Why does for loop using a double fail to terminate

Question

I\'m looking through old exam questions (currently first year of uni.) and I\'m wondering if someone could explain a bit more thoroughly why the following for l

Answer 1

The number 0.1 cannot be exactly represented in binary, much like 1/3 cannot be exactly represented in decimal, as such you cannot guarantee that:

0.1+0.1+0.1+0.1+0.1+0.1+0.1+0.1+0.1+0.1==1

This is because in binary:

0.1=(binary)0.00011001100110011001100110011001....... forever

However a double cannot contain an infinite precision and so, just as we approximate 1/3 to 0.3333333 so must the binary representation approximate 0.1.

Expanded decimal analogy

In decimal you may find that

1/3+1/3+1/3
=0.333+0.333+0.333
=0.999

This is exactly the same problem. It should not be seen as a weakness of floating point numbers as our own decimal system has the same difficulties (but for different numbers, someone with a base-3 system would find it strange that we struggled to represent 1/3). It is however an issue to be aware of.

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