I have a homework assignment that is asking to shift a decimal number by a specified amount of digits. More clearly this bash script will take two input arguments, the first is the number(maximum 9 digits) that the shift will be performed on and the second is the number(-9 to 9) of digits to shift. Another requirement is that when a digit is shifted off the end, it should be attached to the other end of the number. One headache of a requirement is that we cannot use control statements of any kind: no loops, no if, and switch cases.
Example: 12345 3 should come out to 345000012 and 12345 -3 should be 12345000
I know that if I mod 12345 by 10^3 I get 345 and then if I divide 12345 by 10^3 I get 12 and then I can just concatenate those two variables together to get 34512. I am not quite sure if that is exactly correct but that is the closest I can get as of now. As far as the -3 shift, I know that 10^-3 is .001 and would work however when I try using 10^-3 in bash I get an error.
I am just lost at this point, any tips would be greatly appreciated.
EDIT: After several hours of bashing (pun intended) my head against this problem, I finally came up with a script that for the most part works. I would post the code right now but I fear another student hopelessly lost might stumble upon it. I will check back and post what I came up with in a week or two. I was able to do it with mods and division. Thank you all for the responses, it really helped me to open up and think about the problem from different angles.
$# is a special variable in bash , that expands to the number of arguments (positional parameters) i.e. $1, $2 ... passed to the script in question or the shell in case of argument directly passed to the shell e.g. in bash -c '...' .... .
The shift command is one of the Bourne shell built-ins that comes with Bash. This command takes one argument, a number. The positional parameters are shifted to the left by this number, N. The positional parameters from N+1 to $# are renamed to variable names from $1 to $# - N+1.
shift is a bash built-in which kind of removes arguments from the beginning of the argument list. Given that the 3 arguments provided to the script are available in $1 , $2 , $3 , then a call to shift will make $2 the new $1 . A shift 2 will shift by two making new $1 the old $3 .
Here's a hint:
echo ${string:0:3}
echo ${#string}
Edit (2011-02-11):
Here's my solution. I added some additional parameters with defaults.
rotate-string ()
{
local s=${1:-1} p=${2:--1} w=${3:-8} c=${4:-0} r l
printf -vr '%0*d' $w 0 # save $w zeros in $r
r=${r//0/$c}$s # change the zeros to the character in $c, append the string
r=${r: -w} # save the last $w characters of $r
l=${r: -p%w} # get the last part of $r ($p mod %w characters)
echo "$l${r::w-${#l}}" # output the Last part on the Left and the Right part which starts at the beginning and goes for ($w minus the_length_of_the_Left_part) characters
}
usage: rotate-string string positions-to-rotate width fill-character
example: rotate-string abc -4 9 =
result: ==abc====
Arguments can be omitted starting from the end and these defaults will be used:
More examples:
$ rotate-string
00000010
$ rotate-string 123 4
01230000
Fun stuff:
$ for i in {126..6}; do printf '%s\r' "$(rotate-string Dennis $i 20 .)"; sleep .05; done; printf '\n'
$ while true; do for i in {10..1} {1..10}; do printf '%s\r' "$(rotate-string : $i 10 .)"; sleep .1; done; done
$ while true; do for i in {40..2} {2..40}; do printf '%s\r' "$(rotate-string '/\' $i 40 '_')"; sleep .02; done; done
$ d=0; while true; do for i in {1..10} {10..1}; do printf '%s\r' "$(rotate-string $d $i 10 '_')"; sleep .02; done; ((d=++d%10)); done
$ d=0; while true; do for i in {1..10}; do printf '%s\r' "$(rotate-string $d $i 10 '_')"; sleep .2; ((d=++d%10)); done; done
$ shape='▁▂▃▄▅▆▇█▇▆▅▄▃▂▁'; while true; do for ((i=1; i<=COLUMNS; i++)); do printf '%s\r' "$(rotate-string "$shape" $i $COLUMNS ' ')"; done; done
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