[awk] rounding a float number?

Shell Programming and Scripting
1

Heyas

Trying to calculate the total size of a file by reading its bitrate.

Code snippet:

fs_expected() { #
	# Returns the expected filesize in bytes
	#
		pr_str() {
			ff=$(cat $TMP.info)
			d="${ff#*bitrate: }"
			echo "${d%%,*}" | $AWK '{print $1}' | head -n 1
		}
		t_BYTERATE=$(( $(pr_str) * 1024 / 8 ))
		t_TIMES=$( PlayTime | $SED s,":"," ",g)
		
		echo "$t_BYTERATE $t_TIMES"
		
		echo "$t_TIMES" | $AWK '{ (((24*$1*60)+(60*$2)+$3)*BYTES) }' BYTES=$t_BYTERATE
		return $?
	}

Output:

86272 00 44 14.96

Sadly there is no longer any output at all when i remove the bold code line.

Expected output:
When i remove the bold code line, which is just there for debuging, i want to get the proper expected filesize.
As of now, i assume awk fails due to the underlined floating point number?

Playtime does the same as pr_str, just for the duration of the file.
$TMP.info is filled by another prior function.

Any ideas?
Thank you in advance

2

The awk code will not produce output unless you print the result. Try:

echo "$t_TIMES" | $AWK '{ print (((24*$1*60)+(60*$2)+$3)*BYTES) }' BYTES=$t_BYTERATE
1 Like
3

With rounding:

'{ print int ( (24*$1*60+60*$2+$3)*BYTES + 0.5 ) }'

or

'{ printf "%.f\n", (24*$1*60+60*$2+$3)*BYTES }'
1 Like
4

The 'int' made already a good conversion from 2.29049e+08 to 229048709

I see no need/difference by adding + 0.5 .

Thank you

5

The int() function truncates, rounds down. The value is 229048709.12 adding 0.5 yields 229048709.62 In both case the the int value is 229048709

For rounding you could try:

echo "$t_TIMES" | $AWK '{ printf "%.0f\n", (((24*$1*60)+(60*$2)+$3)*BYTES) }' BYTES=$t_BYTERATE
1 Like
6

For rounding, I add 0.5 before truncating it. Decimal places less than 0.5000 will add up to less than 1, and be truncated, higher numbers will add one whole number then truncate.

1 Like
7

In this case the suggested + 0.5 was at the wrong place.
It was $3)*BYTES + 0.5 , where it should have been ($3+0.5) .
Until now i was thinking that the + 0.5 was supposed for the +E conversion, which was already done by the int .

Makes perfectly sense now, thank you.

8

From the point of mathematical "rounding" only my proposal makes sense.
Example:

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

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?

1 Like
10

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... :smiley: