Loop to print odd numbers not printing as many as it should












0














I'm writing a script that prints out a user-provided amount of odd numbers starting from a user-provided number.



So for an example if you would enter that you want to print out 5 numbers starting from the number 3, it would output 3, 5, 7, 9, and 11.



I'm currently trying to use the following code:



echo "Enter how many numbers you want to print"
read n
echo "Enter the first number"
read a

for ((a; a < n; a++)); do
((b = a % 2))
if [ $b -ne 0 ]; then
echo "$a"
fi
done


However, with n=5; a=3, the output is not the expected 3 5 7 9 11 but is instead only 3.










share|improve this question
























  • ...so, what's the part that doesn't actually work? See Minimal, Complete, and Verifiable example guidelines -- a question should have a specific problem (not the larger problem your script tries to solve, but the problem with your script), and the shortest code that lets others see that problem.
    – Charles Duffy
    Nov 22 at 17:31










  • BTW, a good place to start is logging with set -x. See your code running at ideone.com/hgCZDP, with a log in the "stderr" section.
    – Charles Duffy
    Nov 22 at 17:35






  • 2




    ...in the current case, what you have isn't a bash-the-language bug, but a thinking-about-your-problem bug: You're comparing a < n, but a doesn't start at 0, it starts at the value the user entered, so it doesn't print n numbers; instead, it just prints odd numbers greater than or equal to a and less than n. For the examples n=5 and a=3, the only odd number that meets that criteria is 3, so that's all it prints.
    – Charles Duffy
    Nov 22 at 17:36


















0














I'm writing a script that prints out a user-provided amount of odd numbers starting from a user-provided number.



So for an example if you would enter that you want to print out 5 numbers starting from the number 3, it would output 3, 5, 7, 9, and 11.



I'm currently trying to use the following code:



echo "Enter how many numbers you want to print"
read n
echo "Enter the first number"
read a

for ((a; a < n; a++)); do
((b = a % 2))
if [ $b -ne 0 ]; then
echo "$a"
fi
done


However, with n=5; a=3, the output is not the expected 3 5 7 9 11 but is instead only 3.










share|improve this question
























  • ...so, what's the part that doesn't actually work? See Minimal, Complete, and Verifiable example guidelines -- a question should have a specific problem (not the larger problem your script tries to solve, but the problem with your script), and the shortest code that lets others see that problem.
    – Charles Duffy
    Nov 22 at 17:31










  • BTW, a good place to start is logging with set -x. See your code running at ideone.com/hgCZDP, with a log in the "stderr" section.
    – Charles Duffy
    Nov 22 at 17:35






  • 2




    ...in the current case, what you have isn't a bash-the-language bug, but a thinking-about-your-problem bug: You're comparing a < n, but a doesn't start at 0, it starts at the value the user entered, so it doesn't print n numbers; instead, it just prints odd numbers greater than or equal to a and less than n. For the examples n=5 and a=3, the only odd number that meets that criteria is 3, so that's all it prints.
    – Charles Duffy
    Nov 22 at 17:36
















0












0








0







I'm writing a script that prints out a user-provided amount of odd numbers starting from a user-provided number.



So for an example if you would enter that you want to print out 5 numbers starting from the number 3, it would output 3, 5, 7, 9, and 11.



I'm currently trying to use the following code:



echo "Enter how many numbers you want to print"
read n
echo "Enter the first number"
read a

for ((a; a < n; a++)); do
((b = a % 2))
if [ $b -ne 0 ]; then
echo "$a"
fi
done


However, with n=5; a=3, the output is not the expected 3 5 7 9 11 but is instead only 3.










share|improve this question















I'm writing a script that prints out a user-provided amount of odd numbers starting from a user-provided number.



So for an example if you would enter that you want to print out 5 numbers starting from the number 3, it would output 3, 5, 7, 9, and 11.



I'm currently trying to use the following code:



echo "Enter how many numbers you want to print"
read n
echo "Enter the first number"
read a

for ((a; a < n; a++)); do
((b = a % 2))
if [ $b -ne 0 ]; then
echo "$a"
fi
done


However, with n=5; a=3, the output is not the expected 3 5 7 9 11 but is instead only 3.







bash






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 22 at 17:41









Charles Duffy

172k25193249




172k25193249










asked Nov 22 at 17:26









Frostbytee

11




11












  • ...so, what's the part that doesn't actually work? See Minimal, Complete, and Verifiable example guidelines -- a question should have a specific problem (not the larger problem your script tries to solve, but the problem with your script), and the shortest code that lets others see that problem.
    – Charles Duffy
    Nov 22 at 17:31










  • BTW, a good place to start is logging with set -x. See your code running at ideone.com/hgCZDP, with a log in the "stderr" section.
    – Charles Duffy
    Nov 22 at 17:35






  • 2




    ...in the current case, what you have isn't a bash-the-language bug, but a thinking-about-your-problem bug: You're comparing a < n, but a doesn't start at 0, it starts at the value the user entered, so it doesn't print n numbers; instead, it just prints odd numbers greater than or equal to a and less than n. For the examples n=5 and a=3, the only odd number that meets that criteria is 3, so that's all it prints.
    – Charles Duffy
    Nov 22 at 17:36




















