How can I use mach_absolute_time without overflowing?
You're worrying about overflow when multiplying/dividing with values from the mach_timebase_info struct, which is used for conversion to nanoseconds. So, while it may not fit your exact needs, there are easier ways to get a count in nanoseconds or seconds.
All solutions below are using mach_absolute_time internally (and NOT the wall clock).
Use double instead of uint64_t
(supported in Objective-C and Swift)
double tbInSeconds = 0;
mach_timebase_info_data_t tb;
kern_return_t kError = mach_timebase_info(&tb);
if (kError == 0) {
tbInSeconds = 1e-9 * (double)tb.numer / (double)tb.denom;
}
(remove the 1e-9 if you want nanoseconds)
Usage:
uint64_t start = mach_absolute_time();
// do something
uint64_t stop = mach_absolute_time();
double durationInSeconds = tbInSeconds * (stop - start);
Use ProcessInfo.processInfo.systemUptime
(supported in Objective-C and Swift)
It does the job in double seconds directly:
CFTimeInterval start = NSProcessInfo.processInfo.systemUptime;
// do something
CFTimeInterval stop = NSProcessInfo.processInfo.systemUptime;
NSTimeInterval durationInSeconds = stop - start;
For reference, source code of systemUptime just does something similar as previous solution:
struct mach_timebase_info info;
mach_timebase_info(&info);
__CFTSRRate = (1.0E9 / (double)info.numer) * (double)info.denom;
__CF1_TSRRate = 1.0 / __CFTSRRate;
uint64_t tsr = mach_absolute_time();
return (CFTimeInterval)((double)tsr * __CF1_TSRRate);
Use QuartzCore.CACurrentMediaTime()
(supported in Objective-C and Swift)
Same as systemUptime, but without being open source.
Use Dispatch.DispatchTime.now()
(supported in Swift only)
Another wrapper around mach_absolute_time(). Base precision is nanoseconds, backed with UInt64.
DispatchTime start = DispatchTime.now()
// do something
DispatchTime stop = DispatchTime.now()
TimeInterval durationInSeconds = Double(end.uptimeNanoseconds - start.uptimeNanoseconds) / 1_000_000_000
For reference, source code of DispatchTime.now() says it basically simply returns a struct DispatchTime(rawValue: mach_absolute_time()). And the calculation for uptimeNanoseconds is:
(result, overflow) = result.multipliedReportingOverflow(by: UInt64(DispatchTime.timebaseInfo.numer))
result = overflow ? UInt64.max : result / UInt64(DispatchTime.timebaseInfo.denom)
So it just discards results if the multiplication can't be stored in an UInt64.
Most-precise (best) answer
Perform the arithmetic at 128-bit precision to avoid the overflow!
// Returns monotonic time in nanos, measured from the first time the function
// is called in the process.
uint64_t monotonicTimeNanos() {
uint64_t now = mach_absolute_time();
static struct Data {
Data(uint64_t bias_) : bias(bias_) {
kern_return_t mtiStatus = mach_timebase_info(&tb);
assert(mtiStatus == KERN_SUCCESS);
}
uint64_t scale(uint64_t i) {
return scaleHighPrecision(i - bias, tb.numer, tb.denom);
}
static uint64_t scaleHighPrecision(uint64_t i, uint32_t numer,
uint32_t denom) {
U64 high = (i >> 32) * numer;
U64 low = (i & 0xffffffffull) * numer / denom;
U64 highRem = ((high % denom) << 32) / denom;
high /= denom;
return (high << 32) + highRem + low;
}
mach_timebase_info_data_t tb;
uint64_t bias;
} data(now);
return data.scale(now);
}
A simple low-resolution answer
// Returns monotonic time in nanos, measured from the first time the function
// is called in the process. The clock may run up to 0.1% faster or slower
// than the "exact" tick count.
