How to construct a list from a variable in bash?

Question

Possible Duplicate:
How do I iterate over a range of numbers in bash?

I am trying to print "Hello World!" 10 times via the code below, but it is printed only one time. Where am I missing the correct syntax?

RUNS=10

for RUN in {1..$RUNS}
do
    echo "Hello World!"
done

score 2 · Answer 1 · edited May 23 '17 at 12:18

2

In brief, I would suggest you to use:

RUNS=10
for RUN in $(seq 1 $RUNS)
do
    echo "Hello World!"
done

since it will more likely work on other shells too.

If you want to avoid the overhead of the subshell, you can use:

RUNS=10
i=0;
while [ $i -lt $RUNS ]
do
    echo "Hello World!"
    i=$(($i+1))
done

edited May 23 '17 at 12:18

Community

answered Nov 06 '12 at 08:33

mrucci

this fragment spawns child process _seq_, quite an overhead for a simple for-loop – bobah Nov 06 '12 at 08:38
That's the way to do it for maximum compatibility with POSIX shell. Have a look [here](https://wiki.ubuntu.com/DashAsBinSh) and [here](http://mywiki.wooledge.org/Bashism) for other forms of bashisms. – mrucci Nov 06 '12 at 08:46
this question is about Bash, neither about POSIX shell, nor about compatibility with POSIX shell – bobah Nov 06 '12 at 08:50
1

Absolutely. Also the question is not a about performance... – mrucci Nov 06 '12 at 09:00

score 2 · Accepted Answer · edited Nov 06 '12 at 08:45

2

You want to do a brace expansion, but bash does not do double-expansion (it needs to expand $RUNS). You can force double-expansion by

for RUN in $(eval echo {1..$RUNS})
do 
    echo "Hello World!"
done

But I suggest you avoid this mess like the plague and just do

for RUN in $(seq 1 $RUNS) 
do 
    echo "Hello World!"
done

or

for ((RUN=1; RUN<RUNS; RUN++))
do 
    echo "Hello World!"
done

edited Nov 06 '12 at 08:45

choroba

answered Nov 06 '12 at 08:38

Rody Oldenhuis

score 0 · Answer 3 · answered Nov 06 '12 at 08:35

0

{A..B} syntax does not seem to be substituting variables

for ((i=0; i < $RUNS; ++i)); do echo "hello world"; done

answered Nov 06 '12 at 08:35

bobah

1

More specifically, brace expansion occurs before parameter expansion. As brace expansion requires constants, the whole construct is treated as a single string rather than being expanded. – chepner Nov 06 '12 at 13:15

3 Answers3