How to use non printable characters in swift string variable? - ios

I need to use non printable characters in a String constant, but Xcode shows error in my swift file as "Unprintable ASCII character found in source file"
My Simple code is below
let unprintableCharInString = "12345"
You could see the non printable characters at prefix and suffix of above string value, If you just copy paste my above code in Sublime text or some other text editor which supports to show Unprintable characters.
But if you paste the above code in Xcode swift file, you will see the compiler error "Unprintable ASCII character found in source file".
And if I use the same string in Objective C as like below, there is no error.
NSString *unprintableCharInString = #"12345";
So how to use non printable characters in Swift string variable directly as like above Objective C code?
Note:
As the body text box trims those non printable chars while saving my question, you can't see those chars if you copy paste the code from here. Instead of that try to copy the above code by editing my question. So you can get those chars in Body text box during edit.
Screenshot from Sublime Text editor:
Thanks in advance!

To display space characters you can use XCode Editor > Show Invisibles. But I'm not really sure will it help in your case.

Based on the #Alladinian's suggestion in Comment above,
Answer is: We need to add the unprintable ASCII characters manually in source code while declaring string value.
Example:
let unprintableCharInString = "\u{02}123\u{1A}"
Here \u{02} is Hex value of "START OF TEXT (STX)" and \u{1A} is Hex value of "SUBSTITUTE (SUB)"
Thanks #Alladinian!

Related

How to remove ANSI codes from a string?

I am working on string manipulation using LUA and having trouble with the following problem.
Using this as an example of the original data I am given -
"[0;1;36m(Web): You say, "Text here."[0;37m"
I want to keep the string intact except for removing the ANSI codes.
I have been pointed toward using gsub with the LUA pattern matching but I cannot seem to get the pattern correct. I am also unsure how to reference exactly the escape character sent.
text:gsub("[\27\[([\d\;]+)m]", "")
or
text:gsub("%x%[[%d+;+]m", "")
If successful, all I want to be left with, using the above example, would be:
(Web): You say, "Text here."
Your string example is missing the escape character, ASCII 27.
Here's one way:
s = '\x1b[0;1;36m(Web): You say, "Text here."\x1b[0;37m'
s = s:gsub('\x1b%[%d+;%d+;%d+;%d+;%d+m','')
:gsub('\x1b%[%d+;%d+;%d+;%d+m','')
:gsub('\x1b%[%d+;%d+;%d+m','')
:gsub('\x1b%[%d+;%d+m','')
:gsub('\x1b%[%d+m','')
print(s)

SWIFT string with special characters without escape

How to print all special characters without inserting escape sign before every of them? I have very large textiles with many special characters and I'm looking for something like # in c# which prints string literally as it is
What you're referring to, is called a verbatim string literal in C# and that concept does not translate exactly to Swift.
However, with the introduction of multiline string Literals in Swift 4, you can get close.
let multilineString = """
Here you can use \ and newline characters.
Also single " or double "" are allowed.
"""
For reference, find the grammar of a Swift String literal here.

