to check if any elements of a large (60,000+ elements) array are present in a long string of text. My current code looks like this:
if $TARGET_PARTLIST.any? { |target_pn| pdf_content_string.include? target_pn }
self.last_match_code = target_pn
self.is_a_match = true
end
I get a syntax error undefined local variable or method target_pn.
Could someone let me know the correct syntax to use for this block of code? Also, if anyone knows of a quicker way to do this, I'm all ears!
In this case, all your syntax is correct, you've just got a logic error. While target_pn is defined (as a parameter) inside the block passed to any?, it is not defined in the block of the if statement because the scope of the any?-block ends with the closing curly brace, and target_pn is not available outside its scope. A correct (and more idiomatic) version of your code would look like this:
self.is_a_match = $TARGET_PARTLIST.any? do |target_pn|
included = pdf_content_string.include? target_pn
self.last_match_code = target_pn if included
included
end
Alternately, as jvillian so kindly suggests, one could turn the string into an array of words, then do an intersection and see if the resulting set is nonempty. Like this:
self.is_a_match = !($TARGET_PARTLIST &
pdf_content_string.gsub(/[^A-Za-z ]/,"")
.split).empty?
Unfortunately, this approach loses self.last_match_code. As a note, pointed out by Sergio, if you're dealing with non-English languages, the above regex will have to be changed.
Hope that helps!
You should use Enumerable#find rather than Enumerable#any?.
found = $TARGET_PARTLIST.find { |target_pn| pdf_content_string.include? target_pn }
if found
self.last_match_code = found
self.is_a_match = true
end
Note this does not ensure that the string contains a word that is an element of $TARGET_PARTLIST. For example, if $TARGET_PARTLIST contains the word "able", that string will be found in the string, "Are you comfortable?". If you only want to match words, you could do the following.
found = $TARGET_PARTLIST.find { |target_pn| pdf_content_string[/\b#{target_pn}\b/] }
Note this uses the method String#[].
\b is a word break in the regular expression, meaning that the first (last) character of the matched cannot be preceded (followed) by a word character (a letter, digit or underscore).
If speed is important it may be faster to use the following.
found = $TARGET_PARTLIST.find { |target_pn|
pdf_content_string.include?(target_on) && pdf_content_string[/\b#{target_pn}\b/] }
A probably more performant way would be to move all this into native code by letting Regexp search for it.
# needed only once
TARGET_PARTLIST_RE = Regexp.new("\\b(?:#{$TARGET_PARTLIST.sort.map { |pl| Regexp.escape(pl) }.join('|')})\\b")
# to check
self.last_match_code = pdf_content_string[TARGET_PARTLIST_RE]
self.is_a_match = !self.last_match_code.nil?
A much more performant way would be to build a prefix tree and create the regexp using the prefix tree (this optimises the regexp lookup), but this is a bit more work :)
I have a script in bash that calls a TCL script for each element on my network which performs some actions based on the type of the element. This is part of the code that checks whether or not the hostname contains a specific pattern(e.g. *CGN01) and then gives the appropriate command to that machine.
if {[string match "{*CGN01}" $hostname] || $hostname == "AthMet1BG01"} {
expect {
"*#" {send "admin show inventory\r"; send "exit\r"; exp_continue}
eof
}
}
With the code i quoted above i get no error BUT when the hostname is "PhiMSC1CGN01" then the code inside the if is not executed which means that the expression is not correct.
I have tried everything (use of "()" or "{}" or"[]" inside the if) but when i dont put "" on the pattern i get an error like:
invalid bareword "string"
in expression "(string match {*DR0* *1TS0* *...";
should be "$string" or "{string}" or "string(...)" or ...
(parsing expression "(string match {*DR0* *...")
invoked from within
"if {$hostname == "AthMar1BG03" || [string match *CGN01 $hostname]...
or this:
expected boolean value but got "[string match -nocase "*CGN01" $hostname]==0"
while executing
"if {$hostname == "AthMar1BG03" || {[string match -nocase "*CGN01" $hostname]==0}...
when i tried to use ==0 or ==1 on the expression.
My TCL-Version is 8.3 and i cant update it because the machine has no internet connecticity :(
Please help me i am trying to fix this for over a month...
If you want to match a string that is either exactly AthMet1BG01 or any string that ends with CGN01, you should use
if {[string match *CGN01 $hostname] || $hostname == "AthMet1BG01"} {
(For Tcl 8.5 or later, use eq instead of ==.)
Some comments on your attempts:
(The notes about the expression language used by if go for expr and while as well. It is fully described in the documentation for expr.)
To invoke a command inside the condition and substitute its result, it needs to be enclosed in brackets ([ ]). Parentheses (( )) can be used to set the priority of subexpressions within the condition, but don't indicate a command substitution.
Normally, inside the condition strings need to be enclosed in double quotes or braces ({ }). This is because the expression language that is used to express the condition needs to distinguish between e.g. numbers and strings, which Tcl in general doesn't. Inside a command substitution within a condition, you don't need to use quotes or braces, as long as there are no characters in the string that you need to quote.
The string {abc} contains the characters abc. The string "{abc}" contains the characters {abc}, because the double quotes make the braces normal characters (the reverse also holds). [string match "{*bar}" $str] matches the string {foobar} (with the braces as part of the text), but not foobar.
If you put braces around a command substitution, {[incr foo]}, it becomes just the string [incr foo], i.e. the command isn't invoked and no substitution is made. If you use {[incr foo]==1} you get the string [incr foo]==1. The correct way to write this within an expression is [incr foo]==1, with optional whitespace around the ==.
All this is kind of hard to grok, but when you have it is really easy to use. Tcl is stubborn as a mule about interpreting strings, but carries heavy loads if you treat her right.
ETA an alternate matcher (see comments)
You can write your own alternate string matcher:
proc altmatch {patterns string} {
foreach pattern $patterns {
if {[string match $pattern $string]} {
return 1
}
}
return 0
}
If any of the patterns match, you get 1; if none of the patterns match, you get 0.
% altmatch {*bar f?o} foobar
1
% altmatch {*bar f?o} fao
1
% altmatch {*bar f?o} foa
0
For those who have a modern Tcl version, you can actually add it to the string ensemble so it works like other string commands. Put it in the right namespace:
proc ::tcl::string::altmatch {patterns string} {
... as before ...
and install it like this:
% set map [namespace ensemble configure string -map]
% dict set map altmatch ::tcl::string::altmatch
% namespace ensemble configure string -map $map
Documentation:
expr,
string,
Summary of Tcl language syntax
This command:
if {[string match "{*CGN01}" $hostname] || $hostname == "AthMet1BG01"} {
is syntactically valid but I really don't think that you want to use that pattern with string match. I'd guess that you really want:
if {[string match "*CGN01" $hostname] || $hostname == "AthMet1BG01"} {
The {braces} inside that pattern are not actually meaningful (string match only does a subset of the full capabilities of a glob match) so with your erroneous pattern you're actually trying to match a { at the start of $hostname, any number of characters, and then CGN01} at the end of $hostname. With the literal braces. Simply removing the braces lets PhiMSC1CGN01 match.
These parse and execute fine:
"=".scan(/=/)
"=".scan (/=/)
This causes "unterminated regexp meets end of file":
"=".scan /=/
If I insert something before the = the error goes away:
"=".scan /^=/
What's going on?
I'm guessing that you're hitting this in the parser:
case '/':
if (IS_BEG()) {
lex_strterm = NEW_STRTERM(str_regexp, '/', 0);
return tREGEXP_BEG;
}
if ((c = nextc()) == '=') {
set_yylval_id('/');
lex_state = EXPR_BEG;
return tOP_ASGN;
}
Note the nextc() check in the second if. For reference, tOP_ASGN is:
%token <id> tOP_ASGN /* +=, -= etc. */
so it is used for operator-assign tokens.
This suggests that that /=/ in
'='.scan /=/
is being seen as the divide-assign operator (/=) followed by a start-regex-literal (/).
You'll have trouble (of a slightly different sort) with this:
' ='.scan / =/
but not this:
' ='.scan(/ =/)
There is often ambiguity when a method call doesn't have parentheses. In this case, I think operator precedence rules apply and that's not what you're expecting.
I tend to put parentheses on all my method calls because I'm too old and cranky to want to worry about how the parser is going to behave.
I actually have a string called "cond". This is the content of that string:
"20 < 50"
I would like to insert it into a condition like this: (example)
if 20 < 50
return "Hello"
But that condition is a string, so I can't write this:
if cond
return "Hello"
So I would like to know if it is possible to convert a string to a condition to set in an "if" condition. And if it is possible, how can I do it ?
Thank you.
eval might just be your friend here:
>> eval('20 < 50')
=> true
However, eval will execute the arbitrary code inside its argument; you should be sure that your cond can't contain anything detrimental to your system's health!
One alternative to using eval is perhaps to write an evaluator (or use/modify an existing one, like this one by Sterling Camden).
As is his code requires you to write lt, gt, eq, and so on, instead of <, >, ==, ... . As noted in a comment in calc.rb:
# Equality and its clan (note we cannot use '==' or other two-character
# non-word operators, because of the way we parse the string. Non-word
# characters come in one at a time.
If you know the condition will always be basic like the example you provided, you can do this:
left, op, right = "20 < 50".split
cond = left.to_i.send(op.to_sym, right.to_i)
I am currently working on a Ruby Problem quiz but I'm not sure if my solution is right. After running the check, it shows that the compilation was successful but i'm just worried it is not the right answer.
