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.
Related
I have the byte representation of a character in a string, let's say the character is 'H', which has the byte value of 72. My string is therefore "72".
How do I go about converting this string ("72") into its corresponding character value ('H') based on the byte value (72) represented in my string using python 3.6?
Psuedo code:
str = 72
print(decode_as_byte_value(str))
Expected result:
H
ord('H')
chr(72)
Its as simple as that. Remember that chr() only takes int and ord() only takes str
Please do not use this community for such Syntex based questions.
Still your ans is:
# Get the ASCII number of a character
number = ord(char)
# Get the character given by an ASCII number
char = chr(number)
If this is your answer tick mark this response.
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?
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
I have the following piece of code which converts 1 char to a hex at a time. I want to convert two chars to a hex. ie 99ab should be treated as '99', 'ab' to be converted to its equivalent hex.
Current implementation is as follows
$final =~ s/(.)/sprintf("0x%X ",ord($1))/eg;
chop($final);
TIA
Your question doesn't make much sense. Hex is a string representation of a number. You can't convert a string to hex.
You can convert individual characters of a string to hex since characters are merely numbers, but that's clearly not what you want. (That's what your code does.)
I think you are trying to convert from from hex to chars.
6 chars "6a6b0a" ⇒ 3 chars "\x6a\x6b\x0a"
If so, you can use your choice of
$final =~ s/(..)/ chr(hex($1)) /seg;
or
$final = pack 'H*', $final;
The other possibility I can think of is that you want to unpack 16-bit integers.
6 chars "6a6b" ⇒ 13 chars "0x6136 0x6236" (LE byte order)
-or-
6 chars "6a6b" ⇒ 13 chars "0x3661 0x3662" (BE byte order)
If so, you can use
my #nums = unpack 'S<*', $packed; # For 16-bit ints, LE byte order
-or-
my #nums = unpack 'S>*', $packed; # For 16-bit ints, BE byte order
my $final = join ' ', map sprintf('0x%04X', $_), #nums;
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))