I'm walking around syntax tree in Go, trying to find all calls to some particular function and then get its string argument (it's a file name, and should be string literal, not any other identifier). I'm succeeded with this, and now I have ast.BasicLit node with Kind == token.STRING, but its value is Go code, not a value of the string that it should have.
I found question that answers how to do reverse transformation - from string to go code it represents: golang: given a string, output an equivalent golang string literal
But I want the opposite - something like eval function (but just for Go string literals).
You can use the strconv.Unquote() to do the conversion (unquoting).
One thing you should be aware of is that strconv.Unquote() can only unquote strings that are in quotes (e.g. start and end with a quote char " or a back quote char `), so you have to manually append that if it's not in quotes.
Example:
fmt.Println(strconv.Unquote("Hi")) // Error: invalid syntax
fmt.Println(strconv.Unquote(`Hi`)) // Error: invalid syntax
fmt.Println(strconv.Unquote(`"Hi"`)) // Prints "Hi"
fmt.Println(strconv.Unquote(`"Hi\x21"`)) // Prints "Hi!"
// This will print 2 lines:
fmt.Println(strconv.Unquote(`"First line\nSecondline"`))
Output (try it on the Go Playground):
invalid syntax
invalid syntax
Hi <nil>
Hi! <nil>
First line
Secondline <nil>
Related
I'm new to basic.
Does the following line mean the string ":PULS:WIDT1 " concatenates with a string variable named Te?
":PULS:WIDT1 "&VAL$(Te);
Does the following line mean the string ":PULS:WIDT1 " concatenates with a string variable named Te?
The line is incomplete. Even if this were an accepted syntax and concatenation somehow took place, where would the result go?
The ampersand character (&) is indeed a string concatenation operator in BASIC, but so is the plus character (+). Not every BASIC allows both, but most will allow at least one.
Inserted in a syntactically correct statement, your expression
":PULS:WIDT1 "&VAL$(Te)
would concatenate the ":PULS:WIDT1 " string litteral with the contents of an element of the string array VAL$() indexed by the numerical Te variable.
Although VAL happens to be the name of a built-in function, many BASICs don't mind that you name one or more user variables the same as a keyword.
In Swift I can create a String variable such as this:
let s = "Hello\nMy name is Jack!"
And if I use s, the output will be:
Hello
My name is Jack!
(because the \n is a linefeed)
But what if I want to programmatically obtain the raw characters in the s variable? As in if I want to actually do something like:
let sRaw = s.raw
I made the .raw up, but something like this. So that the literal value of sRaw would be:
Hello\nMy name is Jack!
and it would literally print the string, complete with literal "\n"
Thank you!
The newline is the "raw character" contained in the string.
How exactly you formed the string (in this case from a string literal with an escape sequence in source code) is not retained (it is only available in the source code, but not preserved in the resulting program). It would look exactly the same if you read it from a file, a database, the concatenation of multiple literals, a multi-line literal, a numeric escape sequence, etc.
If you want to print newline as \n you have to convert it back (by doing text replacement) -- but again, you don't know if the string was really created from such a literal.
You can do this with escaped characters such as \n:
let secondaryString = "really"
let s = "Hello\nMy name is \(secondaryString) Jack!"
let find = Character("\n")
let r = String(s.characters.split(find).joinWithSeparator(["\\","n"]))
print(r) // -> "Hello\nMy name is really Jack!"
However, once the string s is generated the \(secondaryString) has already been interpolated to "really" and there is no trace of it other than the replaced word. I suppose if you already know the interpolated string you could search for it and replace it with "\\(secondaryString)" to get the result you want. Otherwise it's gone.
I've got a String template looking like this:
val template = "Something %s something else %s. The first was %1$s, the second was %2$s"
works fine with Java. How do I use this reoccurring String values with Kotlin?
Looks like %1$s is not possible.
Compiler warning: unresolved reference: s
String literals in Kotlin are capable of string interpolation, and the dollar sign is the start of a string template expression. If you need the literal dollar sign in a string instead, you should escape it using a backslash: \$. So your template (which I assume you're passing to String.format) becomes:
val template = "Something %s something else %s. The first was %1\$s, the second was %2\$s"
As Alexander Udalov's answer say, $ can be used for String Templates.
Apart from use backslash to escape the char $, you also can use ${'$'} to escape it. This syntax will be more useful when you want to escape the $ in a raw string, where backslash escaping is not supported.
val template = "Something %s something else %s. The first was %1${'$'}s, the second was %2${'$'}s"
I saw the operator r#"" in Rust but I can't find what it does. It came in handy for creating JSON:
let var1 = "test1";
let json = r#"{"type": "type1", "type2": var1}"#;
println!("{}", json) // => {"type2": "type1", "type2": var1}
What's the name of the operator r#""? How do I make var1 evaluate?
