For example, if I were to do this:
print(“\”)
It would say: `unfinished string near: ‘“”’
instead of my expected output of: ‘\’
How would I print this? I have searched on google yet still have yet to find an answer.
The backslash (\) is escaping the following character, being the double quote ("), causing the string to be unfinished.
To include an actual backslash in your string, you escape it with another backslash:
print("\\")
From Lua 5.4 Reference Manual, §3.1 (emphasis mine):
A short literal string can be delimited by matching single or double quotes, and can contain the following C-like escape sequences: '\a' (bell), '\b' (backspace), '\f' (form feed), '\n' (newline), '\r' (carriage return), '\t' (horizontal tab), '\v' (vertical tab), '\\' (backslash), '"' (quotation mark [double quote]), and ''' (apostrophe [single quote]). [...]
Related
I tried to split by "\", but this character is so special in Lua, even if I use escape character "%", the IDE shows an error Unterminated String constant
local index = string.find("lua. is \wonderful", "%\", 1)
To insert backslash \ into a quoted string, escape it with itself: "\\". \ is the escape character in regular quoted strings, so it is escaped with \. Or you can use the long string syntax, which doesn't allow escape sequences, as already pointed out: [[\]].
Percent is only an escape character in a string that is being used as a pattern, so it is used before the magical characters ^$()%.[]*+-? in the second argument to string.find, string.match, string.gmatch, and string.gsub, and %% represents % in the third argument to string.gsub.
The percent is still there in the string that is stored in memory, but backslash escape sequences are replaced with the corresponding character. \\ becomes \ when the string is stored in memory, and if you count the number of backslashes in a string "\\" using string.gsub, it will only find one: select(2, string.gsub("\\", "\\", "")) returns 1.
I use this text text\r2. And I want to print this in debug and get result:
text\r2
but I get this:
text
2
Try to escape the backslash with another backslash: text\\r2.
The \r will otherwise be interpreted as a line break.
\r in a String literal is a special character and represents a carriage return
See Special Characters in String Literals
String literals can include the following special characters:
* The escaped special characters \0 (null character), \\ (backslash), \t (horizontal tab), \n (line feed), \r (carriage return), \" (double quotation mark) and \' (single quotation mark)
* An arbitrary Unicode scalar, written as \u{n}, where n is a 1–8 digit hexadecimal number with a value equal to a valid Unicode code point (Unicode is discussed in Unicode below)
If you want to use in a String Literal a backslash you have to escape it using \\.
So you'll have to write
print("text\\r2")
to get text\r2
On The Lua Interpreter
>print("This is a string
>>spread over multiline")
stdin:1: unfinished string near '"This is a'
Since we know on the Lua interpreter we can finish a statement over mulitline
For eg
>a=2
>a=a+
>>1
This works perfectly
Again:
>print([[This is a multiline
>>string]])
This is a multiline
string
This works fine!! then why display error in the first print() statement??
Read the fine Reference Manual:
3.1 – Lexical Conventions
[…]
A short literal string can be delimited by matching single or double
quotes, and can contain the following C-like escape sequences: '\a' (bell),
'\b' (backspace), '\f' (form feed), '\n' (newline), '\r' (carriage
return), '\t' (horizontal tab), '\v' (vertical tab), '\\' (backslash),
'\"' (quotation mark [double quote]), and '\'' (apostrophe [single
quote]). A backslash followed by a line break results in a newline in the
string. The escape sequence '\z' skips the following span of white-space
characters, including line breaks; it is particularly useful to break and
indent a long literal string into multiple lines without adding the newlines
and spaces into the string contents. A short literal string cannot contain
unescaped line breaks nor escapes not forming a valid escape sequence.
[…]
Literal strings can also be defined using a long format enclosed by long
brackets. We define an opening long bracket of level n as an opening
square bracket followed by n equal signs followed by another opening
square bracket. So, an opening long bracket of level 0 is written as [[,
an opening long bracket of level 1 is written as [=[, and so on. A
closing long bracket is defined similarly; for instance, a closing long
bracket of level 4 is written as ]====]. A long literal starts with an
opening long bracket of any level and ends at the first closing long
bracket of the same level. It can contain any text except a closing
bracket of the same level. Literals in this bracketed form can run for
several lines, do not interpret any escape sequences, and ignore long
brackets of any other level. Any kind of end-of-line sequence (carriage
return, newline, carriage return followed by newline, or newline followed
by carriage return) is converted to a simple newline.
