Reputation: 5247
Instead of printf I wrote the output to a file, but still it gives me output1 as 2.
my $ramount = 2.01;
$ramount = int($ramount*100)/100;
printf "output1: $ramount";
Upvotes: 2
Views: 2579
Reputation: 37146
From perldoc perlfaq4
:
Why is int() broken?
Your
int()
is most probably working just fine. It's the numbers that aren't quite what you think. First, see the answer to "Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)?".For example, this
print int(0.6/0.2-2), "\n";
will in most computers print 0, not 1, because even such simple numbers as 0.6 and 0.2 cannot be presented exactly by floating-point numbers. What you think in the above as 'three' is really more like 2.9999999999999995559.
Upvotes: 1
Reputation: 86774
I think this sequence will answer your question:
DB<1> $a=2.01
DB<2> p $a
2.01
DB<3> printf "%20.10f\n", $a
2.0100000000
DB<4> printf "%20.16f\n", $a
2.0099999999999998
DB<5> printf "%20.16f\n", ($a*100)
200.9999999999999716
DB<6> printf "%20.16f\n", ($a*100)/100
2.0099999999999998
DB<7> printf "%20.16f\n", int($a*100)
200.0000000000000000
DB<8> printf "%20.16f\n", int($a*100)/100
2.0000000000000000
DB<9>
Essentially (and this has been answered many times on SO), 2.01 cannot be represented EXACTLY as a floating point number. The closest possible float is, as you see above, 2.009999999999999716...
As to padding, try
printf "%04d", $number
The leading zero in the format tells printf
(or sprintf
) to left-pad with zero.
Upvotes: 6