How to produce a range with step n in bash generate a sequence of numbers with increments

Question

The way to iterate over a range in bash is  for i in  0  10   do echo  i  done   What would be the syntax for iterating over the sequence with a step  Say  I would like to get only even number in the above example

User · Answer

Pure Bash  without an extra process   for    COUNTER 0  COUNTER lt  10  COUNTER  2     do     echo  COUNTER done

User · Answer

Bash 4 s brace expansion has a step feature   for  0  10  2   do      done   No matter if Bash 2 3  C-style for loop  see answers above  or Bash 4  I would prefer anything over the  seq  command

User · Answer

seq 4 1 2 3 4    seq 2 5 2 3 4 5    seq 4 2 12 4 6 8 10 12    seq -w 4 2 12 04 06 08 10 12    seq -s  4 2 12 4 6 8 10 12

User · Answer

I d do  for i in  seq 0 2 10   do echo  i  done    though of course seq 0 2 10 will produce the same output on its own    Note that seq allows floating-point numbers  e g   seq  5  25 3 5  but bash s brace expansion only allows integers

User · Answer

bin bash for i in   seq 1 2 10  do    echo  skip by 2 value  i  done

[bash] How to produce a range with step n in bash? (generate a sequence of numbers with increments)

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