uint64_t monotonicTimeNanos() {
uint64_t now = mach_absolute_time();
static struct Data {
Data(uint64_t bias_) : bias(bias_) {
kern_return_t mtiStatus = mach_timebase_info(&tb);
assert(mtiStatus == KERN_SUCCESS);
if (tb.denom > 1024) {
double frac = (double)tb.numer/tb.denom;
tb.denom = 1024;
tb.numer = tb.denom * frac + 0.5;
assert(tb.numer > 0);
}
}
mach_timebase_info_data_t tb;
uint64_t bias;
} data(now);
return (now - data.bias) * data.tb.numer / data.tb.denom;
}
A fiddly solution using low-precision arithmetic but using continued fractions to avoid loss of accuracy
// This function returns the rational number inside the given interval with
// the smallest denominator (and smallest numerator breaks ties; correctness
// proof neglects floating-point errors).
static mach_timebase_info_data_t bestFrac(double a, double b) {
if (floor(a) < floor(b))
{ mach_timebase_info_data_t rv = {(int)ceil(a), 1}; return rv; }
double m = floor(a);
mach_timebase_info_data_t next = bestFrac(1/(b-m), 1/(a-m));
mach_timebase_info_data_t rv = {(int)m*next.numer + next.denum, next.numer};
return rv;
}
// Returns monotonic time in nanos, measured from the first time the function
// is called in the process. The clock may run up to 0.1% faster or slower
// than the "exact" tick count. However, although the bound on the error is
// the same as for the pragmatic answer, the error is actually minimized over
// the given accuracy bound.
uint64_t monotonicTimeNanos() {
uint64_t now = mach_absolute_time();
static struct Data {
Data(uint64_t bias_) : bias(bias_) {
kern_return_t mtiStatus = mach_timebase_info(&tb);
assert(mtiStatus == KERN_SUCCESS);
double frac = (double)tb.numer/tb.denom;
uint64_t spanTarget = 315360000000000000llu; // 10 years
if (getExpressibleSpan(tb.numer, tb.denom) >= spanTarget)
return;
for (double errorTarget = 1/1024.0; errorTarget > 0.000001;) {
mach_timebase_info_data_t newFrac =
bestFrac((1-errorTarget)*frac, (1+errorTarget)*frac);
if (getExpressibleSpan(newFrac.numer, newFrac.denom) < spanTarget)
break;
tb = newFrac;
errorTarget = fabs((double)tb.numer/tb.denom - frac) / frac / 8;
}
assert(getExpressibleSpan(tb.numer, tb.denom) >= spanTarget);
}
mach_timebase_info_data_t tb;
uint64_t bias;
} data(now);
return (now - data.bias) * data.tb.numer / data.tb.denom;
}
The derivation
We aim to reduce the fraction returned by mach_timebase_info to one that is essentially the same, but with a small denominator. The size of the timespan that we can handle is limited only by the size of the denominator, not the numerator of the fraction we shall multiply by:
uint64_t getExpressibleSpan(uint32_t numer, uint32_t denom) {
// This is just less than the smallest thing we can multiply numer by without
// overflowing. ceilLog2(numer) = 64 - number of leading zeros of numer
uint64_t maxDiffWithoutOverflow = ((uint64_t)1 << (64 - ceilLog2(numer))) - 1;
return maxDiffWithoutOverflow * numer / denom;
}
If denom=33333335 as returned by mach_timebase_info, we can handle differences of up to 18 seconds only before the multiplication by numer overflows. As getExpressibleSpan shows, by calculating a rough lower bound for this, the size of numer doesn't matter: halving numer doubles maxDiffWithoutOverflow. The only goal therefore is to produce a fraction close to numer/denom that has a smaller denominator. The simplest method to do this is using continued fractions.
The continued fractions method is rather handy. bestFrac clearly works correctly if the provided interval contains an integer: it returns the least integer in the interval over 1. Otherwise, it calls itself recursively with a strictly larger interval and returns m+1/next. The final result is a continued fraction that can be shown by induction to have the correct property: it's optimal, the fraction inside the given interval with the least denominator.
Finally, we reduce the fraction Darwin passes us to a smaller one to use when rescaling the mach_absolute_time to nanoseconds. We may introduce an error here because we can't reduce the fraction in general without losing accuracy. We set ourselves the target of 0.1% error, and check that we've reduced the fraction enough for common timespans (up to ten years) to be handled correctly.
Arguably the method is over-complicated for what it does, but it handles correctly anything the API can throw at it, and the resulting code is still short and extremely fast (bestFrac typically recurses only three or four iterations deep before returning a denominator less than 1000 for random intervals [a,a*1.002]).