Actually, if I add
b.Log("time per op",float64(b.Elapsed().Microseconds())/float64(N_ORDERS*2))
I get a different value than what benchmark reports, so I am certain setting
b.N is not correct.
> On Jul 8, 2024, at 3:08 PM, 'Robert Engels' via golang-nuts
> <golang-nuts@googlegroups.com> wrote:
>
> Given this code (which I know is not “correct”):
>
> func BenchmarkOrders(b *testing.B) {
> var ob = orderBook{}
> var inst = Equity{}
> var ex = testExchangeClient{}
>
> const N_ORDERS = 1000000
>
> b.ResetTimer()
>
> b.N = N_ORDERS
>
> for i:=0;i<N_ORDERS;i++ {
> var o1 = LimitOrder(inst, Buy, NewDecimal("100"), NewDecimal("10"))
> o1.ExchangeId = fmt.Sprint(i)
> var s1 = sessionOrder{ex, o1, time.Now()}
> ob.add(s1)
> }
> for i:=0;i<N_ORDERS;i++ {
> var o1 = LimitOrder(inst, Sell, NewDecimal("100"), NewDecimal("10"))
> o1.ExchangeId = "S"+fmt.Sprint(i)
> var s1 = sessionOrder{ex, o1, time.Now()}
> ob.add(s1)
> }
>
> }
>
> Instead of setting b.N, is there a better way to get a “benchmark” for the
> logical operation.
>
> In this case, the number of operations is 2 * N_ORDERS, so I would like to
> time the entire process and output in the benchmark format the time per “op”.
>
> It seems to work setting b.N, but I’m sure this isn’t ideal.
>
> In Java JMH, you have the ability to set the “number of operations” performed
> during the test - I’m looking for something similar.
>
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