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[awk] rounding a float number?

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Top Forums Shell Programming and Scripting [awk] rounding a float number?
# 8  
Old 03-25-2015
From the point of mathematical "rounding" only my proposal makes sense.
Example:
Code:
echo 00 44 14.96 | awk '{ print int ( (24*$1*60+60*$2+$3)*BYTES + 0.5 ) }' BYTES=8627
22904340
echo 00 44 14.96 | awk '{ printf "%.f\n",(24*$1*60+60*$2+$3)*BYTES }' BYTES=8627
22904340
echo 00 44 14.96 | awk '{ print int ( (24*$1*60+60*$2+$3)*BYTES ) }' BYTES=8627
22904339

# 9  
Old 03-26-2015
Out of sheer curiosity: bitrates usually are measured per second, and, by the conversion algorithm you apply, play time seems to be counted in minutes. For the correct calculation of the file size, shouldn't a factor of 60 be considered?
This User Gave Thanks to RudiC For This Post:
sea
# 10  
Old 03-27-2015
Quote:
Originally Posted by RudiC
Out of sheer curiosity: bitrates usually are measured per second, and, by the conversion algorithm you apply, play time seems to be counted in minutes. For the correct calculation of the file size, shouldn't a factor of 60 be considered?
True, but PlayTime returns 3 numbers, hours ($1 -- 00) minutes ($2 -- 44) and seconds ($3 - 14.97).

But you got a point, i accidently used 24 instead of a 2nd 60 for the hours... Smilie
Last edited by sea; 03-27-2015 at 02:31 AM..
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