I'm trying to create a string from hex values in an array, but whenever a hex in the array starts with a zero it disappears in the resulting string as well.
I use String(value:radix:uppercase) to create the string.
An example:
Here's an array: [0x13245678, 0x12345678, 0x12345678, 0x12345678].
Which gives me the string: 12345678123456781234567812345678 (32 characters)
But the following array: [0x02345678, 0x12345678, 0x02345678, 0x12345678] (notice that I replaced two 1's with zeroes).
Gives me the string: 234567812345678234567812345678 (30 characters)
I'm not sure why it removes the zeroes. I know the value is correct; how can I format it to keep the zero if it was there?
The number 0x01234567 is really just 0x1234567. Leading zeros in number literals don't mean anything (unless you are using the leading 0 for octal number literals).
Instead of using String(value:radix:uppercase), use String(format:).
let num = 0x1234567
let str = String(format: "%08X", num)
Explanation of the format:
The 0 means to pad the left end of the string with zeros as needed.
The 8 means you want the result to be 8 characters long
The X means you want the number converted to uppercase hex. Use x if you want lowercase hex.
Related
I have a string of format
"'Year'-'Month'-'Day'T'Hour':'Minute':'Second'Z" for example '2020-11-26T16:56:09.676Z'
(Note the milliseconds are considered part of the second)
I would like to convert it to the format:
t1 = 1x6
2020 11 26 16 56 09.676
Or in other words a 1x6 array.
Note: This is to be completed using MatLab.
You can
Use regexp with the 'match' input flag to detect the numbers as one or more digits (\d+) optionally followed by a decimal point (\.?) and then zero or more digits (\d*). This will give a cell array of strings.
Apply str2double to convert the strings to numbers. This will give a numeric row vector.
s = '2020-11-26T16:56:09.676Z';
result = str2double(regexp(s, '\d+\.?\d*', 'match'));
I am working with Swift's Decimal type, trying to ensure that an user-entered String is a valid Decimal.
I have two String values, each including a letter, within my Playground file. One of the values contains a letter at the start, while the other contains a letter at the end. I initialize a Decimal using each value, and only one Decimal initialization fails; the Decimal initialized with the value that contains the letter at the beginning.
Why does the Decimal initialized with a value that contains a letter at the end return a valid Decimal? I expect nil to be returned.
Attached is a screenshot from my Playground file.
It works this way because Decimal accepts any number values before the letters. The letters act as a terminator for any numbers that comes after it. So in your example:
12a = 12 ( a is the terminator in position 3 )
a12 = nil ( a is the terminator in position 1 )
If wanting both to be invalid if the string contains a letter then you could use Float instead.
I have a string and I need two characters to be returned.
I tried with strsplit but the delimiter must be a string and I don't have any delimiters in my string. Instead, I always want to get the second number in my string. The number is always 2 digits.
Example: 001a02.jpg I use the fileparts function to delete the extension of the image (jpg), so I get this string: 001a02
The expected return value is 02
Another example: 001A43a . Return values: 43
Another one: 002A12. Return values: 12
All the filenames are in a matrix 1002x1. Maybe I can use textscan but in the second example, it gives "43a" as a result.
(Just so this question doesn't remain unanswered, here's a possible approach: )
One way to go about this uses splitting with regular expressions (MATLAB's strsplit which you mentioned):
str = '001a02.jpg';
C = strsplit(str,'[a-zA-Z.]','DelimiterType','RegularExpression');
Results in:
C =
'001' '02' ''
In older versions of MATLAB, before strsplit was introduced, similar functionality was achieved using regexp(...,'split').
If you want to learn more about regular expressions (abbreviated as "regex" or "regexp"), there are many online resources (JGI..)
In your case, if you only need to take the 5th and 6th characters from the string you could use:
D = str(5:6);
... and if you want to convert those into numbers you could use:
E = str2double(str(5:6));
If your number is always at a certain position in the string, you can simply index this position.
In the examples you gave, the number is always the 5th and 6th characters in the string.
filename = '002A12';
num = str2num(filename(5:6));
Otherwise, if the formating is more complex, you may want to use a regular expression. There is a similar question matlab - extracting numbers from (odd) string. Modifying the code found there you can do the following
all_num = regexp(filename, '\d+', 'match'); %Find all numbers in the filename
num = str2num(all_num{2}) %Convert second number from str
i donot know why there is error in this coding:
hex_str1 = '5'
bin_str1 = dec2bin(hex2dec(hex_str1))
hex_str2 = '4'
bin_str2 = dec2bin(hex2dec(hex_str2))
c=xor(bin_str1,bin_str2)
the value of c is not correct when i transform the hex to binary by using the xor function.but when i used the array the value of c is correct.the coding is
e=[1 1 1 0];
f=[1 0 1 0];
g=xor(e,f)
what are the mistake in my first coding to xor of hec to binary value??anyone can help me find the solution...
Your mistake is applying xor on two strings instead of actual numerical arrays.
For the xor command, logical "0"s are represented by actual zero elements. Any non-zero elements are interpreted as logical "1"s.
When you apply xor on two strings, the numerical value of each character (element) is its ASCII value. From xor's point of view, the zeroes in your string are not really zeroes, but simply non-zero values (being equal to the ASCII value of the character '0'), which are interpreted as logical "1"s. The bottom line is that in your example you're xor-ing 111b and 111b, and so the result is 0.
The solution is to convert your strings to logical arrays:
num1 = (bin_str1 == '1');
num2 = (bin_str2 == '1');
c = xor(num1, num2);
To convert the result back into a string (of a binary number), use this:
bin_str3 = sprintf('%d', c);
... and to a hexadecimal string, add this:
hex_str3 = dec2hex(bin2dec(bin_str3));
it is really helpful, and give me the correct conversion while forming HMAC value in matlab...
but in matlab you can not convert string of length more than 52 character using bin2dec() function and similarly hex2dec() can not take hexadecimal character string more than 13 length.
Is there a format string to truncate a number to a specific number of digits?
For example, any number greater than 5 digits i would like to truncate to 3 digits.
132456 -> 132
5000000 -> 500
#Erik : Format specifiers like %2d are specific to a language? I actually want to use it in javascript
Pseudo-Code
Function returning a String, receiving a String representing a Number as a parameter
IF the String has more than 5 characters
RETURN a substring containing the first 3 characters.
ELSE
RETURN the string received as a parameter
END IF
END Function
I assume you refer to printf format strings. I couldn't find anything that will truncate an integer argument (i.e. %d). But you can specify the maximum length of a string by referring to a string format string and specifying lengths via "%<MinLength>.<MaxLength>s".
So in your case you could turn your number arguments into strings and then use "%3.3s".