Consider the following pattern:
err1 := foo()
err2 := bar()
err3 := baz()
return multierr.New(err1, err2, err3)
If err1, err2, and err3 are all nil, then multierr.New should not allocate.
Thus, modify the logic of New to count the number of distinct error values
and allocate the exactly needed slice. This also speeds up non-empty error
situation since repeatedly growing with append is slow.
Performance:
name old time/op new time/op delta
Empty-24 41.8ns ± 2% 6.4ns ± 1% -84.73% (p=0.000 n=10+10)
NonEmpty-24 120ns ± 3% 69ns ± 1% -42.01% (p=0.000 n=9+10)
name old alloc/op new alloc/op delta
Empty-24 64.0B ± 0% 0.0B -100.00% (p=0.000 n=10+10)
NonEmpty-24 168B ± 0% 88B ± 0% -47.62% (p=0.000 n=10+10)
name old allocs/op new allocs/op delta
Empty-24 1.00 ± 0% 0.00 -100.00% (p=0.000 n=10+10)
NonEmpty-24 3.00 ± 0% 2.00 ± 0% -33.33% (p=0.000 n=10+10)
Signed-off-by: Joe Tsai <joetsai@digital-static.net>
Joe Tsai
350aab05e5