255/1000 is a periodic number in binary just like 1/3 is a periodic number in decimal. It would take infinite storage to store it as a floating point number, so it is stored approximately. 

$ perl -e'printf "%.20e\n", 1.255'
1.25499999999999989342e+00

$ perl -e'printf "%.20e\n", (1.255 * 100 + 0.5) * 10000000'
1.25999999999999976158e+09

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Is there a way to make perl behave as a humble user of arithmetics would have expected

No. If you're using floats, there's no way to make both of these true since the input is the same: 

int(1259999999.999999)              = 1259999999
int((1.255 * 100 + 0.5) * 10000000) = 1256000000

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Said humble user will be disappointed in one of the two cases. But maybe you're ok with the user being disappointed in the former case. Just apply a bit of rounding at the desired tolerance. 

( $j = sprintf("%.6f", $j) ) =~ s/\..*//s;

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In reply to Re: How to use the int-function? by ikegami
in thread How to use the int-function? by Anonymous Monk

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