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problem in awk int() function

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# 1  
Old 01-02-2009
problem in awk int() function
awk -vwgt=$vWeight -vfac=$vFactor '
BEGIN {
printf("wgt:" wgt "\n");
printf("factor:" fac "\n");
total = sprintf("%.0f", wgt * fac);
total2 = sprintf("%.0f", int(wgt * fac));
printf("total:" total "\n");
printf("total2:" total2 "\n");
}
'
if
vWeight=326.4
vFactor=100

the result would be:
total:32640
total2:32639

Could anyone know how is 32639 calucated?

thanks!
# 2  
Old 01-02-2009
Rounding
Rounding!

Try this...

awk 'BEGIN {printf("%.50f\n", 326.4*100)}'

and you should see something like...

32639.99999999999636202119290828704833984375000000000000

Including the int function simply rounds the output down to 32639. 'int' does not round to the nearest integer

printf (or sprintf) rounds by default when using %f, so without the int, it will display 32640.

HTH.

Jerry
# 3  
Old 01-02-2009
Rounding numbers in awk
I send you an attached file with the awk source of a simple function for rounding numbers in any scale. Of course there are some limits, but you can edit the code and make it better (if you can).
Have fun.
round.txt (438 Bytes) 
Last edited by Panos1962; 01-02-2009 at 10:04 AM..
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