  • ...so, what's the part that doesn't actually work? See Minimal, Complete, and Verifiable example guidelines -- a question should have a specific problem (not the larger problem your script tries to solve, but the problem with your script), and the shortest code that lets others see that problem.
    – Charles Duffy
    Nov 22 at 17:31










  • BTW, a good place to start is logging with set -x. See your code running at ideone.com/hgCZDP, with a log in the "stderr" section.
    – Charles Duffy
    Nov 22 at 17:35






  • 2




    ...in the current case, what you have isn't a bash-the-language bug, but a thinking-about-your-problem bug: You're comparing a < n, but a doesn't start at 0, it starts at the value the user entered, so it doesn't print n numbers; instead, it just prints odd numbers greater than or equal to a and less than n. For the examples n=5 and a=3, the only odd number that meets that criteria is 3, so that's all it prints.
    – Charles Duffy
    Nov 22 at 17:36


















...so, what's the part that doesn't actually work? See Minimal, Complete, and Verifiable example guidelines -- a question should have a specific problem (not the larger problem your script tries to solve, but the problem with your script), and the shortest code that lets others see that problem.
– Charles Duffy
Nov 22 at 17:31




...so, what's the part that doesn't actually work? See Minimal, Complete, and Verifiable example guidelines -- a question should have a specific problem (not the larger problem your script tries to solve, but the problem with your script), and the shortest code that lets others see that problem.
– Charles Duffy
Nov 22 at 17:31












BTW, a good place to start is logging with set -x. See your code running at ideone.com/hgCZDP, with a log in the "stderr" section.
– Charles Duffy
Nov 22 at 17:35




BTW, a good place to start is logging with set -x. See your code running at ideone.com/hgCZDP, with a log in the "stderr" section.
– Charles Duffy
Nov 22 at 17:35




2




2




...in the current case, what you have isn't a bash-the-language bug, but a thinking-about-your-problem bug: You're comparing a < n, but a doesn't start at 0, it starts at the value the user entered, so it doesn't print n numbers; instead, it just prints odd numbers greater than or equal to a and less than n. For the examples n=5 and a=3, the only odd number that meets that criteria is 3, so that's all it prints.
– Charles Duffy
Nov 22 at 17:36






...in the current case, what you have isn't a bash-the-language bug, but a thinking-about-your-problem bug: You're comparing a < n, but a doesn't start at 0, it starts at the value the user entered, so it doesn't print n numbers; instead, it just prints odd numbers greater than or equal to a and less than n. For the examples n=5 and a=3, the only odd number that meets that criteria is 3, so that's all it prints.
– Charles Duffy
Nov 22 at 17:36














1 Answer
1






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oldest

votes


















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This is a logic error, rather than a problem using bash. If you want to print n numbers, the easiest way to make sure that happens is to iterate from 0 to n, as follows:



#!/usr/bin/env bash
n=5; a=3 # of course, you can also read from the user.

if ((a % 2 == 0)); then # if our starting number is even...
(( ++a )) # add 1 to make it odd.
fi

for ((i=0; i<n; i++)); do # iterate from 0 to n...
echo "$((a + i*2))" # ...emitting 2*i+a each time.
done





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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0














    This is a logic error, rather than a problem using bash. If you want to print n numbers, the easiest way to make sure that happens is to iterate from 0 to n, as follows:



    #!/usr/bin/env bash
    n=5; a=3 # of course, you can also read from the user.

    if ((a % 2 == 0)); then # if our starting number is even...
    (( ++a )) # add 1 to make it odd.
    fi

    for ((i=0; i<n; i++)); do # iterate from 0 to n...
    echo "$((a + i*2))" # ...emitting 2*i+a each time.
    done





    share|improve this answer




























      0














      This is a logic error, rather than a problem using bash. If you want to print n numbers, the easiest way to make sure that happens is to iterate from 0 to n, as follows:



      #!/usr/bin/env bash
      n=5; a=3 # of course, you can also read from the user.

      if ((a % 2 == 0)); then # if our starting number is even...
      (( ++a )) # add 1 to make it odd.
      fi

      for ((i=0; i<n; i++)); do # iterate from 0 to n...
      echo "$((a + i*2))" # ...emitting 2*i+a each time.
      done





      share|improve this answer


























        0












        0








        0






        This is a logic error, rather than a problem using bash. If you want to print n numbers, the easiest way to make sure that happens is to iterate from 0 to n, as follows:



        #!/usr/bin/env bash
        n=5; a=3 # of course, you can also read from the user.

        if ((a % 2 == 0)); then # if our starting number is even...
        (( ++a )) # add 1 to make it odd.
        fi

        for ((i=0; i<n; i++)); do # iterate from 0 to n...
        echo "$((a + i*2))" # ...emitting 2*i+a each time.
        done





        share|improve this answer














        This is a logic error, rather than a problem using bash. If you want to print n numbers, the easiest way to make sure that happens is to iterate from 0 to n, as follows:



        #!/usr/bin/env bash
        n=5; a=3 # of course, you can also read from the user.

        if ((a % 2 == 0)); then # if our starting number is even...
        (( ++a )) # add 1 to make it odd.
        fi

        for ((i=0; i<n; i++)); do # iterate from 0 to n...
        echo "$((a + i*2))" # ...emitting 2*i+a each time.
        done






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 22 at 18:27


























        community wiki





        2 revs, 2 users 92%
        Charles Duffy































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