GO基准测试

如何确定N的循环次数

不变量：时间，一般为1s，可以外部运行时指定

关键代码

func (b *B) launch() {
	// Signal that we're done whether we return normally
	// or by FailNow's runtime.Goexit.
	defer func() {
		b.signal <- true
	}()

	// Run the benchmark for at least the specified amount of time.
	if b.benchTime.n > 0 {
		// We already ran a single iteration in run1.
		// If -benchtime=1x was requested, use that result.
		// See https://golang.org/issue/32051.
		if b.benchTime.n > 1 {
			b.runN(b.benchTime.n)
		}
	} else {
		d := b.benchTime.d
		for n := int64(1); !b.failed && b.duration < d && n < 1e9; {
			last := n
			// Predict required iterations.
			goalns := d.Nanoseconds()
			prevIters := int64(b.N)
			prevns := b.duration.Nanoseconds()
			if prevns <= 0 {
				// Round up, to avoid div by zero.
				prevns = 1
			}
			// Order of operations matters.
			// For very fast benchmarks, prevIters ~= prevns.
			// If you divide first, you get 0 or 1,
			// which can hide an order of magnitude in execution time.
			// So multiply first, then divide.
			n = goalns * prevIters / prevns 
			// Run more iterations than we think we'll need (1.2x).
			n += n / 5
			// Don't grow too fast in case we had timing errors previously.
			n = min(n, 100*last)
			// Be sure to run at least one more than last time.
			n = max(n, last+1)
			// Don't run more than 1e9 times. (This also keeps n in int range on 32 bit platforms.)
			n = min(n, 1e9)
			b.runN(int(n))
		}
	}
	b.result = BenchmarkResult{b.N, b.duration, b.bytes, b.netAllocs, b.netBytes, b.extra}
}

1. 先用总执行时间*上次执行次数 / 上次执行时间
2. 增加 20%的次数
3. 每次增加执行次数不操作上次的 100 倍
4. 至少要比上次多 1 次
5. 不能操作 1的9次方 10亿次