Swift: What Does Backslash Dot "\." Mean

Swift: What does backslash dot \. mean?

tl;dr: We take a look at the Swift language reference, and sure enough, the usage of this backslash-dot notation is called a key-path-expression.

(The question has been sufficiently answered, by this point.)

A more hands-on approach on how to get to that piece of buried documentation:

As you can see from the code you posted, the User class contains a property named email.

Notice that, assuming you're using Xcode, if you replace return \.email with return \, you get the compile-error "Expected expression path in Swift key path", so this is a hint that this backslash-dot notation might have to do with something called a key path.

From that documentation on key-path, we see that we could also have written \User.email (and you can try it out in Xcode with no compiler error).

Understanding the greater context of what's going on in that code:

So, semantically, to understand the meaning of the usernameKey declaration you're looking at, we might want to understand what a WritableKeyPath is. In simple, from the documentation, we see that a WritableKeyPath is: "A key path that supports reading from and writing to the resulting value."

So, we see that the usernameKey declaration takes in a WritableKeyPath object and returns a String that is User.email.

Furthermore, it's apparent that the User class needs this usernameKey property in order to conform to the PasswordAuthenticatable protocol, which was imported on the first line with import Authentication (if you care to explore there, take a look at Dependencies > Auth 2.0.0 > Authentication > Basic > BasicAuthenticatable.swift).

What does backslash do in Swift?

The backslash has a few different meanings in Swift, depending on the context. In your case, it means string interpolation:

print("The total cost of my meal is \(dictionary["pizza"]! + dictionary["ice cream"]!)")

...is the same as:

print("The total cost of my meal is " + String(dictionary["pizza"]! + dictionary["ice cream"]!))

But the first form is more readable. Another example:

print("Hello \(person.firstName). You are \(person.age) years old")

Which may print something like Hello John. You are 42 years old. Much clearer than:

print("Hello " + person.firstName + ". You are " + String(person.age) + " years old")

What is the backslash(\) used for in SwiftUI?

\.self it's a identity keypath that apple added for:

Add the ability to reference the identity key path, which refers to the entire input value it is applied to.

More info in proposal.

(Swift) how to print \ character in a string?

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.

Difference between \n and \r?

In terms of ascii code, it's 3 -- since they're 10 and 13 respectively;-).

But seriously, there are many:

• in Unix and all Unix-like systems, \n is the code for end-of-line, \r means nothing special
• as a consequence, in C and most languages that somehow copy it (even remotely), \n is the standard escape sequence for end of line (translated to/from OS-specific sequences as needed)
• in old Mac systems (pre-OS X), \r was the code for end-of-line instead
• in Windows (and many old OSs), the code for end of line is 2 characters, \r\n, in this order
• as a (surprising;-) consequence (harking back to OSs much older than Windows), \r\n is the standard line-termination for text formats on the Internet
• for electromechanical teletype-like "terminals", \r commands the carriage to go back leftwards until it hits the leftmost stop (a slow operation), \n commands the roller to roll up one line (a much faster operation) -- that's the reason you always have \r before \n, so that the roller can move while the carriage is still going leftwards!-) Wikipedia has a more detailed explanation.
• for character-mode terminals (typically emulating even-older printing ones as above), in raw mode, \r and \n act similarly (except both in terms of the cursor, as there is no carriage or roller;-)

In practice, in the modern context of writing to a text file, you should always use \n (the underlying runtime will translate that if you're on a weird OS, e.g., Windows;-). The only reason to use \r is if you're writing to a character terminal (or more likely a "console window" emulating it) and want the next line you write to overwrite the last one you just wrote (sometimes used for goofy "ascii animation" effects of e.g. progress bars) -- this is getting pretty obsolete in a world of GUIs, though;-).

What characters are allowed in an email address?

See RFC 5322: Internet Message Format and, to a lesser extent, RFC 5321: Simple Mail Transfer Protocol.

