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.
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 am writing a function that takes stringA and stringB as parameters and compares the first character of stringB with the last character of StringA. If they are equal, then the function returns true, else false is returned.
I have nearly the whole function ready, however I can't find a way to take the last character of stringA because its length is unknown. I checked the documentation and I found nothing. Any suggestions?
(cond
[(string=? (substring stringA ???) (substring stringB 0 2))"True"]
[else "False"])
You can get the last character position of a string using string-length (or rather one less than):
(string-ref str (sub1 (string-length str)))
Note that a character is different from a string of length 1. Thus the correct way to extract a character is with string-ref or the like, rather than substring.
It seems Chris answered your question. Just a reminder, to use the string-ref, which returns a character, you should use the comparison function char=? (or equal?).
I'd like to add another solution which I find more elaborate, but requires to download a collection from the planet racket (after installing package collections). Using the collections package, you can use the same function with any collection rather then just strings, using the (last ..) and (first ..) functions of the module.
(require data/collection)
(let ([stringA "abcd"]
[stringB "dcba"])
(cond
[(equal? (last stringA)
(first stringB)) "True"]
[else "False"]))
You could also use the SRFI-13 function string-take-right, which returns the last n characters of the argument string as a string.
every language has a length function for a string. in Racket I found this :
https://docs.racket-lang.org/reference/strings.html#%28def.%28%28quote.~23~25kernel%29._string-length%29%29
there is this : string-length str
so just run that it will give you the length and then you can extract the last character
I need to convert an int to its equivalent char using the Char.chr-function, but why does the function return every char in the form of #"\^A" instead of just #"A" (that's how I want it to be).
What you see there is just the way control characters (ASCII code 0-31) are pretty-printed by the interactive toplevel. For example, #"\^A" is equivalent to #"\001". The SML system presumably uses its own Char.toString function to print values of type char. Try chr 65, which should be printed as #"A".
Imagine I have String in C#: "I Don’t see ya.."
I want to remove (replace to nothing or etc.) these "’" symbols.
How do I do this?
That 'junk' looks a lot like someone interpreted UTF-8 data as ISO 8859-1 or Windows-1252, probably repeatedly.
’ is the sequence C3 A2, E2 82 AC, E2 84 A2.
UTF-8 C3 A2 = U+00E2 = â
UTF-8 E2 82 AC = U+20AC = €
UTF-8 E2 84 A2 = U+2122 = ™
We then do it again: in Windows 1252 this sequence is E2 80 99, so the character should have been U+2019, RIGHT SINGLE QUOTATION MARK (’)
You could make multiple passes with byte arrays, Encoding.UTF8 and Encoding.GetEncoding(1252) to correctly turn the junk back into what was originally entered. You will need to check your processing to find the two places that UTF-8 data was incorrectly interpreted as Windows-1252.
"I Don’t see ya..".Replace( "’", string.Empty);
How did that junk get in there the first place? That's the real question.
By removing any non-latin character you'll be intentionally breaking some internationalization support.
Don't forget the poor guy who's name has a "â" in it.
This looks disturbingly familiar to a character encoding issue dealing with the Windows character set being stored in a database using the standard character encoding. I see someone voted Will down, but he has a point. You may be solving the immediate issue, but the combinations of characters are limitless if this is the issue.
If you really have to do this, regular expressions are probably the best solution.
I would strongly recommend that you think about why you have to do this, though - at least some of the characters your listing as undesirable are perfectly valid and useful in other languages, and just filtering them out will most likely annoy at least some of your international users. As a swede, I can't emphasize enough how much I hate systems that can't handle our å, ä and ö characters correctly.
Consider Regex.Replace(your_string, regex, "") - that's what I use.
