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  1. Practice
  2. Linux Shell
  3. Bash
  4. Looping and Skipping
  5. Discussions

Looping and Skipping

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  • kevineid_eid + 3 comments

    I used "for x in {1..99..2}".

    • aiden_p + 1 comment

      This to me seems like the most concise, elegant way of implementing the for loop, but it seems like it's only a BASH4.0v+ feature.

      • PRASHANTB1984Challenge Author + 1 comment

        As per our environment the version of bash we support is 4.3.11

        • shukla_manas2 + 1 comment

          but what if we are programming unix shell script ?

          • ddruva_boomer + 0 comments

            unix shell is bash

    • icasimpan + 2 comments

      Nice to know {1..99..22} instead of using $(seq 1 99) which is an external command. Thanks for sharing @kevineid_eid

      • papa1980 + 0 comments

        seq is always faster

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    • jellybeanfiend + 1 comment

      Just for fun, you can also skip the for loop altogether with: echo -e 1 "\n"{3..99..2}. I'm trying to figure out if there's a way to avoid having to put the 1 in there separately to avoid having the newline character first.

      • cloaked1 + 0 comments

        I didn't have to put the "1" in there:

        $ echo -e "\n"{1..99..2}

1 
3 
5 
7 
.
.
.

        Maybe bash has changed since you posted this. The problem is the leading lf before the "1" so:

        echo -e "\n"{1..99..2} | grep -P '\d'

        works...but it's totally not what they're probably desiring for a solution.

  • dvt32 + 1 comment
    for number in {1..99..2}
do
    echo $number
done

    1 is the low end.
    99 is the high end.
    2 is the step after each iteration.

    • hex_cab + 0 comments

      Looks so pythonic :)

  • leonardo_neves + 2 comments

    why not only this code?: seq 1 2 99

    • cli__ + 0 comments

      same with you

    • nbkhope + 1 comment

      That is pretty cool, but I think the purpose of the assignment is for people to learn the for loop in Bash and apply it to the challenge.

      • rveeblefetzer + 0 comments

        you'd think so from the problem statement, but if you click on Editorial, the problem tester's code is a while loop

  • miraldo + 0 comments
    for a in {1..50}
do
    echo $((2*$a-1))
done

  • joshua_lotriet + 2 comments
    seq 1 2 99

    That's all you need.

    • thekkohli + 0 comments

      Nice!

    • lipe_and3 + 1 comment

      OMG, look what I did:

      LIMIT=100
prime=0
for ((a=1; a <= LIMIT ; a++))
do
    let "prime= $a % 2"
    if [ $prime -eq 1 ]
    then
    echo $a
    fi
done

      Totally unnecessary T.T

      • zulu_ya + 0 comments

        Ha, you're not the only one. 

        #!/bin/bash

for NUM in {1..99} 
do
    evenNUM=$(expr $NUM % 2)
    if [ $evenNUM = 1 ] 
    then
        echo $NUM
    fi 
done

  • himanhsu54 + 3 comments

    whats wrong??

     for i in {1..100}
do
   echo $i
   i = $((i+2))
 done
    • AbhishekVermaIIT + 1 comment

      Three things :


      1) i = $((i+2)) will give an error because of extra space around "=". It can be corrected to : i=$((i+2))


      2) The above mentioned line i=$((i+2)) is actually unnecessary because the updated value of i is never used, and overwritten by for in next iteration. Hence, this line can be completely removed.


      3) Use for i in {1..100..2} instead of for i in {1..100} to get increment of 2.

      • himanhsu54 + 1 comment

        what if i want to increment inside the loop?

        • AbhishekVermaIIT + 0 comments

          Then you can use while [ $i -le 100 ], also make sure to initialize i=1 in this case.

          You can also use C like construct : for (( i=1; i<=100; i+=2 )), but then again, this won't allow you to increment inside loop.

    • Papa13 + 0 comments

      for i in {1..99}; do echo $i; done

      it doesn't work? do you know Bash? Don't do theses exercices for suckers !

    • hitieshri + 1 comment

      you should add that echo statment after i = $((i+2))

      • AbhishekVermaIIT + 0 comments

        No, that won't work! As mentioned in my previous comment, value of i will be over-written by for loop in the next iteration.

        Doing this will print all numbers from 3 to 102 instead. Try it yourself.

  • vibhorsingh1983 + 1 comment

    use seq command to generate the list seq 100 o/p: 1 2 3....100

    or simply return the odd numbers like seq 1 2 100 o/p: 1 3 5...99

    • thameera + 0 comments

      I'd avoid this because it calls the external command 'seq'.

      Using in-built ranges (like for x in {1..99..2}) is much faster.

  • MasterFoo + 0 comments

    Three solutions; the last one is the " real" solution since it is checking for odd numbers according to the problem instructions.

    # Solution (1)
#for n in {1..99..2} # start at 1, count up to 99 by incrementing by 2
#do
#    echo $n
#done
#    
# Solution (2)      # same here
#for (( n = 1; n <= 99; n += 2 ))
#do
#    echo $n
#done
#
# Solution (3)      # by checking for odd numbers
for n in {1..99}
do
    if (( $n%2 != 0 ))
    then
        echo $n
    fi
done

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  • sparsh_maheswari + 0 comments

    I did using for number in {1..99..2};do echo $number; done

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