bash simple math (division) question

what's wrong with my scripting here?

PROCESSORS=6
FRAMES=23715
INCREMENTS=($FRAMES / $PROCESSORS)

echo $INCREMENTS

i'm expecting a return of 3952 for $INCREMENTS.
instead i keep getting either a blank space or
something like the text $FRAMES.

thanks,
BabaG

[me@linuxbox me]$ echo $((2+2))
4

As you can see, when you surround an arithmetic expression with the double parentheses, the shell will perform arithmetic evaluation.

Try , something like...
hours=$((seconds / 3600))

Gentoo

thanks prad77!

Note that bash only does integer mat internally. If you want to do floating point arithmatic, you will need to used some sort of external program. Commonly used is bc, for example:

$ cat t.sh
#!/bin/bash

result=$(echo "7 / 3" |bc -l)
echo "result = $result"

$ ./t.sh
result = 2.33333333333333333333

The modulo (%) operator is often useful when doing integer division. This routine converts total lapsed seconds to a more human comprehensible format:

#!/bin/bash
# Convert Arg 1 (total seconds) to human readable string of weeks, days, hours, minutes, and seconds
TOTALSECONDS=$1

MINUTES=$((TOTALSECONDS / 60))
SECONDSREMAINING=$((TOTALSECONDS % 60))

HOURS=$((MINUTES / 60))
MINUTESREMAINING=$((MINUTES % 60))

DAYS=$((HOURS / 24))
HOURSREMAINING=$((HOURS % 24))

WEEKS=$((DAYS / 7))
DAYSREMAINING=$((DAYS % 7))

echo $TOTALSECONDS secs = $WEEKS wks $DAYSREMAINING days $HOURSREMAINING hrs $MINUTESREMAINING mins $SECONDSREMAINING

Originally Posted by babag what's wrong with my scripting here? PROCESSORS=6 FRAMES=23715 INCREMENTS=($FRAMES / $PROCESSORS) echo $INCREMENTS i'm expecting a return of 3952 for $INCREMENTS. instead i keep getting either a blank space or something like the text $FRAMES. thanks, BabaG

I think you got'er figured out but yeah this typ O'

INCREMENTS=($FRAMES / $PROCESSORS)
should be
INCREMENTS=$(($FRAMES / $PROCESSORS))

example, CLI

userx@manjaro:~
$ g=8
 
$ d=2
 
$ v=$(($g/$d))
 
$ echo $v
4

By the way, you don’t need ‘$’ inside of ‘$((…))’. Some even claim that it’s less error prone. Compare:

]$ a=2+2
$ echo $((a * 2))
8
$ echo $(($a * 2))
6

Originally Posted by mina86 By the way, you don’t need ‘$’ inside of ‘$((…))’. Some even claim that it’s less error prone. Compare: ]$ a=2+2 $ echo $((a * 2)) 8 $ echo $(($a * 2)) 6

i wouldn't say less error prone - it seems both use cases are treated differently:
the first, counts 2+2=4 BEFORE multiplying it, the second uses the literal string 2+2.
while the second is mathematically correct, you cannot say that either of them is wrong because of the unique way bash (and all shells i believe) can treat strings as numbers and vice versa.

i would never use a construct like 'a=2+2' inside a bash script. if anything, that's where the error is, it's confusing mathematical operations with variable assigment.

PS: i'm perfectly aware this is a necrobumped thread, but why not...