[Help-bash] Can `declare -i` increase the performance of integer arithmetic?

[Peng Yu]

Hi,

I tried `declare -i`. But it doesn't see to improve performance. Is it
supposed to improve the performance? Thanks.

$ time ./script2.sh; time ./script.sh

real 0m6.764s
user 0m6.733s
sys 0m0.015s

real 0m6.810s
user 0m6.782s
sys 0m0.010s

==> script.sh <==
#!/usr/bin/env bash
# vim: set noexpandtab tabstop=2:

declare -i i
for ((i=0;i<1000000;++i))
do
:
done

==> script2.sh <==
#!/usr/bin/env bash
# vim: set noexpandtab tabstop=2:

for ((i=0;i<1000000;++i))
do
:
done

[Eduardo Bustamante]

Post by Peng Yu
Hi,
I tried `declare -i`. But it doesn't see to improve performance. Is it
supposed to improve the performance? Thanks.

No, it's not meant to improve performance.

[Peng Yu]

Post by Eduardo Bustamante
No, it's not meant to improve performance.

Then, what is the purpose of `declare -i`?

[Pierre Gaston]

Post by Peng Yu

Post by Eduardo Bustamante
No, it's not meant to improve performance.

Then, what is the purpose of `declare -i`?
--
Regards,
Peng

It can save you a couple of $(( )):

declare -i sum
sum=2+3
echo $sum # prints 5
sum+=1
echo $sum # prints 6

[Chet Ramey]

Post by Peng Yu

Post by Eduardo Bustamante
No, it's not meant to improve performance.

Then, what is the purpose of `declare -i`?

The purpose is as documented.

For assignment statements:

"If the variable has its integer attribute set, then value
is evaluated as an arithmetic expression even if the $((...)) expansion
is not used (see Arithmetic Expansion below)."

This has implications for how += is treated when applied to a variable
with the integer attribute set.

[Chet Ramey]

Is a variable always stored as a string whether `declare -i` is used?

Yes.

It sounds like if `declare -i` is used on a variable, it is content
should be saved as an integer instead of a string as it is likely that
subsequent arithmetic operations will be performed on the variable. Is
it?

No.