I can't find what it does
It has to do with string literals and raw strings. I think it is explained pretty well in this part of the documentation, in the code block that is posted there you can see what it does:
"foo"; r"foo"; // foo
"\"foo\""; r#""foo""#; // "foo"
"foo #\"# bar";
r##"foo #"# bar"##; // foo #"# bar
"\x52"; "R"; r"R"; // R
"\\x52"; r"\x52"; // \x52
It negates the need to escape special characters inside the string.
The r character at the start of a string literal denotes a raw string literal. It's not an operator, but rather a prefix.
In a normal string literal, there are some characters that you need to escape to make them part of the string, such as " and \. The " character needs to be escaped because it would otherwise terminate the string, and the \ needs to be escaped because it is the escape character.
In raw string literals, you can put an arbitrary number of # symbols between the r and the opening ". To close the raw string literal, you must have a closing ", followed by the same number of # characters as there are at the start. With zero or more # characters, you can put literal \ characters in the string (\ characters do not have any special meaning). With one or more # characters, you can put literal " characters in the string. If you need a " followed by a sequence of # characters in the string, just use the same number of # characters plus one to delimit the string. For example: r##"foo #"# bar"## represents the string foo #"# bar. The literal doesn't stop at the quote in the middle, because it's only followed by one #, whereas the literal was started with two #.
To answer the last part of your question, there's no way to have a string literal that evaluates variables in the current scope. Some languages, such as PHP, support that, but not Rust. You should consider using the format! macro instead. Note that for JSON, you'll still need to double the braces, even in a raw string literal, because the string is interpreted by the macro.
fn main() {
let var1 = "test1";
let json = format!(r#"{{"type": "type1", "type2": {}}}"#, var1);
println!("{}", json) // => {"type2": "type1", "type2": test1}
}
If you need to generate a lot of JSON, there are many crates that will make it easier for you. In particular, with serde_json, you can define regular Rust structs or enums and have them serialized automatically to JSON.
The first time I saw this weird notation is in glium tutorials (old crate for graphics management) and is used to "encapsulate" and pass GLSL code (GL Shading language) to shaders of the GPU
https://github.com/glium/glium/blob/master/book/tuto-02-triangle.md
As far as I understand, it looks like the content of r#...# is left untouched, it is not interpreted in any way. Hence raw string.
This question already has answers here:
What is the syntax for a multiline string literal?
(5 answers)
Closed 1 year ago.
Is it possible to write something like:
fn main() {
let my_string: &str = "Testing for new lines \
might work like this?";
}
If I'm reading the language reference correctly, then it looks like that should work. The language ref states that \n etc. are supported (as common escapes, for inserting line breaks into your string), along with "additional escapes" including LF, CR, and HT.
Another way to do this is to use a raw string literal:
Raw string literals do not process any escapes. They start with the
character U+0072 (r), followed by zero or more of the character U+0023
(#) and a U+0022 (double-quote) character. The raw string body can
contain any sequence of Unicode characters and is terminated only by
another U+0022 (double-quote) character, followed by the same number
of U+0023 (#) characters that preceded the opening U+0022
(double-quote) character.
All Unicode characters contained in the raw string body represent
themselves, the characters U+0022 (double-quote) (except when followed
by at least as many U+0023 (#) characters as were used to start the
raw string literal) or U+005C (\) do not have any special meaning.
Examples for string literals:
"foo"; r"foo"; // foo
"\"foo\""; r#""foo""#; // "foo"
"foo #\"# bar";
r##"foo #"# bar"##; // foo #"# bar
"\x52"; "R"; r"R"; // R
"\\x52"; r"\x52"; // \x52
If you'd like to avoid having newline characters and extra spaces, you can use the concat! macro. It concatenates string literals at compile time.
let my_string = concat!(
"Testing for new lines ",
"might work like this?",
);
assert_eq!(my_string, "Testing for new lines might work like this?");
The accepted answer with the backslash also removes the extra spaces.
Every string is a multiline string in Rust.
But if you have indents in your text like:
fn my_func() {
const MY_CONST: &str = "\
Hi!
This is a multiline text!
";
}
you will get unnecessary spaces. To remove them you can use indoc! macros from indoc crate to remove all indents: https://github.com/dtolnay/indoc
There are two ways of writing multi-line strings in Rust that have different results. You should choose between them with care depending on what you are trying to accomplish.
Method 1: Dangling whitespace
If a string starting with " contains a literal line break, the Rust compiler will "gobble up" all whitespace between the last non-whitespace character of the line and the first non-whitespace character of the next line, and replace them with a single .
Example:
fn test() {
println!("{}", "hello
world");
}
No matter how many literal (blank space) characters (zero or a hundred) appear after hello, the output of the above will always be hello world.
Method 2: Backslash line break
This is the exact opposite. In this mode, all the whitespace before a literal \ on the first line is preserved, and all the subsequent whitespace on the next line is also preserved.
Example:
fn test() {
println!("{}", "hello \
world");
}
In this example, the output is hello world.
Additionally, as mentioned in another answer, Rust has "raw literal" strings, but they do not enter into this discussion as in Rust (unlike some other languages that need to resort to raw strings for this) supports literal line breaks in quoted content without restrictions, as we can see above.