Question

int main(int argc, char *argv[])
{
    uint64_t length = 0x4f56aa5d4b2d8a80;
    uint64_t new_length = 0;

    new_length = length + 119.000000;

    printf("new length  0x%"PRIx64"\n",new_length);

    new_length = length + 238.000000;

    printf("new length  0x%"PRIx64"\n",new_length);

    return 0;
}

With the above code. I am adding two different double values to a unsigned 64-bit integer.I am getting the exact same result in both the cases.The output of the program is show below

$./a.out
new length  0x4f56aa5d4b2d8c00
new length  0x4f56aa5d4b2d8c00

I would expect two different results but that is not the case.I have also tried type-casting the uint64_t value to a double as in

new_length = (double)length + 119.000000;

But this too doesn’t seem to help.Any idea on what might be the problem?

Answer 1

Floating point arithmetic is not precise. As numbers get bigger, the accuracy of lower digits is reduced.

0x4f56aa5d4b2d8a80 is a Very Large Number.

What is happening in

new_length = length + 119.000000;

Is that length + 119.000000 is getting cast to a double, to do the addition. That double is rounded, rather dramatically, because it’s so large. It is then cast again to the integral type uint64_t when it is assigned to new_length.

When you call

new_length = length + 238.000000; 

It happens that the rounded result ends up being the same.

What you really want to do is

new_length = length + (uint64_t)238.0; 

That will give you the answer you want. It will initially cast the double to an integral type, which is added precisely.