Thread: What is wrong with this code of perfect number generation

Originally Posted by megafiddle

I am wondering if the sqrt function is guaranteed to return an integral value when the square root
is a perfect divisor, eg, 1000.00000000 returned as sqrt of 1000000.

It apparently worked in this case, but not sure if it always would.

I think the answer is no due to floating point inaccuracy, but you can always add some small value to the result before casting to the integer type to avoid truncating to less than the integer square root.

The answer is no due to floating point inaccuracy (and the fact that floating point variables can't represent as many integral values as an integral type of the same or larger size).

It's generally not a good idea to use sqrt() when working with integers, anyway. There are plenty of algorithms around for computing the integer square root (the largest integer which, when multiplied by itself, does not exceed the value).