Using floating point number the following 'then' branch will not necessarily execute:

{ double x,y; x = 5.0; y = sqrt(25.0); //or another operation resulting in '5' if ( (x/y) == 1.0) { printf("This message may appear or may not.\n"); } }

The reason is due to the imprecision in floating point number representation. Technicaly, the real numbers numbers 0.99999....... and 1.0 are the same, but their finite representation as floating point numbers differ and thus would not pass the test above. For a detailed explanation see: http://citeseer.ist.psu.edu/goldberg91what.html I added the g_warning code, as suggested, and this is the result (which I expected):

** (inkscape:10949): WARNING **: expecting element to be equal to 1, but found 1

After that I added the following warning, to print the difference between the two numbers:

g_warning("\tdifference is %30g\n", 1.0 - u[last]);

The result is:

** (inkscape:13405): WARNING **: difference is 1.11022e-16

So there is the imprecision.

I still recommend against using == to compare floating point numbers.

By the way, I forgot to describe my environment:

Athlon 64, Linux 2.6.11, gcc 3.4.3

Hope this clears things a bit.

João

On 7/12/05, Peter Moulder <Peter.Moulder@...38...> wrote:

On Mon, Jul 11, 2005 at 10:11:44PM -0300, João Rafael Nicola wrote:

...

This is a snipped of chord_length_parameterize:

gdouble tot_len = u[len - 1];
g_return_if_fail( tot_len != 0 );
if (finite(tot_len)) {
    for (unsigned i = 1; i < len; ++i) {
        u[i] /= tot_len;
    }
} else {
    /* We could do better, but this probably never happens anyway. */
    for (unsigned i = 1; i < len; ++i) {
        u[i] = i / (gdouble) ( len - 1 );
    }
}

In both cases, u[len-1] should be very near 1.0, but not necessarily equals 1.0.

Surely x / x is exactly equal to 1.0 when x is a positive integer? On what platform does this `else' case fail?

The `if' case is another `x / x' case, but with weaker constraints on x: we've tested above that it's non-zero and (in the 0.41 version you've pasted above) not +/-inf.

tot_len is read from u[], which is presumably in memory, so we should be safe from issues from architectures like x86 where registers are wider than `double's in memory.

I would guess that the bug is rather that someone has erroneously replaced the original `isfinite' test with `finite', which differs for NaN. This bug has already been corrected in CVS.

I have previously come across a problem under windows where there was a number with strange properties; I think the windows libc printf showed it as something like `0.#' or something like that (involving a zero and I think a hash character).

João, rather than forcing the last element to be 1.0, can you change the assertion code to something like:

if (u[len - 1] != 1) { g_warning("expecting last element to equal 1, but found %17g"); }

and report here what it gives? Btw, I suggest trying current CVS, which is near to being released as 0.42.

Meanwhile, I'll try to check that the current code still respects documented preconditions.

pjrm.

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I'll try.

Thanks,

João

On 7/14/05, Peter Moulder <Peter.Moulder@...38...> wrote:

On Tue, Jul 12, 2005 at 12:50:35PM -0300, João Rafael Nicola wrote:

...

Using floating point number the following 'then' branch will not necessarily execute:

{   double x,y;
    x = 5.0;
    y = sqrt(25.0); //or another operation resulting in '5'
    if ( (x/y) == 1.0) {
        printf("This message may appear or may not.\n");
    }
}

The above code isn't a good example because the most likely reason for inequality would be sqrt resulting in a number not exactly equal to 5.

Anyway, I've now added some code to bezier-utils.cpp intended to deal with the case that x / x != 1.0 (despite x being isFinite and non-zero).

Can you cvs up and tell me if it still produces any messages (and what the message says, if it does produce a message) ?

Thanks,

pjrm.