The functions insert's example is demonstrated as:
(with-temp-buffer
(insert "hello" ?\s "world" ?\n)
(buffer-string))
What does ? mean here?
?\s and ?\n are read syntax for characters. The first is read syntax for the space character; the second for the character NEWLINE (also called Control J, and which sometimes appears as ^J, where the ^J is a newline character.
In Emacs Lisp characters are positive integers. They can be written as numerals for their ASCII codes (values) and they can be read by the Lisp reader when written that way: 32 (for \s) and 10 (for \n). But Elisp also offers a character read syntax introduced by ?, and these two characters have their own ? syntax. Other, more ordinary characters can just be prefixed by ?: ?s is read as the character s, and ?# is read as the character #.
The ? read syntax is more readable for humans than using just integers: 115 (for ?s) and 64 for ?#. But you can always use just the integers -- characters are integers.
Your example: (insert "hello" ?\s "world" ?\n). Function insert accepts strings and chars as its arguments. In this case you pass string "hello", character ?\s (space char), string "world", and character ?\n (newline char, Control J).
See the Elisp manual, node Basic Char Syntax.
Related
How do we use an emoji with a rune literal that is beyond I think
code point U+265F?
a1 := '\u2665'
this works
a2 := '\u1F3A8'
this gives error invalid character literal, more that one character.
Is there a way to represent higher positioned emojis as rune literals?
https://unicode.org/emoji/charts/full-emoji-list.html
You may use the \U sequence followed by 8 hex digits which is the hexadecimal representation of the Unicode codepoint. This is detailed in Spec: Rune literals:
There are four ways to represent the integer value as a numeric constant: \x followed by exactly two hexadecimal digits; \u followed by exactly four hexadecimal digits; \U followed by exactly eight hexadecimal digits, and a plain backslash \ followed by exactly three octal digits. In each case the value of the literal is the value represented by the digits in the corresponding base.
For example:
a1 := '\u2665'
fmt.Printf("%c\n", a1)
a2 := '\U0001F3A8'
fmt.Printf("%c\n", a2)
Which outputs (try it on the Go Playground):
♥
🎨
Note (response to #torek):
I believe the Go authors chose to require exactly 4 and 8 hex digits because this allows to use the exact same form, the exact same rune literals inside interpreted string literals. E.g. if you want a string that contains 2 runes, one having code point 0x0001F3A8 and another rune being 4, it could look like this:
s := "\U0001F3A84"
If the spec would not require exactly 8 hex digits, it would be ambiguous whether the last '4' is part of the code point or is an individual rune of the string, so you would have to break the string to a concatenation like "\U1F3A8" + "4".
Spec: String literals:
Interpreted string literals are character sequences between double quotes, as in "bar". Within the quotes, any character may appear except newline and unescaped double quote. The text between the quotes forms the value of the literal, with backslash escapes interpreted as they are in rune literals (except that \' is illegal and \" is legal), with the same restrictions. The three-digit octal (\nnn) and two-digit hexadecimal (\xnn) escapes represent individual bytes of the resulting string; all other escapes represent the (possibly multi-byte) UTF-8 encoding of individual characters. Thus inside a string literal \377 and \xFF represent a single byte of value 0xFF=255, while ÿ, \u00FF, \U000000FF and \xc3\xbf represent the two bytes 0xc3 0xbf of the UTF-8 encoding of character U+00FF.
I've been looking into this but searching seems to lead to nothing.
It might be too simple to be described, but here I am, scratching my head...
Any help would be appreciated.
Verilog knows about "strings".
A single ASCII character requires 8 bits. Thus to store 8 characters you need 64 bits:
wire [63:0] string8;
assign string8 = "12345678";
There are some gotchas:
There is no End-Of-String character (like the C null-character)
The most RHS character is in bits 7:0.
Thus string8[7:0] will hold 8h'38. ("8").
To walk through a string you have to use e.g.: string[ index +: 8];
As with all Verilog vector assignments: unused bits are set to zero thus
assign string8 = "ABCD"; // MS bit63:32 are zero
You can not use two dimensional arrays:
wire [7:0] string5 [0:4]; assign string5 = "Wrong";
You are probably mislead by a misconception about characters. There are no such thing as a character in hardware. There are only sets of bits or codes. The only thing which converts binary codes to characters is your terminal. It interprets codes in a certain way and forming letters for you to se. So, all the printfs in 'c' and $display in verilog only send the codes to the terminal (or to a file).
The thing which converts characters to the codes is your keyboard, which you also use to type in the program. The compiler then interprets your program. Verilog (as well as the 'c') compiler represents strings in double quotes (which you typed in) as a set of bytes directly. Verilog, as well as 'c' use ascii-8 encoding for such character strings, meaning that the code for 'a' is decimal 97 and 'b' is 98, .... Every character is 8-bit wide and the quoted string forms a concatenation of bytes of ascii codes.