Test each character in turn to see if it is a valid alphabetic or numeric character and if not then remove it from the string. The character test is very simple, just use...
char.IsLetterOrDigit;
Please there are various others such as...
char.IsSymbol;
char.IsControl;
Regex.Replace("The string", "[^a-zA-Z ]","");
That's how you'd do it in C#, although that regular expression ([^a-zA-Z ]) should work in most languages.
[Edited: forgot the space in the regex]
The ASCII / Integer code for these characters would be out of the normal alphabetic Ranges. Seek and replace with empty characters. String has a Replace method I believe.
Either use a blacklist of stuff you do not want, or preferably a white list (set). With a white list you iterate over the string and only copy the letters that are in your white list to the result string. You said remove, and the way you do that is having two pointers one you read from (R) and one you write to (W):
I Donââ‚
W R
if comma is in your whitelist then you would in this case read the comma and write it where à is then advance both pointers. UTF-8 is a multi-byte encoding, so you advancing the pointer may not just be adding to the address.
With C an easy to way to get a white list by using one of the predefined functions (or macros): isalnum, isalpha, isascii, isblank, iscntrl, isdigit, isgraph, islower, isprint, ispunct, isspace, isupper, isxdigit. In this case you send up with a white list function instead of a set of course.
Usually when I see data like you have I look for memory corruption, or evidence to suggest that the encoding I expect is different than the one the data was entered with.
/Allan
I had the same problem with extraneous junk thrown in by adobe in an EXIF dump. I spent an hour looking for a straight answer and trying numerous half-baked suggestions which did not work here.
This thread more than most I have read was replete with deep, probing questions like 'how did it get there?', 'what if somebody has this character in their name?', 'are you sure you want to break internationalization?'.
There were some impressive displays of erudition positing how this junk could have gotten here and explaining the evolution of the various character encoding schemes. The person wanted to know how to remove it, not how it came to be or what the standards orgs are up to, interesting as this trivia may be.
I wrote a tiny program which gave me the right answer. Instead of paraphrasing the main concept, here is the entire, self-contained, working (at least on my system) program and the output I used to nuke the junk:
#!/usr/local/bin/perl -w
# This runs in a dos window and shows the char, integer and hex values
# for the weird chars. Install the HEX values in the REGEXP below until
# the final test line looks normal.
$str = 's: “Brian'; # Nuke the 3 werid chars in front of Brian.
#str = split(//, $str);
printf("len str '$str' = %d, scalar \#str = %d\n",
length $str, scalar #str);
$ii = -1;
foreach $c (#str) {
$ii++;
printf("$ii) char '$c', ord=%03d, hex='%s'\n",
ord($c), unpack("H*", $c));
}
# Take the hex characters shown above, plug them into the below regexp
# until the junk disappears!
($s2 = $str) =~ s/[\xE2\x80\x9C]//g; # << Insert HEX values HERE
print("S2=>$s2<\n"); # Final test
Result:
M:\new\6s-2014.1031-nef.halloween>nuke_junk.pl
len str 's: GÇ£Brian' = 11, scalar #str = 11
0) char 's', ord=115, hex='73'
1) char ':', ord=058, hex='3a'
2) char ' ', ord=032, hex='20'
3) char 'G', ord=226, hex='e2'
4) char 'Ç', ord=128, hex='80'
5) char '£', ord=156, hex='9c'
6) char 'B', ord=066, hex='42'
7) char 'r', ord=114, hex='72'
8) char 'i', ord=105, hex='69'
9) char 'a', ord=097, hex='61'
10) char 'n', ord=110, hex='6e'
S2=>s: Brian<
It's NORMAL!!!
One other actionable, working suggestion I ran across:
iconv -c -t ASCII < 6s-2014.1031-238246.halloween.exf.dif > exf.ascii.dif
If String having the any Junk date , This is good to way remove those junk date
string InputString = "This is grate kingdom¢Ã‚¬â";
string replace = "’";
string OutputString= Regex.Replace(InputString, replace, "");
//OutputString having the following result
It's working good to me , thanks for looking this review.