I have a series of octal values I'd like to output in their ASCII character equivalents. CHR does not seem to recognize an octal character when it sees it. Is there any straightforward way of doing this with ColdFusion?
Use InputBaseN to convert from Octal to Decimal, then you can use Chr to output the character.
For example:
Chr(InputBaseN( 101 , 8 )) => A
To go back the other way, you can reverse the process with Asc and FormatBaseN:
FormatBaseN( Asc('A') , 8 ) => 101
Related
I pipe different values to bc.
If the value is a number, it works fine. If it's a string with lowercase letters, it returns 0 which makes sense to me, but if it's uppercase letters, bc converts it to 9 as the length of the input characters:
echo 1 | bc
1
echo aaa | bc
0
echo AAA | bc
999
echo FO | bc
99
echo null | bc
0
echo NULL | bc
9999
Why does bc have this behavior? What's the best way to work with unexpected string values?
According to https://www.gnu.org/software/bc/manual/html_mono/bc.html
(emphasis by me):
A simple expression is just a constant. bc converts constants into internal decimal numbers using the current input base, specified by the variable ibase. (There is an exception in functions.) The legal values of ibase are 2 through 16. Assigning a value outside this range to ibase will result in a value of 2 or 16. Input numbers may contain the characters 0-9 and A-F. (Note: They must be capitals. Lower case letters are variable names.) Single digit numbers always have the value of the digit regardless of the value of ibase. (i.e. A = 10.) For multi-digit numbers, bc changes all input digits greater or equal to ibase to the value of ibase-1. This makes the number FFF always be the largest 3 digit number of the input base.
So, assuming that your ibase is 10 your observation is explained.
It is unrelated to "unexected string values" or "the length of the input characters". bc does consider them (somewhat odd) attempts to provide numeric values and converts and uses them according to the quoted rule.
I would like to find all the number combinaitions without having 3 zero's in between.
There might be some delimiters (max 2 characters) in between the numbers.
I'm using python and I would like to perform this search with the regex.
Accepted numbers
This is number 1234 which should be accepted.
12-45
1 2 0 0 3 4 5
not accepted numbers:
1
12
123
1000
1000-2000
30000-31000
21 000-32 000-50 000
21 00 03 00 00
The regex with which I could come up is:
([\s\-]{0,2}\d(?!000)){4,}
My regex can find all the accepted numbers but it doesn't filter out all the excepted numbers.
See the results in regex
Actually this regex is used in python to remove the matched numbers from the text:
See python code
p.s. Delimiters are not only space but should be at least \s and dash.
p.s.s. The numbers might be in the middle of the string. So I think I cannot use ^ and $ in my regex.
You could assert not 3 zeroes in a row while matching optional delimiters in between.
\b(?![\d\s-]*?0(?:[\s-]*0){2})\d(?:[\s-]*\d){3,}\b
Explanation
\b A word boundary
(?! Negative lookahead, assert what is at the right is not
[\d\s-]*? Match any of a digit, whitespace char or - as least as possible
0(?:[\s-]*0){2} - ) Match a zere followed by 2 times a zero with optional delimiters in between
\d Match a digit
(?:[\s-]*\d){3,} Repeat 3 or more times matching a digit with optional delimiters in between
\b A word boundary
Regex demo
I am confused about the following code snippet:
#!/bin/bash
H=$(date +%H);
if (( 10#$H > 5 ))
then
# do something
else
# do something else
fi
What does the (( 10#$H > 5 )) mean in above code snippet?
The 10#$H means to expand the number using base 10.
This is probably done to remove any leading zeros from the date due to the fact that bash will interpret the number in base 8 (octal).
Example:
$ echo "$(( 08 < 5 ))"
bash: 08: value too great for base (error token is "08")
ARITHMETIC EVALUATION: Constants with a leading 0 are interpreted as octal numbers. A leading 0x or 0X denotes hexadecimal. Otherwise, numbers take the form [base#]n, where base is a decimal number between 2 and 64 representing
the arithmetic base and n is a number in that base. If base# is omitted, then base 10 is used. The digits greater than 9 are represented by the lowercase letters, the uppercase letters, #, and _, in that order. If base is less than or equal to 36, lowercase and uppercase letters may be used interchangeably to represent numbers between 10 and 35.
source: man bash
Now I use this code to read data from standard input:
print =: 1!:2&2
read =: 1!:1[3
in =. (read-.LR)-.CR
But it returns just a sequence of numbers, e.g. input:
2
3
4
5
Output:
2345
Number of numbers is unknown, but each is in the separate line
When reading with (1!:1) you read a stream of characters. You have to manipulate the stream to get your desired input.
For example. If you want to enter a list of line separated integers, you would read the list, then split it by LF, remove LF and then convert to integer. You can achieve the first two steps using cut (;._2) and the conversion using do (".):
in =: ".;._2 (1!:1) 3
If you want to enter a list of space separated integers, you would just use do, the splitting would be implied by the spaces:
in =: ". LF -.~ (1!:1) 3
trailing LF (if present) has to be removed before applying ". because do can't convert special characters.
After a copy-paste from Wikipedia into Vim, I get this:
1 A
2
3 [+] Métier agricole<200e> – 44 P • 2 C
4 [×] Métier de l'ameublement<200e> – 10 P
5 [×] Métier de l'animation<200e> – 5 P
6 [+] Métier en rapport avec l'art<200e> – 11 P • 4 C
7 [×] Métier en rapport avec l'automobile<200e> – 10 P
8 [×] Métier de l'aéronautique<200e> – 15 P
The problem is that <200e> is only a char.
I'd like to know how to put it in a search/replace (via the / or :).
Check the help for \%u:
/\%d /\%x /\%o /\%u /\%U E678
\%d123 Matches the character specified with a decimal number. Must be
followed by a non-digit.
\%o40 Matches the character specified with an octal number up to 0377.
Numbers below 040 must be followed by a non-octal digit or a non-digit.
\%x2a Matches the character specified with up to two hexadecimal characters.
\%u20AC Matches the character specified with up to four hexadecimal
characters.
\%U1234abcd Matches the character specified with up to eight hexadecimal
characters.
These are sequences you can use. Looks like you have two bytes, so \%u200e
should match it. Anyway, it's pretty strange. 20 in UTF-8 / ASCII is the space
character, and 0e is ^N. Check your encoding settings.
replace ^#
:%s/\%x00//g
replace ^L
// Enter the ^L using ctrl-V ctrl-L
:%s/^L//g
refers:
gvim - How to remove this symbol "^#" with vim? - Super User
vim - Deleting form feed ^L characters - Stack Overflow
If you want to quickly select this extraneous character everywhere and replace it / get rid of it, you could:
isolate one of the strange characters by adding a space before and after it, so it becomes a "word"
use the * command to search for the word under the cursor. If you have set hlsearch on, you should then see all of the occurrences of the extraneous character highlighted.
replace last searched item by something else, globally:
:%s//something else/