So, answering you question, you can convert an ascii codes to characters by sending them to the terminal via $display (or other) function, using the %s modifier.
So, an example:
module A;
reg[8*5-1:0] hello;
reg[8*3 - 1: 0] bye;
initial begin
hello = "hello"; // 5 bytes of characters
bye = {8'd98, 8'd121, 8'd101}; // 3 bytes 'b' 'y' 'e'
$display("hello=%s bye=%s", hello, bye);
end
endmodule
I'd like to translate a character in Emacs to its numeric ascii code, similar to casting char a = 'a'; int i = (int)a in c. I've tried string-to-number and a few other functions, but none seem to make Emacs read the char as a number in the end.
What's the easiest way to do this?
To get the ascii-number which represents the character --as Drew said-- put a question mark before the character and evaluate that expression
?a ==> 97
Number appears in minibuffer, with C-u it's written behind expression.
Also the inverse works
(insert 97) will insert an "a" in the buffer.
BTW In some cases the character should be quoted
?\" will eval to 34
A character is a whole number in Emacs Lisp. There is no separate character data type.
Function string-to-char is built-in, and does what you want. (string-to-char "foo") is equivalent to (aref "foo" 0), which is #abo-abo's answer --- but it is coded in C.
String is an array.
(aref "foo" 0)
What is the opposite of Regexp.escape ?
> Regexp.escape('A & B')
=> "A\\ &\\ B"
> # do something, to get the next result: (something like Regexp.unescape(A\\ &\\ B))
=> "A & B"
How can I get the original value?
replaces = Hash.new { |hash,key| key } # simple trick to return key if there is no value in hash
replaces['t'] = "\t"
replaces['n'] = "\n"
replaces['r'] = "\r"
replaces['f'] = "\f"
replaces['v'] = "\v"
rx = Regexp.escape('A & B')
str = rx.gsub(/\\(.)/){ replaces[$1] }
Also make sure to #puts output in irb, because #inspect escapes characters by default.
Basically escaping/quoting looks for meta-characters, and prepends \ character (which has to be escaped for string interpretation in source code). But if we find any control character from list: \t, \n, \r, \f, \v, then quoting outputs \ character followed by this special character translated to ascii.
UPDATE:
My solution had problems with special characters (\n, \t ans so on), I updated it after investigating source code for rb_reg_quote method.
UPDATE 2:
replaces is hash, which converts escaped characters (thats why it is used in block attached to gsub) to unescaped ones. It is indexed by character without escape character (second character in sequence) and searches for unescaped value. The only defined values are control-characters, but there is also default_proc attached (block attached to Hash.new), which returns key if there is no value found in hash. So it works like this:
for "n" it returns "\n", the same for all other escaped control characters, because it is value associated with key
for "(" it returns "(", because there is no value associated with "(" key, hash calls #default_proc, which returns key itself
The only characters escaped by Regexp.escape are meta characters and control characters, so we don't have to worry about alphanumerics.
Take a look at http://ruby-doc.org/core-2.0.0/Hash.html#method-i-default_proc for documentation on #defoult_proc
You can perhaps use something like this?
def unescape(s)
eval %Q{"#{s}"}
end
puts unescape('A\\ &\\ B')
Credits to this question.
codepad demo
If you are okay with a regex solution, you can use this:
res = s.gsub(/\\(?!\\)|(\\)\\/, "\\1")
codepad demo
Try this
>> r = Regexp.escape("A & B (and * c [ e] + )")
# => "A\\ &\\ B\\ \\(and\\ \\*\\ c\\ \\[\\ e\\]\\ \\+\\ \\)"
>> r.gsub("\\(","(").gsub("\\)",")").gsub("\\[","[").gsub("\\]","]").gsub("\\{","{").gsub("\\}","}").gsub("\\.",".").gsub("\\?","?").gsub("\\+","+").gsub("\\*","*").gsub("\\ "," ")
# => "A & B (and * c [ e] + )"
Basically, these (, ), [, ], {, }, ., ?, +, * are the meta characters in regex. And also \ which is used as an escape character.
The chain of gsub() calls replace the escaped patterns with corresponding actual value.
I am sure there is a way to DRY this up.
Update: DRY version as suggested by user2503775
>> r.gsub("\\","")
Update:
following are the special characters in regex
[,],{,},(,),|,-,*,.,\\,?,+,^,$,<space>,#,\t,\f,\v,\n,\r
using a regex replace using \\(?=([\\\*\+\?\|\{\[\(\)\^\$\.\#\ ]))\
should give you the string unescaped, you would only have to replace \r\n sequences with there CrLf counterparts.
