kevineid_eid I used "for x in {1..99..2}".
aiden_p 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.
PRASHANTB1984 As per our environment the version of bash we support is 4.3.11
shukla_manas2 but what if we are programming unix shell script ?
ddruva_boomer unix shell is bash
icasimpan Nice to know {1..99..22} instead of using $(seq 1 99) which is an external command. Thanks for sharing @kevineid_eid
papa1980 seq is always faster
Davidrobert yes this discussion has good information shared here. visit https://printerchatsupport.co.uk/epson-printer-support/ for more such information.
jellybeanfiend 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 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.
leonardo_neves why not only this code?: seq 1 2 99
cli__ same with you
nbkhope 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 you'd think so from the problem statement, but if you click on Editorial, the problem tester's code is a while loop
miraldo for a in {1..50} do echo $((2*$a-1)) done
joshua_lotriet seq 1 2 99
That's all you need.
thekkohli Nice!
lipe_and3 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 $a fi done
Totally unnecessary T.T
zulu_ya Ha, you're not the only one.
#!/bin/bash for NUM in {1..99} do evenNUM=$(expr $NUM % 2) if [ $evenNUM = 1 ] then echo $NUM fi done
himanhsu54 whats wrong??
for i in {1..100} do echo $i i = $((i+2)) done
AbhishekVermaIIT 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 linei=$((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) Usefor i in {1..100..2}instead offor i in {1..100}to get increment of 2.himanhsu54 what if i want to increment inside the loop?
AbhishekVermaIIT Then you can use
while [ $i -le 100 ], also make sure to initializei=1in 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 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 you should add that echo statment after i = $((i+2))
AbhishekVermaIIT No, that won't work! As mentioned in my previous comment, value of i will be over-written by for loop in the next iteration.
Doing this will print all numbers from 3 to 102 instead. Try it yourself.
dolmen man seq
vibhorsingh1983 use seq command to generate the list seq 100 o/p: 1 2 3....100
or simply return the odd numbers like seq 1 2 100 o/p: 1 3 5...99
thameera I'd avoid this because it calls the external command 'seq'.
Using in-built ranges (like for x in {1..99..2}) is much faster.
saharanrenu02 "#!/bin/bash for (( i=1; i<=99; i+=2)) do echo "$i" done "
simple is always best.
a111997kumarvin1 for i in {1..50} do echo i - 1)) done
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