woah your bash is legit good. I thought numeric pretexts needed $(( blah )), but you’re ommiting the $ like an absolute madman. How is this wizardy possible
The difference is that (( is a “compound command”, similar to [[ (evaluate conditional expression), while $(( )) is “aritmetic expansion”. They behave in almost exactly the same way but are used in different contexts - the former uses “exit codes” while the latter returns a string, so the former would be used where you would expect a command, while the latter would be used where you expect an expression.
This is similar to how there is ( compound command (run in a subshell), and $( ) (command substitution). You can actually use the former to define a function too (as it’s a compound command):
real_exit() { exit 1; }
fake_exit() ( exit 1 )
Calling real_exit will exit from the shell, while calling fake_exit will do nothing as the exit 1 command is executed in a separate subshell. Notice how you can also do the same in a command substition (because it runs in a subshell):
echo $(echo foo; exit 1)
Will run successfully and output foo.
It is another one of those unknown, very rarely useful features of bash.
If your codebase is closed source there’s no risk of that happening, if it’s open source there’s nothing you can do about it.
Either way there’s no use worrying.
I prefer good ole regex test of a binary num
function isEven(number){ binary=$(echo "obase=2; $number" | bc) if [ "${binary:-1}" = "1" ]; then return 255 fi return 0 }@tetris11 @Aedis More like:
isEven() { case "$1" in *[02468]) return 0;; *) return 1;; esac;}(If all the line breaks are gone from this code snippet, blame Lemmy. It looks fine here.)
Amateur! I can read and understand that almost right away. Now I present a better solution:
even() ((($1+1)&1))(I mean, it’s funny cause it’s unreadable, but I suspect this is also one of the most efficient bash implementations possible)(Actually the obvious one is a slight bit faster. But this impl for
oddis the fastest one as far as I can tellodd() (($1&1)))I’m waiting for a code golf style solution now.
woah your bash is legit good. I thought numeric pretexts needed
$(( blah )), but you’re ommiting the $ like an absolute madman. How is this wizardy possibleSee:
man bash, “Compound Commands” and “Shell Function Definitions”Oh I see it, but for some reason I was taught to always use
$(( arith ))instead of(( arith ))and I guess I’m just wondering what the difference isThe difference is that
((is a “compound command”, similar to[[(evaluate conditional expression), while$(( ))is “aritmetic expansion”. They behave in almost exactly the same way but are used in different contexts - the former uses “exit codes” while the latter returns a string, so the former would be used where you would expect a command, while the latter would be used where you expect an expression.This is similar to how there is
(compound command (run in a subshell), and$( )(command substitution). You can actually use the former to define a function too (as it’s a compound command):real_exit() { exit 1; } fake_exit() ( exit 1 )Calling
real_exitwill exit from the shell, while callingfake_exitwill do nothing as theexit 1command is executed in a separate subshell. Notice how you can also do the same in a command substition (because it runs in a subshell):echo $(echo foo; exit 1)Will run successfully and output
foo.It is another one of those unknown, very rarely useful features of bash.
@tetris11 @Aedis
Never touch my codebase please
If your codebase is closed source there’s no risk of that happening, if it’s open source there’s nothing you can do about it.
Either way there’s no use worrying.