Rounding errors using doubles?

[Dima Volodin]

Sam Lantinga wrote:

> On a Pentium II, the following code gives an incorrect result when using
> doubles, but a correct result when using floats:
>
> #include <stdio.h>
>
> #ifdef USE_FLOAT
> /* This works */
> typedef float percentage_type;
> #else
> /* This doesn't */
> typedef double percentage_type;
> #endif
>
> main()
> {
>         int value = 86;
>         percentage_type percentage;
>
>         printf("Starting Percentage is: %d\n", value);
>         percentage = ((percentage_type)value)/100.0;
>         printf("Ending Percentage is: %d\n", (int)(percentage * 100.0));

Your code is broken here. What you need is

printf("Ending Percentage is: %d\n", (int) floor(percentage * 100.0 + 0.5));

And, BTW, converting to int is _not_ rounding.

>         -Sam Lantinga

Dima