英文:
How to use time value of benchmark
问题
我已经用Go语言为我的国际象棋引擎编写了一个基准测试：
func BenchmarkStartpos(b *testing.B) {
board := ParseFen(startpos)
for i := 0; i < b.N; i++ {
Perft(&board, 5)
}
}
当运行时，我看到了以下输出：
goos: darwin
goarch: amd64
BenchmarkStartpos-4 10 108737398 ns/op
PASS
ok _/Users/dylhunn/Documents/go-chess 1.215s
我想使用每次执行的时间（在这个例子中为 108737398 ns/op）来计算另一个值，并将其作为基准测试的结果输出。具体来说，我想输出每秒节点数，即Perft调用的结果除以每次调用的时间。
我该如何获取基准测试的执行时间，以便我可以打印出自己计算得到的结果？
英文:
I have written a benchmark for my chess engine in Go:
func BenchmarkStartpos(b *testing.B) {
board := ParseFen(startpos)
for i := 0; i < b.N; i++ {
Perft(&board, 5)
}
}
I see this output when it runs:
goos: darwin
goarch: amd64
BenchmarkStartpos-4 10 108737398 ns/op
PASS
ok _/Users/dylhunn/Documents/go-chess 1.215s
I want to use the time per execution (in this case, 108737398 ns/op) to compute another value, and also print it as a result of the benchmark. Specifically, I want to output nodes per second, which is given as the result of the Perft call divided by the time per call.
How can I access the time the benchmark took to execute, so I can print my own derived results?
答案1
得分: 10
你可以使用testing.Benchmark()函数来手动测量/基准测试“基准测试”函数（其签名为func(*testing.B)），并将结果作为testing.BenchmarkResult的值返回，该结构包含了你所需的所有细节：
type BenchmarkResult struct {
N int // 迭代次数
T time.Duration // 总共花费的时间
Bytes int64 // 每次迭代处理的字节数
MemAllocs uint64 // 总的内存分配次数
MemBytes uint64 // 总的字节分配数
}
每次执行的时间可以通过BenchmarkResult.NsPerOp()方法返回，你可以根据需要进行处理。
看下面这个简单的例子：
func main() {
res := testing.Benchmark(BenchmarkSleep)
fmt.Println(res)
fmt.Println("Ns per op:", res.NsPerOp())
fmt.Println("Time per op:", time.Duration(res.NsPerOp()))
}
func BenchmarkSleep(b *testing.B) {
for i := 0; i < b.N; i++ {
time.Sleep(time.Millisecond * 12)
}
}
输出结果为（在Go Playground上尝试一下）：
100 12000000 ns/op
Ns per op: 12000000
Time per op: 12ms
英文:
You may use the testing.Benchmark() function to manually measure / benchmark "benchmark" functions (that have the signature of func(*testing.B)), and you get the result as a value of testing.BenchmarkResult, which is a struct with all the details you need:
type BenchmarkResult struct {
N int // The number of iterations.
T time.Duration // The total time taken.
Bytes int64 // Bytes processed in one iteration.
MemAllocs uint64 // The total number of memory allocations.
MemBytes uint64 // The total number of bytes allocated.
}
The time per execution is returned by the BenchmarkResult.NsPerOp() method, you can do whatever you want to with that.
See this simple example:
func main() {
res := testing.Benchmark(BenchmarkSleep)
fmt.Println(res)
fmt.Println("Ns per op:", res.NsPerOp())
fmt.Println("Time per op:", time.Duration(res.NsPerOp()))
}
func BenchmarkSleep(b *testing.B) {
for i := 0; i < b.N; i++ {
time.Sleep(time.Millisecond * 12)
}
}
Output is (try it on the Go Playground):
100 12000000 ns/op
Ns per op: 12000000
Time per op: 12ms
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