The problem:
A string S consisting only of characters '(' and ')' is called properly nested if:
S is empty,
S has the form "(U)" where
U is a properly nested string,
S has
the form "VW" where V and W are
properly nested strings.
For example, "(()(())())" is properly nested and "())" isn't.
Write a function
def nesting(s)
that given a string S returns 1 if S
is properly nested and 0 otherwise.
Assume that the length of S does not
exceed 1,000,000. Assume that S
consists only of characters '(' and
')'.
For example, given S = "(()(())())"
the function should return 1 and given
S = "())" the function should return
0, as explained above.
Solution:
def nesting ( s )
# write your code here
if s == '(()(())())' && s.length <= 1000000
return 1
elsif s == ' ' && s.length <= 1000000
return 1
elsif
s == '())'
return 0
end
end
Here are descriptions of two algorithms that should accomplish the goal. I'll leave it as an exercise to the reader to turn them into code (unless you explicitly ask for a code solution):
Start with a variable set to 0 and loop through each character in the string: when you see a '(', add one to the variable; when you see a ')', subtract one from the variable. If the variable ever goes negative, you have seen too many ')' and can return 0 immediately. If you finish looping through the characters and the variable is not exactly 0, then you had too many '(' and should return 0.
Remove every occurrence of '()' in the string (replace with ''). Keep doing this until you find that nothing has been replaced (check the return value of gsub!). If the string is empty, the parentheses were matched. If the string is not empty, it was mismatched.
You're not supposed to just enumerate the given examples. You're supposed to solve the problem generally. You're also not supposed to check that the length is below 1000000, you're allowed to assume that.
The most straight forward solution to this problem is to iterate through the string and keep track of how many parentheses are open right now. If you ever see a closing parenthesis when no parentheses are currently open, the string is not well-balanced. If any parentheses are still open when you reach the end, the string is not well-balanced. Otherwise it is.
Alternatively you could also turn the specification directly into a regex pattern using the recursive regex feature of ruby 1.9 if you were so inclined.
My algorithm would use stacks for this purpose. Stacks are meant for solving such problems
Algorithm
Define a hash which holds the list of balanced brackets for
instance {"(" => ")", "{" => "}", and so on...}
Declare a stack (in our case, array) i.e. brackets = []
Loop through the string using each_char and compare each character with keys of the hash and push it to the brackets
Within the same loop compare it with the values of the hash and pop the character from brackets
In the end, if the brackets stack is empty, the brackets are balanced.
def brackets_balanced?(string)
return false if string.length < 2
brackets_hash = {"(" => ")", "{" => "}", "[" => "]"}
brackets = []
string.each_char do |x|
brackets.push(x) if brackets_hash.keys.include?(x)
brackets.pop if brackets_hash.values.include?(x)
end
return brackets.empty?
end
You can solve this problem theoretically. By using a grammar like this:
S ← LSR | LR
L ← (
R ← )
The grammar should be easily solvable by recursive algorithm.
That would be the most elegant solution. Otherwise as already mentioned here count the open parentheses.
Here's a neat way to do it using inject:
class String
def valid_parentheses?
valid = true
self.gsub(/[^\(\)]/, '').split('').inject(0) do |counter, parenthesis|
counter += (parenthesis == '(' ? 1 : -1)
valid = false if counter < 0
counter
end.zero? && valid
end
end
> "(a+b)".valid_parentheses? # => true
> "(a+b)(".valid_parentheses? # => false
> "(a+b))".valid_parentheses? # => false
> "(a+b))(".valid_parentheses? # => false
You're right to be worried; I think you've got the very wrong end of the stick, and you're solving the problem too literally (the info that the string doesn't exceed 1,000,000 characters is just to stop people worrying about how slow their code would run if the length was 100times that, and the examples are just that - examples - not the definitive list of strings you can expect to receive)
I'm not going to do your homework for you (by writing the code), but will give you a pointer to a solution that occurs to me:
The string is correctly nested if every left bracket has a right-bracket to the right of it, or a correctly nested set of brackets between them. So how about a recursive function, or a loop, that removes the string matches "()". When you run out of matches, what are you left with? Nothing? That was a properly nested string then. Something else (like ')' or ')(', etc) would mean it was not correctly nested in the first place.
Define method:
def check_nesting str
pattern = /\(\)/
while str =~ pattern do
str = str.gsub pattern, ''
end
str.length == 0
end
And test it:
>ruby nest.rb (()(())())
true
>ruby nest.rb (()
false
>ruby nest.rb ((((()))))
true
>ruby nest.rb (()
false
>ruby nest.rb (()(((())))())
true
>ruby nest.rb (()(((())))()
false
Your solution only returns the correct answer for the strings "(()(())())" and "())". You surely need a solution that works for any string!
As a start, how about counting the number of occurrences of ( and ), and seeing if they are equal?