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 {
Related
In sales column i have values with pound sign £1200. It is not readable by Data frame in scala, please help me for the same. i want column value in double, 1200. I am using below method but its not working.
def getRemovedDollarValue = udf(
(actualSales: String) => {
val actualSalesDouble = actualSales
.replace(",", "")
.replace("$", "")
.replace("\\u00A3","")
.replace("\\U00A3","")
.replaceAll("\\s", "_").trim().toDouble
java.lang.Double.parseDouble(actualSalesDouble.toString)
}
)
You need write: .replace("\u00A3","") instead of escaping .replace("\\u00A3","").
But I prefer just: .replace("£", "") - it is more readable.
I think the proposed solutions and comments all work but don't address the confusion behind why your code isn't working.
From the Pattern docs:
Thus the strings "\u2014" and "\\u2014", while not equal, compile into the same pattern, which matches the character with hexadecimal value 0x2014.
replace and replaceAll are both replacing all occurrences in a String, but only replaceAll is taking in a regular expression. You're passing in "\\u00A3" which will work as a pattern, but not a unicode literal due to the added backslash. As already suggested, either use replace with a unicode literal or the actual symbol, or change to replaceAll.
All I want to do is convert a single Character to uppercase without the overhead of converting to a String and then calling .uppercased(). Is there any built-in way to do this, or a way for me to call the toupper() function from C without any bridging? I really don't think I should have to go out of my way for something so simple.
To call the C toupper() you need to get the Unicode code point of the Character. But Character has no method for getting its code point (a Character may consist of multiple code points), so you have to convert the Character into a String to obtain any of its code points.
So you really have to convert to String to get anywhere. Unless you store the character as a UnicodeScalar instead of a Character. In this case you can do this:
assert(unicodeScalar.isASCII) // toupper argument must be "representable as an unsigned char"
let uppercase = UnicodeScalar(toupper(CInt(unicodeScalar.value)))
But this isn't really more readable than simply using String:
let uppercase = Character(String(character).uppercased())
just add this to your program
extension Character {
//converts a character to uppercase
func convertToUpperCase() -> Character {
if(self.isUppercase){
return self
}
return Character(self.uppercased())
}
}
I have the code snippet below (as suggested in this previous Stack Overflow answer ... Deleting all special characters from a string in progress 4GL) which is attempting to remove all extended characters from a string so that I may transmit it to a customer's system which will not accept any extended characters.
do v-int = 128 to 255:
assign v-string = replace(v-string,chr(v-int),"").
end.
It is working perfectly with one exception (which makes me fear there may be others I have not caught). When it gets to 255, it will replace all 'y's in the string.
If I do the following ...
display chr(255) = chr(121). /* 121 is asc code of y */
I get true as the result.
And therefore, if I do the following ...
display replace("This is really strange",chr(255),"").
I get the following result:
This is reall strange
I have verified that 'y' is the only character affected by running the following:
def var v-string as char init "abcdefghijklmnopqrstuvwxyz".
def var v-int as int.
do v-int = 128 to 255:
assign v-string = replace(v-string,chr(v-int),"").
end.
display v-string.
Which results in the following:
abcdefghijklmnopqrstuvwxz
I know I can fix this by removing 255 from the range but I would like to understand why this is happening.
Is this a character collation set issue or am I missing something simpler?
Thanks for any help!
This is a bug. Here's a Progress Knowledge Base article about it:
http://knowledgebase.progress.com/articles/Article/000046181
The workaround is to specify the codepage in the CHR() statement, like this:
CHR(255, "UTF-8", "1252")
Here it is in your example:
def var v-string as char init "abcdefghijklmnopqrstuvwxyz". def var v-int as int.
do v-int = 128 to 255:
assign v-string = replace(v-string, chr(v-int, "UTF-8", "1252"), "").
end.
display v-string.
You should now see the 'y' in the output.
This seems to be a bug!
The REPLACE() function returns an unexpected result when replacing character CHR(255) (ÿ) in a String.
The REPLACE() function modifies the value of the target character, but additionally it changes any occurrence of characters 'Y' and 'y' present in the String.
This behavior seems to affect only the character ÿ. Other characters are correctly changed by REPLACE().
Using default codepage ISO-8859-1
Link to knowledgebase
I am a bit confused about Scala string split behaviour as it does not work consistently and some list elements are missing. For example, if I have a CSV string with 4 columns and 1 missing element.
"elem1, elem2,,elem 4".split(",") = List("elem1", "elem2", "", "elem4")
Great! That's what I would expect.
On the other hand, if both element 3 and 4 are missing then:
"elem1, elem2,,".split(",") = List("elem1", "elem2")
Whereas I would expect it to return
"elem1, elem2,,".split(",") = List("elem1", "elem2", "", "")
Am I missing something?
As Peter mentioned in his answer, "string".split(), in both Java and Scala, does not return trailing empty strings by default.
You can, however, specify for it to return trailing empty strings by passing in a second parameter, like this:
String s = "elem1,elem2,,";
String[] tokens = s.split(",", -1);
And that will get you the expected result.
You can find the related Java doc here.
I believe that trailing empty spaces are not included in a return value.
JavaDoc for split(String regex) says: "This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array."
So in your case split(String regex, int limit) should be used in order to get trailing empty string in a return value.
I have the following code:
val z: String = tree.symbol.toString
z match {
case "method +" | "method -" | "method *" | "method ==" =>
println("no special op")
false
case "method /" | "method %" =>
println("we have the special div operation")
true
case _ =>
false
}
Is it possible to create a match for the primitive operations in Scala:
"method *".matches("(method) (+-*==)")
I know that the (+-*) signs are used as quantifiers. Is there a way to match them anyway?
Thanks from a avidly Scala scholar!
Sure.
val z: String = tree.symbol.toString
val noSpecialOp = "method (?:[-+*]|==)".r
val divOp = "method [/%]".r
z match {
case noSpecialOp() =>
println("no special op")
false
case divOp() =>
println("we have the special div operation")
true
case _ =>
false
}
Things to consider:
I choose to match against single characters using [abc] instead of (?:a|b|c).
Note that - has to be the first character when using [], or it will be interpreted as a range. Likewise, ^ cannot be the first character inside [], or it will be interpreted as negation.
I'm using (?:...) instead of (...) because I don't want to extract the contents. If I did want to extract the contents -- so I'd know what was the operator, for instance, then I'd use (...). However, I'd also have to change the matching to receive the extracted content, or it would fail the match.
It is important not to forget () on the matches -- like divOp(). If you forget them, a simple assignment is made (and Scala will complain about unreachable code).
And, as I said, if you are extracting something, then you need something inside those parenthesis. For instance, "method ([%/])".r would match divOp(op), but not divOp().
Much the same as in Java. To escape a character in a regular expression, you prefix the character with \. However, backslash is also the escape character in standard Java/Scala strings, so to pass it through to the regular expression processing you must again prefix it with a backslash. You end up with something like:
scala> "+".matches("\\+")
res1 : Boolean = true
As James Iry points out in the comment below, Scala also has support for 'raw strings', enclosed in three quotation marks: """Raw string in which I don't need to escape things like \!""" This allows you to avoid the second level of escaping, that imposed by Java/Scala strings. Note that you still need to escape any characters that are treated as special by the regular expression parser:
scala> "+".matches("""\+""")
res1 : Boolean = true
Escaping characters in Strings works like in Java.
If you have larger Strings which need a lot of escaping, consider Scala's """.
E. g. """String without needing to escape anything \n \d"""
If you put three """ around your regular expression you don't need to escape anything anymore.