How to convert hexadecimal to decimal in a substitution statement? - vim

I have a line like this:
pad (2) = 0x0041
I wanna change the hex into decimal and the expected result is
pad (2) = 65
I just tried :%s/\(.*\) = \(.*\)/\1 = \=printf("%d", submatch(2)), but it failed.
Would you help to solve this?

Vim has a str2nr() function to convert different number representations to their decimal values. To convert hex values you could use it like this:
s/0x[0-9a-fA-F]\+/\=str2nr(submatch(0), 16)

Your code is almost ok, but, according to the documentation:
When the substitute string starts with "\=" the remainder is interpreted as an
expression. This does not work recursively: a substitute() function inside
the expression cannot use "\=" for the substitute string.
So, you may change your code to
%s/\(.*\) = \(.*\)/\=submatch(1)." = ".printf("%d", submatch(2))

Related

Padding a hexadecimal string with zeros to a 6 character length

I have this:
function dec2hex(IN)
local OUT
OUT = string.format("%x",IN)
return OUT
end
and need IN to have padded zeros to string length of 6.
I can't use String.Utils or PadLeft. It's within an app called Watchmaker which uses a cut down version of Lua.
String formats in Lua work mostly just like in C. So to pad a number with zeros, just use %0n where n is the number of places. For example
print(string.format("%06x", 16^4-1))
will print 00ffff.
See chapter 20 The String Library of “Programming in Lua”, the reference of string.format, and the C reference for the printf family of functions for details.
If you store your format string locally you can call the format method on to the format string and the example of #Henri results in ("%06x"):format(0xffff)
print(("%06x"):format(0xffff)) -- Prints `00ffff`
You can write numbers in hex format. It is the same as C.

How can I convert a character code to a string character in Lua?

How can I convert a character code to a string character in Lua?
E.g.
d = 48
-- this is what I want
str_d = "0"
You are looking for string.char:
string.char (···)
Receives zero or more integers. Returns a string with length equal to the number of arguments, in which each character has the internal numerical code equal to its corresponding argument.
Note that numerical codes are not necessarily portable across platforms.
For your example:
local d = 48
local str_d = string.char(d) -- str_d == "0"
For ASCII characters, you can use string.char.
For UTF-8 strings, you can use utf8.char(introduced in Lua 5.3) to get a character from its code point.
print(utf8.char(48)) -- 0
print(utf8.char(29790)) -- 瑞

AS3 - "\u2605" NOT the same as "\\u"+"2605"?

Trying to make a textfield where people write the unicode without the backslash. I want to add the backslash after they typed it. So the user types u2605 and the code converts it to "\u2605", i then convert this to a unicode character and insert it in textflow.
My code:
this works:
span.text = publicFunctions.htmlUnescape(he.encode("\u2605"))
this doesn't work:
span.text = publicFunctions.htmlUnescape(he.encode("\\u"+"2605"))
how to make a string that acts as a unicode string?
Tried all sorts of things, escape(unescape()), convert to number, "\u", "\u" ... nothing helps.
trace("\u2605" == "\u"+"2605") ... will return false. So will
trace("\u2605" == "\u"+"2605")
"\u2605" is a string with a single character, the character with the code point 2605, while "\\u" + "2605" is a string with 6 characters (the backslash, the u and the four digit number).
If you want to construct a unicode character from just the four digits, you should be able to use String.fromCharCode. The thing is just that the escape sequence uses a hexadecimal number, while the method obviously takes a decimal number. So if the user enters a hexadecimal string, you will have to convert that first:
trace(String.fromCharCode(parseInt('2605', 16)) == '\u2605'));
That's an interesting issue! I don't think you can concatenate a string literal and achieve what you're trying to do. The relevant character escaping happens when the string literal is originally formed, which means that you need the whole sequence together in the first place.
But you should be able to take the user-supplied number and dynamically generate a Unicode string with String.fromCharCode(...).
http://help.adobe.com/en_US/FlashPlatform/reference/actionscript/3/String.html#fromCharCode()

Multiline string literal in Matlab?

Is there a multiline string literal syntax in Matlab or is it necessary to concatenate multiple lines?
I found the verbatim package, but it only works in an m-file or function and not interactively within editor cells.
EDIT: I am particularly after readbility and ease of modifying the literal in the code (imagine it contains indented blocks of different levels) - it is easy to make multiline strings, but I am looking for the most convenient sytax for doing that.
So far I have
t = {...
'abc'...
'def'};
t = cellfun(#(x) [x sprintf('\n')],t,'Unif',false);
t = horzcat(t{:});
which gives size(t) = 1 8, but is obviously a bit of a mess.
EDIT 2: Basically verbatim does what I want except it doesn't work in Editor cells, but maybe my best bet is to update it so it does. I think it should be possible to get current open file and cursor position from the java interface to the Editor. The problem would be if there were multiple verbatim calls in the same cell how would you distinguish between them.
I'd go for:
multiline = sprintf([ ...
'Line 1\n'...
'Line 2\n'...
]);
Matlab is an oddball in that escape processing in strings is a function of the printf family of functions instead of the string literal syntax. And no multiline literals. Oh well.
I've ended up doing two things. First, make CR() and LF() functions that just return processed \r and \n respectively, so you can use them as pseudo-literals in your code. I prefer doing this way rather than sending entire strings through sprintf(), because there might be other backslashes in there you didn't want processed as escape sequences (e.g. if some of your strings came from function arguments or input read from elsewhere).
function out = CR()
out = char(13); % # sprintf('\r')
function out = LF()
out = char(10); % # sprintf('\n');
Second, make a join(glue, strs) function that works like Perl's join or the cellfun/horzcat code in your example, but without the final trailing separator.
function out = join(glue, strs)
strs = strs(:)';
strs(2,:) = {glue};
strs = strs(:)';
strs(end) = [];
out = cat(2, strs{:});
And then use it with cell literals like you do.
str = join(LF, {
'abc'
'defghi'
'jklm'
});
You don't need the "..." ellipses in cell literals like this; omitting them does a vertical vector construction, and it's fine if the rows have different lengths of char strings because they're each getting stuck inside a cell. That alone should save you some typing.
Bit of an old thread but I got this
multiline = join([
"Line 1"
"Line 2"
], newline)
I think if makes things pretty easy but obviously it depends on what one is looking for :)

transform string/char to uint8

Why does the expression:
test = cast(strtrim('3'), 'uint8')
produce 51?
This is also true for:
test = cast(strtrim('3'), 'int8')
Thanks.
Because 51 is the ASCII code for the character '3'.
If you want to transform the string to numeric 3, you should use
uint8(str2double('3'))
Note that str2double will ignore trailing spaces, so that strtrim isn't necessary.
EDIT
When a string is used in an numeric operation, Matlab automatically converts it to its ASCII value. For example
>> '1'+1
ans =
50
Because 51 is the ASCII value for the character '3'.
This is because '3' is seen as an ASCII character to matlab. By casting as a signed or unsigned integer (8 bits in this case) you are asking Matlab to convert an ASCII '3' to a decimal number. In this case the decimal number is 51. If you want to look at more conversions here is a basic document.

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