RFC 822 also covers email addresses, but it deals mostly with its structure:

 addr-spec   =  local-part "@" domain        ; global address     
 local-part  =  word *("." word)             ; uninterpreted
                                             ; case-preserved
 
 domain      =  sub-domain *("." sub-domain)     
 sub-domain  =  domain-ref / domain-literal     
 domain-ref  =  atom                         ; symbolic reference

And as usual, Wikipedia has a decent article on email addresses:

The local-part of the email address may use any of these ASCII characters:

• uppercase and lowercase Latin letters A to Z and a to z;
• digits 0 to 9;
• special characters !#$%&'*+-/=?^_`{|}~;
• dot ., provided that it is not the first or last character unless quoted, and provided also that it does not appear consecutively unless quoted (e.g. John..Doe@example.com is not allowed but "John..Doe"@example.com is allowed);
• space and "(),:;<>@[\] characters are allowed with restrictions (they are only allowed inside a quoted string, as described in the paragraph below, and in addition, a backslash or double-quote must be preceded by a backslash);
• comments are allowed with parentheses at either end of the local-part; e.g. john.smith(comment)@example.com and (comment)john.smith@example.com are both equivalent to john.smith@example.com.

In addition to ASCII characters, as of 2012 you can use international characters above U+007F, encoded as UTF-8 as described in the RFC 6532 spec and explained on Wikipedia. Note that as of 2019, these standards are still marked as Proposed, but are being rolled out slowly. The changes in this spec essentially added international characters as valid alphanumeric characters (atext) without affecting the rules on allowed & restricted special characters like !# and @:.

For validation, see Using a regular expression to validate an email address.

The domain part is defined as follows:

The Internet standards (Request for Comments) for protocols mandate that component hostname labels may contain only the ASCII letters a through z (in a case-insensitive manner), the digits 0 through 9, and the hyphen (-). The original specification of hostnames in RFC 952, mandated that labels could not start with a digit or with a hyphen, and must not end with a hyphen. However, a subsequent specification (RFC 1123) permitted hostname labels to start with digits. No other symbols, punctuation characters, or blank spaces are permitted.

Regular expression for floating point numbers

TL;DR

Use [.] instead of \. and [0-9] instead of \d to avoid escaping issues in some languages (like Java).

^{Thanks to the nameless one for originally recognizing this.}

One relatively simple pattern for matching a floating point number in a larger string is:

[+-]?([0-9]*[.])?[0-9]+

This will match:

• 123
• 123.456
• .456

^{See a working example}

If you also want to match 123. (a period with no decimal part), then you'll need a slightly longer expression:

[+-]?([0-9]+([.][0-9]*)?|[.][0-9]+)

^{See pkeller's answer for a fuller explanation of this pattern}

If you want to include a wider spectrum of numbers, including scientific notation and non-decimal numbers such as hex and octal, see my answer to How do I identify if a string is a number?.

If you want to validate that an input is a number (rather than finding a number within the input), then you should surround the pattern with ^ and $, like so:

^[+-]?([0-9]+([.][0-9]*)?|[.][0-9]+)$

Irregular Regular Expressions

"Regular expressions", as implemented in most modern languages, APIs, frameworks, libraries, etc., are based on a concept developed in formal language theory. However, software engineers have added many extensions that take these implementations far beyond the formal definition. So, while most regular expression engines resemble one another, there is actually no standard. For this reason, a lot depends on what language, API, framework or library you are using.

(Incidentally, to help reduce confusion, many have taken to using "regex" or "regexp" to describe these enhanced matching languages. See Is a Regex the Same as a Regular Expression? at RexEgg.com for more information.)

That said, most regex engines (actually, all of them, as far as I know) would accept \.. Most likely, there's an issue with escaping.

The Trouble with Escaping

Some languages have built-in support for regexes, such as JavaScript. For those languages that don't, escaping can be a problem.