"There\ is\ a\ \?\ after\ the\ \(white\)\ car\.\ \r\n\ it\ should\ be\ http://car\.com\?\r\n"
is unescaped to :
"There is a ? after the (white) car. \r\n it should be http://car.com?\r\n"
and removing the \r\n gives you :
There is a ? after the (white) car.
it should be http://car.com?
There is a file that has control B and control C commands separating fields of text. It looks like:
"TEST\003KEY\002TEST\003KEY"
I tried to create a regex that will match this and remove it. I am not sure why this regex is not working:
"TEST\003KEY\002TEST\003KEY".gsub(/\00[23]/, ',')
Try the following:
"TEST\003KEY\002TEST\003KEY".gsub(/\002|\003/, ',')
Here it is demonstrated in irb on my machine:
$ irb
1.9.3p448 :007 > "TEST\003KEY\002TEST\003KEY".gsub(/\002|\003/, ',')
=> "TEST,KEY,TEST,KEY"
The syntax \002|\003 means "match the character literal \002 or the character literal \003". The expression given in the original question \00[23] is not valid: this is the character literal \00 (a null character) followed by the character class [23]: i.e. it matches two-character sequences.
You can also use the [[:cntrl:]] character class to match all control characters:
$ irb
1.9.3p448 :007 > "TEST\003KEY\002TEST\003KEY\005TEST".gsub(/[[:cntrl:]]/, ',')
=> "TEST,KEY,TEST,KEY,TEST"
Here's the deal. First and foremost, computers cannot store characters--they can only store numbers. So when a computer stores a string it converts every character to a number. The numbers for all the basic characters are given by an ascii chart(you can search google for one).
When you tell a computer to print a string, it retrieves the numbers saved for the string and outputs them as characters (using an ascii chart to convert the numbers to characters).
Double quoted strings can contain what are called escape sequences. The most common escape sequence is "\n":
puts "hello\nworld"
--output:--
hello
world
A double quoted string converts the escape sequence "\n" to the ascii code 10:
puts "\n".ord #=>10 (ord() will show you the ascii code for a character)
A double quoted string can also contain escape sequences of the form \ddd, e.g. \002. Escape sequences like that are called octal escape sequences, which means 002 is the octal representation of an ascii code.
In an octal number, the right most digit is the 1's column, and the next digit to the left is the 8's column and the next digit to the left is the 64's column. For instance, this octal number:
\123
is equivalent to 3*1 + 2*8 + 1*64 = 83. It so happens that an "S" has the ascii code 83:
puts "\123" #=>S
Because you also can use octal escape sequences in a double quoted string, that means that instead of using the escape sequence "\n" you could use the octal escape "\012" (2*1 + 1*8 + 0*64 = 10). A double quoted string converts the octal escape sequence "\012" to the ascii code 10, which is the same thing that a double quoted string does to "\n". Here is an example:
puts "hello" + "\012" + "world"
--output:--
hello
world
The final thing to note about octal escape sequences is that you can optionally leave off any leading 0's:
puts "hello" + "\12" + "world"
--output:--
hello
world
Okay, now examine your string:
"TEST\003KEY\002TEST\003KEY"
You can see that it contains three octal escape sequences. A double quoted string converts the octal escape sequence \003 to the ascii code: 3*1 + 0*8 + 0*64 = 3. If you check an ascii chart, the ascii code 3 represents a character called "end of text". A double quoted string converts the octal escape sequence \002 to the ascii code: 2*1 + 0*8 + 0*64 = 2, which represents a character called 'start of text'. I'm not sure where you are getting the "control B" and "control C" names from (maybe those are the key strokes on your keyboard that are mapped to those characters?).
Next, a regex acts like a double quoted string, so
/<in here>/
you can use the same escape sequences as in a double quoted string, and the regex will convert the escape sequences to ascii codes.
Now, in light of all the above, examine your regex:
/\00[23]/
As Richard Cook pointed out, your regex gets interpreted as the octal escape sequence \00 followed by the character class [23]. The octal escape sequence \00 gets converted to the ascii code: 0*1 + 0*8 = 0. And if you look at an ascii chart, the number 0 represents a character called 'null'. So your regex is looking for a null character, followed by either a "2" or a "3", which means your regex is looking for a two character string. But a two character string will never match the octal escape sequence "\003" (or "\002"), which represents only one character.
The main thing to take away from all this is that when you see a string that contains an octal escape sequence:
"hello\012world"
...that string does not contain the characters \, 0, 1, and 2. A double quoted string converts that sequence of characters into one ascii code, which represents ONE character. You can prove that very easily:
puts "hello".length #=>5
puts "hello\012".length #=>6
There are also many other types of escape sequences that can appear in double quoted strings. You would think they would be listed in the String class docs, but they are not.
s = "TEST\003KEY\002TEST\003KEY"
s.split(/[[:cntrl:]]/) * ","
# => "TEST,KEY,TEST,KEY"