Question

I have following function:

s() { c=$1; while((c--)) ; do printf 'x' ; done }; s 2

So x is printed two times.

Is it possible to simplify this code ?

At first I’d like to skip assignation part c=$1; and do:

while((1--))

Do you have any ideas ?

Answer 1

Try this:

s() { for i; do printf 'x' ; done }; s 2 3

Here the for loop takes by default the arguments passed to the function and assigns them to i, so you don’t need c=$1 anymore.

Ok, I haven’t answered your problem directly.

To modify parameters, you must use set -- word, for example:

> set -- 5
> echo $1
5
> set -- $(( $1-1 ))
> echo $1
4

Your function will be :

s() { while (( $1 )); do echo 'x'; set -- $(( ${1}-1 )); done }

But you will lose a little performance and clarity in the script :

>time s 100000 > /dev/null
real    0m1.235s
user    0m1.196s
sys     0m0.040s

While with s() { c=$1; while (( c-- )); do echo 'x'; done }:

>time s 100000 > /dev/null
real    0m0.520s
user    0m0.496s
sys     0m0.024s

I have tried this one :

s1() { for i in $(eval echo {1..${1}}); do printf 'x'; done }

But the main problem is the creation of a child process for the eval statement. The performance is a little faster than above, but still a little difficult to read:

>time s1 100000 > /dev/null
real    0m0.453s
user    0m0.412s
sys     0m0.040s

Hope this helps.