Forums General Discussion $int(1.9999999) = 2

Obviously, I am not a programmer ...

$int(N)
Returns the integer part of a floating point number with no rounding.

$int(1.999999) = 1

$int(1.9999999) = 2

Why would a function which is dropping decimals without rounding do round after the sixth decimal and return a wrong answer?

Why can it simply drop all decimals regardless of the number of decimals?

gemeau50

I'd say you found a bug. ;-)

My guess (and I'm probably right) is $int() does nothing more than return an (int) type cast to the number.

I doubt it's a bug in C/C++ (I've never noticed it when working with > 7 digit decimal floating points/doubles.

Perhaps you post this in the Bug Reports forum.

all $int does is return the number value without the decimal.
its the same as $gettok(1.99999,1,46)

if you want to round the number you must use $round(1.9999,0)

The point is $int(1.9999999) is returning 2, which it shouldn't.

I bet it has something to do with string output - try printf("%g",1.999999); to see what I mean.

Well.. why wouldn't he just do [ws|s]printf("%d",(int)1.999999);

Reproducing it.

//echo -a $int(1.9999999)
2

//echo -a $int(1.9)
1