I am trying to validate a phone number using NSPredicate and regex. The only problem is when setting the regex Swift thinks that I am trying to escape part of it due to the backslashes. How can I get around this?
My code is as follows:
let phoneRegEx = "^((\(?0\d{4}\)?\s?\d{3}\s?\d{3})|(\(?0\d{3}\)?\s?\d{3}\s?\d{4})|(\(?0\d{2}\)?\s?\d{4}\s?\d{4}))(\s?\#(\d{4}|\d{3}))?$"
In Swift regular string literals, you need to double-escape the slashes to define literal backslashes:
let phoneRegEx = "^((\\(?0\\d{4}\\)?\\s?\\d{3}\\s?\\d{3})|(\\(?0\\d{3}\\)?\\s?\\d{3}\\s?\\d{4})|(\\(?0\\d{2}\\)?\\s?\\d{4}\\s?\\d{4}))(\\s?#(\\d{4}|\\d{3}))?$"
Starting from Swift 5, you can use raw string literals and escape regex escapes with a single backslash:
let phoneRegEx = #"^((\(?0\d{4}\)?\s?\d{3}\s?\d{3})|(\(?0\d{3}\)?\s?\d{3}\s?\d{4})|(\(?0\d{2}\)?\s?\d{4}\s?\d{4}))(\s?#(\d{4}|\d{3}))?$"#
Please refer to the Regular Expression Metacharacters table on the ICU Regular Expressions page to see what regex escapes should be escaped this way.
Please mind the difference between the regex escapes (in the above table) and string literal escape sequences used in the regular string literals that you may check, say, at Special Characters in String Literals:
String literals can include the following special characters:
The escaped special characters \0 (null character), \\ (backslash), \t (horizontal tab), \n (line feed), \r (carriage return), \" (double quotation mark) and \' (single quotation mark)
An arbitrary Unicode scalar value, written as \u{n}, where n is a 1–8 digit hexadecimal number (Unicode is discussed in Unicode below)
So, in regular string literals, "\"" is a " string written as a string literal, and you do not have to escape a double quotation mark for the regex engine, so "\"" string literal regex pattern is enough to match a " char in a string. However, "\\\"", a string literal repesenting \" literal string will also match " char, although you can already see how redundant this regex pattern is. Also, "\n" (an LF symbol) matches a newline in the same way as "\\n" does, as "\n" is a literal representation of the newline char and "\\n" is a regex escape defined in the ICU regex escape table.
In raw string literals, \ is just a literal backslash.
I have tried to print it but it just by passes because it's an escaped character.
e.g output should be as follows.
\correct
For that and also future reference:
\0 – Null character (that is a zero after the slash)
\\ – Backslash itself. Since the backslash is used to escape other characters, it needs a special escape to actually print itself.
\t – Horizontal tab
\n – Line Feed
\r – Carriage Return
\” – Double quote. Since the quotes denote a String literal, this is necessary if you actually want to print one.
\’ – Single Quote. Similar reason to above.
Use the following code for Swift 5, Xcode 10.2
let myText = #"This is a Backslash: \"#
print(myText)
Output:
This is a Backslash: \
Now not required to add a double slash to use a single slash in swift 5, even now required slash before some character, for example, single quote, double quote etc.
See this post for latest update about swift 5
https://www.hackingwithswift.com/articles/126/whats-new-in-swift-5-0
var s1: String = "I love my "
let s2: String = "country"
s1 += "\"\(s2)\""
print(s1)
It will print I love my "country"
The backslash character \ acts as an escape character when used in a string. This means you can use, for example, double quotes, in a string by pre-pending them with \. The same also applies for the backslash character itself, which is to say that println("\\") will result in just \ being printed.