I have a C application called Test. Test takes in a single int parameter. I want to run Test with many different parameters, so I made the following bash script:
#!/bin/bash
for i in {0..5}
do
./Test "$i"
done
However, this executes ./Test "$i" and not ./Test 0, ./Test 1 etc. Changing it to ./Test $i simply executes ./Test $i 5 times.What am I doing wrong?
Try
#!/bin/bash
for i in `seq 0 5`
do
./Test $i
done
To create sequences in bash you can use seq, which has an advantage of working in any bourne-compatible shell (that is, seq works everywhere because it's an external program, it's the for loop that becomes more portable).
Yes, it makes no difference with respect to ./Test invocation, but it provides a respectful reason to procrastinate before fixing ./Test.
{} just as well. What's the point of advising to try this and that if you don't understand what's wrong? (It would help if you had at least an idea on what's wrong. Not necessary correct idea, just any idea, without "try random things doing the same" nonsense).
for i in {0..5}
do
./Test " $i"
done
Or
for i in {1..5}; do exec "./Test $i" ; done
./Test $i 5 times, the value of i is still not being replaced..
./Testitself.