I would like to know how to put a line break in below HDF5 attribute string.
DescType_id = H5T.copy ('H5T_C_S1');
H5T.set_size (DescType_id, numel(description));
H5T.set_strpad(DescType_id,'H5T_STR_NULLTERM');
DescAttr_id = H5A.create (g2id, 'description', DescType_id, ...
STimeSpace, 'H5P_DEFAULT');
H5A.write (DescAttr_id, DescType_id, description);
H5T.close(DescType_id);
H5A.close(DescAttr_id);
My description variable would be:
description="Experiment:\nID: 1234\nLocation: London"
I am expecting the line break character to be something like '\n' in above code. Expected Output:
Group '/'
Group '/G1'
Attributes:
'description': 'Experiment:
ID: 1234
Location: London'
Your help is greatly appreciated. Thanks.
You should set description to the correct string:
description = sprintf('Experiment:\nID: 1234\nLocation: London');
One should note that the usual escape sequences are actually interpreted by the I/O functions (i.e. fprintf, sprintf) and not by the MATLAB language parser. That means, for example, that the literal '\n' (in MATLAB) is a char array of two chars, a backslash and an n, while in C the literal "\n" is a const char array of two chars, one being a new-line, and the other being the string terminator NUL.
A non-portable but "equivalent" way of writing the above description would be concatenating the sub-strings with the line separators:
description = ['Experiment:' char(10) 'ID: 1234' char(10) 'Location: London'];
where char(10) is the char that has the UTF-8 code 10, which happens to be the new-line character.
Related
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 wanted a String containing a text \1.
What I did was (the real string was longer but it's not important):
'''
\1
'''
Which resulted in a String containing a unicode 0x1 codepoint.
I think what I should've done is just escape the backslash like this:
'''
\\1
'''
What I don't understand is why Groovy didn't report an error here. I thought unicode escapes are supposed to look like \u1?
Instead of a syntax error I got a runtime exception when I tried to put this String into an XML element:
An invalid XML character (Unicode: 0x1) was found in the element content of the document.
The \ (backward slash) symbol is an escape symbol. If you mean to use it literally, you must escape it itself: \\.
When you escape any character, the character is interpreted to have special meaning. In the case of the \1 sequence, it just happens that this can be interpreted as the 0x01 codepoint.
This is the same in Java Strings.
If you want to not have to escape characters in Groovy, use slashy strings:
def x = /\1/
assert x == "\\1"
which also works as multiline:
def x = /
\1
/
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.
So say I have a string with some underscores like hi_there.
Is there a way to auto-convert that string into "hi there"?
(the original string, by the way, is a variable name that I'm converting into a plot title).
Surprising that no-one has yet mentioned strrep:
>> strrep('string_with_underscores', '_', ' ')
ans =
string with underscores
which should be the official way to do a simple string replacements. For such a simple case, regexprep is overkill: yes, they are Swiss-knifes that can do everything possible, but they come with a long manual. String indexing shown by AndreasH only works for replacing single characters, it cannot do this:
>> s = 'string*-*with*-*funny*-*separators';
>> strrep(s, '*-*', ' ')
ans =
string with funny separators
>> s(s=='*-*') = ' '
Error using ==
Matrix dimensions must agree.
As a bonus, it also works for cell-arrays with strings:
>> strrep({'This_is_a','cell_array_with','strings_with','underscores'},'_',' ')
ans =
'This is a' 'cell array with' 'strings with' 'underscores'
Try this Matlab code for a string variable 's'
s(s=='_') = ' ';
If you ever have to do anything more complicated, say doing a replacement of multiple variable length strings,
s(s == '_') = ' ' will be a huge pain. If your replacement needs ever get more complicated consider using regexprep:
>> regexprep({'hi_there', 'hey_there'}, '_', ' ')
ans =
'hi there' 'hey there'
That being said, in your case #AndreasH.'s solution is the most appropriate and regexprep is overkill.
A more interesting question is why you are passing variables around as strings?
regexprep() may be what you're looking for and is a handy function in general.
regexprep('hi_there','_',' ')
Will take the first argument string, and replace instances of the second argument with the third. In this case it replaces all underscores with a space.
In Matlab strings are vectors, so performing simple string manipulations can be achieved using standard operators e.g. replacing _ with whitespace.
text = 'variable_name';
text(text=='_') = ' '; //replace all occurrences of underscore with whitespace
=> text = variable name
I know this was already answered, however, in my case I was looking for a way to correct plot titles so that I could include a filename (which could have underscores). So, I wanted to print them with the underscores NOT displaying with as subscripts. So, using this great info above, and rather than a space, I escaped the subscript in the substitution.
For example:
% Have the user select a file:
[infile inpath]=uigetfile('*.txt','Get some text file');
figure
% this is a problem for filenames with underscores
title(infile)
% this correctly displays filenames with underscores
title(strrep(infile,'_','\_'))
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.