Escape Unicode Characters for iOS

There are some Unicode arrangements that I want to use in my app. I am having trouble properly escaping them for use.
For instance this Unicode sequence: 🅰
If I escape it using an online tool i get: \ud83c\udd70
But of course this is an invalid sequence per the compiler:
var str = NSString.stringWithUTF8String("\ud83c\udd70")
Also if I do this:
var str = NSString.stringWithUTF8String("\ud83c")
I get an error "Invalid Unicode Scalar"
I'm trying to use these Unicode "fonts":
http://www.panix.com/~eli/unicode/convert.cgi?text=abcdefghijklmnopqrstuvwxyz
If I view the source of this website I see sequences like this:
&#x1D552
Struggling to wrap my head around what is the "proper" way to work with/escape unicode.
And simply need a to figure out a way to get them working on iOS.
Any thoughts?
\ud83c\udd70 is a UTF-16 surrogate pair which encodes the unicode character 🅰 (U+1F170). Swift string literals do not use UTF-16, so that escape sequence doesn't make sense. However, since 1F170 has five digits you can't use a \uXXXX escape sequence (which only accepts four hexadecimal digits). Instead, use a \UXXXXXXXX sequence (note the capital U), which accepts eight:
var str = "\U0001F170" // returns "🅰"
You can also just paste the character itself into your string:
var str = "🅰" // returns "🅰"
Swift is an early Beta, is is broken in many ways. This issue is a Swift bug.
let ringAboveA: String = "\u0041\u030A" is Ã… and is accepted
let negativeSquaredA: String = "\uD83D\uDD70" is 🅰 and produces an error
Both are decomposed UTF16 characters that are accepted by Objective-C. The difference is that the composed character 🅰 is in plane 1.
Note: to get the UTF32 code point either use the OSX Character Viewer or a code snippet:
NSLog(#"utf32: %#", [#"🅰" dataUsingEncoding:NSUTF32BigEndianStringEncoding]);
utf32: <0001f170>
To get the Character Viewer in the Apple Menu go to the "System Preferences", "Keyboard", "Keyboard" tab and select the checkbox: "Show Keyboard & Character Viewers in menu bar". The "Character View" item will be in the menu bar just to the left of the Date.
After entering the character right (control) click on the character in favorites to copy the search results.
Copied information:
🅰
NEGATIVE SQUARED LATIN CAPITAL LETTER A
Unicode: U+1F170 (U+D83C U+DD70), UTF-8: F0 9F 85 B0
Better yet: Add unicode in the list on the left and select it.

Handing strings with binary data in it using java.nio

I am having issues parsing text files that have illegal characters(binary markers) in them. An answer would be something as follows:
test.csv
^000000^id1,text1,text2,text3
Here the ^000000^ is a textual representation of illegal characters in the source file.
I was thinking about using the java.nio to validate the line before I process it. So, I was thinking of introducing a Validator trait as follows:
import java.nio.charset._
trait Validator{
private def encoder = Charset.forName("UTF-8").newEncoder
def isValidEncoding(line:String):Boolean = {
encoder.canEncode(line)
}
}
Do you guys think this is the correct approach to handle the situation?
Thanks
It is too late when you already have a String, UTF-8 can always encode any string*. You need to go to the point where you are decoding the file initially.
ISO-8859-1 is an encoding with interesting properties:
Literally any byte sequence is valid ISO-8859-1
The code point of each decoded character is exactly the same as the value of the byte it was decoded from
So you could decode the file as ISO-8859-1 and just strip non-English characters:
//Pseudo code
str = file.decode("ISO-8859-1");
str = str.replace( "[\u0000-\u0019\u007F-\u00FF]", "");
You can also iterate line-by-line, and ignore each line that contains a character in [\u0000-\u0019\u007F-\u00FF], if that's what you mean by validating a line before processing it.
It also occurred to me that the binary marker could be a BOM. You can use a hex editor to view the values.
*Except those with illegal surrogates which is probably not the case here.
Binary data is not a string. Don't try to hack around input sequences that would be illegal upon conversion to a String.
If your input is an arbitrary sequence of bytes (even if many of them conform to ASCII), don't even try to convert it to a String.

What character encoding are the following German words using?

I'm trying to process a German word list and can't figure out what encoding the file is in. The 'file' unix command says the file is "Non-ISO extended-ASCII text". Most of the words are in ascii, but here are the exceptions:
ANDR\x82
ATTACH\x82
C\x82ZANNE
CH\x83TEAU
CONF\x82RENCIER
FABERG\x82
L\x82VI-STRAUSS
RH\x93NETAL
P\xF2ANGE
Any hints would be great. Thanks!
EDIT: To be clear, the hex codes above are C hex string literals so replace \xXX with the literal hex value XX.
It looks like CP437 or CP852, assuming the \x82 sequences encode single characters, and are not literally four characters. Well, at least everything else does, but the last line is a bit of a puzzle.

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