My bash arithmetic always evaluates to 0, no matter what












0















I type in the script:



result=$(( ($1-32)*(5/9) ))

echo $result


and the result variable always evaluates to zero, no matter what value $1 is. How do I solve this problem?










share|improve this question



























    0















    I type in the script:



    result=$(( ($1-32)*(5/9) ))

    echo $result


    and the result variable always evaluates to zero, no matter what value $1 is. How do I solve this problem?










    share|improve this question

























      0












      0








      0








      I type in the script:



      result=$(( ($1-32)*(5/9) ))

      echo $result


      and the result variable always evaluates to zero, no matter what value $1 is. How do I solve this problem?










      share|improve this question














      I type in the script:



      result=$(( ($1-32)*(5/9) ))

      echo $result


      and the result variable always evaluates to zero, no matter what value $1 is. How do I solve this problem?







      bash






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 9 at 15:59









      ThomasLMahoneyThomasLMahoney

      72




      72






















          1 Answer
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          The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.



          Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas



          $ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
          0


          if you allow * and / to have their natural precedence



          $ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
          1


          because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.



          $ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
          1.65





          share|improve this answer























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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4














            The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.



            Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas



            $ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
            0


            if you allow * and / to have their natural precedence



            $ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
            1


            because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.



            $ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
            1.65





            share|improve this answer




























              4














              The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.



              Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas



              $ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
              0


              if you allow * and / to have their natural precedence



              $ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
              1


              because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.



              $ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
              1.65





              share|improve this answer


























                4












                4








                4







                The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.



                Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas



                $ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
                0


                if you allow * and / to have their natural precedence



                $ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
                1


                because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.



                $ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
                1.65





                share|improve this answer













                The bash shell natively only supports integer arithmetic. In particular, since 5 is less than 9, (5/9) will evaluate to zero regardless of the other terms in your product.



                Depending on your requirements, you might be able to do what you want (still using integer arithmetic) by changing the order of operations i.e. whereas



                $ bash -c 'echo $(( ($1-32)*(5/9) ))' bash 35
                0


                if you allow * and / to have their natural precedence



                $ bash -c 'echo $(( ($1-32)*5/9 ))' bash 35
                1


                because (35-32)*5 is 15, which when integer-divided by 9 yields 1. If you actually want floating point arithmetic then you can use an external program such as bc or awk e.g.



                $ bash -c 'echo "scale = 2; ($1-32)*(5/9)" | bc' bash 35
                1.65






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Jan 9 at 16:37









                steeldriversteeldriver

                66.2k11106179




                66.2k11106179






























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