可以选择的方法
拷贝Bytes
思想:
序列化成[]bytes,然后拷贝
func BenchmarkBigIntCopyBytes(b *testing.B) {
b.ReportAllocs()
old, _ := new(big.Int).SetString("100000000222222222222222222220000000000000000000", 10)
new := new(big.Int)
for i := 0; i < b.N; i++ {
new.SetBytes(old.Bytes())
}
}
反射赋值
思想:
通过反射赋值
func BenchmarkBigIntCopier(b *testing.B) {
b.ReportAllocs()
old, _ := new(big.Int).SetString("100000000222222222222222222220000000000000000000", 10)
new := new(big.Int)
for i := 0; i < b.N; i++ {
// "github.com/jinzhu/copier"
copier.Copy(new, old)
}
}
copier
内部实现使用了reflect
。
+0
思想
new = old = old + 0
func TestCopyByAdd(t *testing.T) {
old, _ := new(big.Int).SetString("100000000222222222222222222220000000000000000000", 10)
new := new(big.Int)
// new = old = old + 0
new.Add(old, new)
if old.Cmp(new) != 0 {
t.FailNow()
}
new.Add(new, big.NewInt(1))
t.Logf("old:%v,new:%v", old, new)
if old.Cmp(new) >= 0 {
t.FailNow()
}
}
Benchmark测试
func BenchmarkBigIntCopyByAdd(b *testing.B) {
b.ReportAllocs()
old, _ := new(big.Int).SetString("100000000222222222222222222220000000000000000000", 10)
new := new(big.Int)
for i := 0; i < b.N; i++ {
// new = old = old + 0
new.Add(old, new)
}
}
性能对比
big.Int = 10
BenchmarkBigIntCopier-8 30000000 62.5 ns/op 0 B/op 0 allocs/op
BenchmarkBigIntCopyBytes-8 30000000 46.7 ns/op 8 B/op 1 allocs/op
BenchmarkBigIntCopyByAdd-8 100000000 20.8 ns/op 0 B/op 0 allocs/op
big.Int = 100000000222222222222222222220000000000000000000
BenchmarkBigIntCopier-8 30000000 60.8 ns/op 0 B/op 0 allocs/op
BenchmarkBigIntCopyBytes-8 20000000 69.1 ns/op 32 B/op 1 allocs/op
BenchmarkBigIntCopyByAdd-8 100000000 22.1 ns/op 0 B/op 0 allocs/op
比较两次运行的结果,发现:
-
BenchmarkBigIntCopyBytes
有额外的内存分配,其它两个方法则没有 - 当
big.Int
值变大时,BenchmarkBigIntCopyBytes
分配的内存增加,性能变差,结果BenchmarkBigIntCopier
接近,甚至还差一点 -
BenchmarkBigIntCopyByAdd
是性能最好的,没有额外的内存分配,且耗时稳定
结论
BenchmarkBigIntCopyByAdd
是最好的选择。
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