Evaluating math Equations with Brackets

Question

I am just trying to Convert secs back to hours n minutes for a day. However am hitting an error in the Minute Equation! its not evaluating the braces in the order! how can we evalvate math equations like the one I have below.

I want to divide the total_secs by 3600 and then multiply the result with 60 then Again find the modulus of the result with 60.

Here s my attempt

#!/bin/ksh  
set -x  
total_secs=3685;  
hour=$(( (total_secs % 3600) ));  
minute=$(( (((total_secs / 3600) * 60) % 60) ));  
seconds=$(( (((total_secs / 3600) * (60) * (60)) % 60) ));  
echo ""$hour":"$minute":"$seconds""  

Thanks In advance

Answer 1

The ordering of operations wasn't what was causing the problem. According to this reference, the Korn shell has the same precedence and associativity as the C programming language, so multiplication, division and remainder happen from left to right. The correct maths would be:

#!/bin/ksh

total_secs=3685
hour=$(( total_secs / 3600 ))
minute=$(( total_secs / 60 % 60  )) # equivalent to (total_secs / 60) % 60
seconds=$(( total_secs % 60 ))
echo "$hour:$minute:$seconds"
# output 1:1:25

If you want to format the output, perhaps you could use printf

printf "%02d:%02d:%02d\n" $hour $minute $seconds
# output 01:01:25

edit: operator precedence

Basic arithmetic operations are carried out in the following order:

1. * / %
2. + -

Multiple operations of the same precedence will be evaluated from left to right as they appear in the expression.

$(( ((x*y) / z) + n ))

is equivalent to either of the following:

$(( x * y / z + n ))
$(( n + x * y / z ))

because in both cases, the * is the highest precedence and nearest to the left of the expression. Second will be the / and finally the +.

There is no harm in inserting additional ( ) to make your intentions clear but remember that the whole expression must be enclosed within $(( )).