Commit f587a43e authored by Deucе's avatar Deucе 👌🏾
Browse files

When a double is cast to an int, but the double has a larger value

than the int supports, it's set to 0x80000000 to indicate overflow.

msclock() is *always* overflowing, and clock_t is only 32-bits on
some platforms (specifically FreeBSD).  To "avoid" problems, just
keep subtracting UIN32_MAX from the value until it's less than INT_MAX
then cast.

This function is, of course, terrible and shouldn't actually be used,
but it should at least sorta kinda workish.
parent 74729f19
Pipeline #2710 passed with stage
in 9 minutes and 31 seconds
......@@ -696,7 +696,14 @@ char* os_cmdshell(void)
#ifdef __unix__
clock_t msclock(void)
return (clock_t)(xp_timer() * 1000);
long double t = roundl(xp_timer() * 1000);
if (sizeof(clock_t) < 8) {
while (t > INT32_MAX)
t -= UINT32_MAX;
return (clock_t)t;
  • From the Linux man page on roundl(): If you want to store the rounded value in an integer type, you probably want to use one of the functions described in lround(3) instead.

  • Can't because it's too big to fit... we don't want to deal with FP exceptions.

  • Interesting. So were you seeing msclock() always returning the same number before?

  • Yep.

  • And this is because we're now greater than some magic number of milliseconds away from the Epoch?

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