Monks,
Came across this case today: 
# int vs float
$str1 = 0.00;
$str2 = 0;
$str1 += $str2;
print "str1 $str1\n";

Result:
str1 0

[download]
Given that Perl is written in C and iirc follows the same rules, I'd expect $str2 to be promototed to float/double during the calc, then, given the tgt is float/double, I'd expect a float/double result, NOT an integer.
Any explanations?
Cheers
Chris
PS: using $str1 = $str1 + $str2; produces the same result

In reply to Integer vs Float during addition by chrism01

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