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Asked: 2026-06-07T04:29:01+00:00

myInt = int( 5 * myRandom() ) myRandom() is a randomly generated float, which

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myInt = int( 5 * myRandom() )

myRandom() is a randomly generated float, which should be 0.2.

So this statement should evaluate to be 1.

My question: is it possible that due to a floating point error it will NOT evaluate to 1?

For example if due to a floating point error something which should be 0.2 could that be LESS than that?
IE, for instance consider the following 3 possibilities:

int(5 * 0.2 )                = 1 //case 1 normal
int(5 * 0.2000000000000001 ) = 1 //case 2 slightly larger, its OK
int(5 * 0.1999999999999999 ) = 0 //case 3 negative, is NOT OK, as int() floors it

Is case3 even possible?, with 0.1999999999999999 be a result of a floating point error? I have never actually seen a negative epsilon so far, only case 2, when its a slightly bit larger, and thats OK, as when it is cast to int(), that ‘floors’ it to the correct result. However with a negative epsilon the ‘flooring’ effect will make the resulting 0.9999999999999996 evaluate to 0.

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1 Answer

  1. Editorial Team

    It is impossible for myRandom to return .2 because .2 is not representable as a float or a double, assuming your target system is using the IEEE 754 binary floating-point standard, which is overwhelmingly the default.

    If myRandom() returns the representable number nearest .2, then myInt will be 1, because the number nearest .2 representable as a float is slightly greater than .2 (it is 0.20000000298023223876953125), and so is the nearest representable double (0.20000000000000001110223024625156540423631668090820312).

    In other cases, this will not be true. E.g., the nearest double to .6 is 0.59999999999999997779553950749686919152736663818359375, so myInt will be 2, not 3.

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