[SOLVED] BASH incrementing a variable

Garrett85 · 01-12-2013, 08:35 PM

In the code below I can not get STARTPOINT to increment by 3 each time through the loop. It starts at 0 and then goes to 3 one time and wont increment anymore after that. The starnge thing to me is that my variable NUMBEROFGROUPS is incrementing by -1 without a problem and the two lines are pretty much identiacal except for the variable name.

Code:

#!/bin/bash
set -x

##############################
### FUNCTION-leading_zeros ###
##############################

leading_zeros() {
    local NUM=${1}
    local NUMBEROFDIGITS=${2}
    local NUMBEROFGROUPS=${3}

    while [[ $(( ${#NUM} / 3 )) -ne ${NUMBEROFGROUPS} ]] || [[ ${#NUM} -lt 3  ]]; do
        NUM="0${NUM}"
    done

    echo ${NUM}
}

read -p "Enter a number no higher than the billions range" NUM

NUMBEROFDIGITS=${#NUM}
NUMBEROFGROUPS=$( expr ${NUMBEROFDIGITS} / 3 )

if [[ ${NUMBEROFDIGITS} -lt 3 ]]; then # {
    NUMBEROFGROUPS=1
    NUM=$(leading_zeros ${NUM} ${NUMBEROFDIGITS} ${NUMBEROFGROUPS})
elif [[ `expr ${NUMBEROFDIGITS} % 3` -ne 0 ]]; then # } {
    let NUMBEROFGROUPS=NUMBEROFGROUPS+1
    NUM=$(leading_zeros ${NUM} ${NUMBEROFDIGITS} ${NUMBEROFGROUPS})
fi # }

while [[ ${NUMBEROFGROUPS} -gt 0 ]]; do # {
    STARTPOINT=0
    CURRENTGROUP=${NUM:${STARTPOINT}:3}
    echo ${CURRENTGROUP}

    let STARTPOINT=$(( ${STARTPOINT} + 3 ))
    let NUMBEROFGROUPS=$(( ${NUMBEROFGROUPS} -1 ))
done # }

konsolebox · 01-12-2013, 08:39 PM

Shouldn't STARTPOINT be outside of the loop?

Code:

STARTPOINT=0

while [[ ${NUMBEROFGROUPS} -gt 0 ]]; do # {
    CURRENTGROUP=${NUM:${STARTPOINT}:3}
    echo ${CURRENTGROUP}

    let STARTPOINT=$(( ${STARTPOINT} + 3 ))
    let NUMBEROFGROUPS=$(( ${NUMBEROFGROUPS} -1 ))
done # }

And I suggest using (( )) instead. It's more efficient and you don't have to use $ with names.

Btw, I think the condition of that loop makes it go infinite.

konsolebox · 01-12-2013, 08:57 PM

I still don't how the functions must go but as a suggestion for alternative format:

Code:

#!/bin/bash
set -x

##############################
### FUNCTION-leading_zeros ###
##############################

leading_zeros() {
    local NUM=${1}
    local NUMBEROFDIGITS=${2}
    local NUMBEROFGROUPS=${3}

    while (( ${#NUM} / 3 != NUMBEROFGROUPS || ${#NUM} < 3 )); do
        NUM="0${NUM}"
    done

    echo "${NUM}"
}

read -p "Enter a number no higher than the billions range: " NUM

(( NUMBEROFDIGITS = ${#NUM}, NUMBEROFGROUPS = NUMBEROFDIGITS / 3 ))

if [[ NUMBEROFDIGITS -lt 3 ]]; then # {
    NUMBEROFGROUPS=1
    NUM=$(leading_zeros "${NUM}" "${NUMBEROFDIGITS}" "${NUMBEROFGROUPS}")
elif (( NUMBEROFDIGITS % 3 != 0 )); then # } {
    (( ++NUMBEROFGROUPS ))
    NUM=$(leading_zeros "${NUM}" "${NUMBEROFDIGITS}" "${NUMBEROFGROUPS}")
fi # }

STARTPOINT=0

while [[ NUMBEROFGROUPS -gt 0 ]]; do # {    
    CURRENTGROUP=${NUM:STARTPOINT:3}
    echo "${CURRENTGROUP}"

    (( STARTPOINT += 3, ++NUMBEROFGROUPS ))
done # }

Garrett85 · 01-12-2013, 09:05 PM

Thanks I see I was kinda stupid putting that in the loop. I saw that myself right after I posted this. Oh, and I though I was using (( )).

Quote:

Originally Posted by konsolebox

Shouldn't STARTPOINT be outside of the loop?

Code:

STARTPOINT=0

while [[ ${NUMBEROFGROUPS} -gt 0 ]]; do # {
    CURRENTGROUP=${NUM:${STARTPOINT}:3}
    echo ${CURRENTGROUP}

    let STARTPOINT=$(( ${STARTPOINT} + 3 ))
    let NUMBEROFGROUPS=$(( ${NUMBEROFGROUPS} -1 ))
done # }

And I suggest using (( )) instead. It's more efficient and you don't have to use $ with names.

Btw, I think the condition of that loop makes it go infinite.