This is because you are basically coding in a language within a language. Java, for example, uses \ as an escape character within it's strings, so if you want to place a literal backslash character within a string, you must escape it:

// creates a single character string: "\"
String x = "\\";

However, regexes also use the \ character for escaping, so if you want to match a literal \ character, you must escape it for the regex engine, and then escape it again for Java:

// Creates a two-character string: "\\"
// When used as a regex pattern, will match a single character: "\"
String regexPattern = "\\\\";

In your case, you have probably not escaped the backslash character in the language you are programming in:

// will most likely result in an "Illegal escape character" error
String wrongPattern = "\.";
// will result in the string "\."
String correctPattern = "\\.";

All this escaping can get very confusing. If the language you are working with supports raw strings, then you should use those to cut down on the number of backslashes, but not all languages do (most notably: Java). Fortunately, there's an alternative that will work some of the time:

String correctPattern = "[.]";

For a regex engine, \. and [.] mean exactly the same thing. Note that this doesn't work in every case, like newline (\\n), open square bracket (\\[) and backslash (\\\\ or [\\]).

A Note about Matching Numbers

(Hint: It's harder than you think)

Matching a number is one of those things you'd think is quite easy with regex, but it's actually pretty tricky. Let's take a look at your approach, piece by piece:

[-+]?

Match an optional - or +

[0-9]*

Match 0 or more sequential digits

\.?

Match an optional .

[0-9]*

Match 0 or more sequential digits

First, we can clean up this expression a bit by using a character class shorthand for the digits (note that this is also susceptible to the escaping issue mentioned above):

[0-9] = \d

I'm going to use \d below, but keep in mind that it means the same thing as [0-9]. (Well, actually, in some engines \d will match digits from all scripts, so it'll match more than [0-9] will, but that's probably not significant in your case.)

Now, if you look at this carefully, you'll realize that every single part of your pattern is optional. This pattern can match a 0-length string; a string composed only of + or -; or, a string composed only of a .. This is probably not what you've intended.

To fix this, it's helpful to start by "anchoring" your regex with the bare-minimum required string, probably a single digit:

\d+

Now we want to add the decimal part, but it doesn't go where you think it might:

\d+\.?\d* /* This isn't quite correct. */

This will still match values like 123.. Worse, it's got a tinge of evil about it. The period is optional, meaning that you've got two repeated classes side-by-side (\d+ and \d*). This can actually be dangerous if used in just the wrong way, opening your system up to DoS attacks.

To fix this, rather than treating the period as optional, we need to treat it as required (to separate the repeated character classes) and instead make the entire decimal portion optional:

\d+(\.\d+)? /* Better. But... */

This is looking better now. We require a period between the first sequence of digits and the second, but there's a fatal flaw: we can't match .123 because a leading digit is now required.

This is actually pretty easy to fix. Instead of making the "decimal" portion of the number optional, we need to look at it as a sequence of characters: 1 or more numbers that may be prefixed by a . that may be prefixed by 0 or more numbers:

(\d*\.)?\d+

Now we just add the sign:

[+-]?(\d*\.)?\d+

Of course, those slashes are pretty annoying in Java, so we can substitute in our long-form character classes:

[+-]?([0-9]*[.])?[0-9]+

Matching versus Validating

This has come up in the comments a couple times, so I'm adding an addendum on matching versus validating.

The goal of matching is to find some content within the input (the "needle in a haystack"). The goal of validating is to ensure that the input is in an expected format.

Regexes, by their nature, only match text. Given some input, they will either find some matching text or they will not. However, by "snapping" an expression to the beginning and ending of the input with anchor tags (^ and $), we can ensure that no match is found unless the entire input matches the expression, effectively using regexes to validate.

The regex described above ([+-]?([0-9]*[.])?[0-9]+) will match one or more numbers within a target string. So given the input:

apple 1.34 pear 7.98 version 1.2.3.4

The regex will match 1.34, 7.98, 1.2, .3 and .4.

To validate that a given input is a number and nothing but a number, "snap" the expression to the start and end of the input by wrapping it in anchor tags:

^[+-]?([0-9]*[.])?[0-9]+$

This will only find a match if the entire input is a floating point number, and will not find a match if the input contains additional characters. So, given the input 1.2, a match will be found, but given apple 1.2 pear no matches will be found.

Note that some regex engines have a validate, isMatch or similar function, which essentially does what I've described automatically, returning true if a match is found and false if no match is found. Also keep in mind that some engines allow you to set flags which change the definition of ^ and $, matching the beginning/end of a line rather than the beginning/end of the entire input. This is typically not the default, but be on the lookout for these flags.

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