See the following run time comparison. {1..1000000} is slower than
$(seq 1000000).
Since seq involves an external program, I'd expect the latter to be
slower. But the comparison shows the opposite.
I guess seq did some optimization?
Can the performance of {1..1000000} be improved so that it is faster
than $(seq 1000000)?
$ time builtin printf %.sx {1..1000000} > /dev/null
real 0m2.614s
user 0m2.361s
sys 0m0.166s
$ time builtin printf %.sx $(seq 1000000) > /dev/null
real 0m1.516s
user 0m1.317s
sys 0m0.158s
--
Regards,
Peng
