remove resolver.UnregisterForTesting usage#417
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| func NewDescriptorProviderReflection(args ReflectionArgs) (DescriptorProvider, error) { | ||
| r, deregisterScheme := GenerateAndRegisterManualResolver() | ||
| defer deregisterScheme() | ||
| r := GenerateAndRegisterManualResolver() |
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We're creating potential memory leak here.
Let's discuss potential solution with a custom resolver, i.e. to register only one resolver, and then reuse it for different services.
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I've looked into two options to address it:
Keep the manual resolver.
We can create the scheme deterministically (I'm thinking scheme = "dest-" + args.Service).
This allows us to use resolver.Get(scheme) to avoid creating unnecessary resolvers, and we also can keep an updated the list of peers in the resolver.
Use passthrough
This results in simpler code, but only allows to specify one peer.
If the peer-list file used contains outdated addresses this could make the reflection fail where otherwise it would have worked.
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| rb := resolver.Get(scheme) | ||
| if rb != nil { | ||
| if r, ok := rb.(*manual.Resolver); ok { |
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Manual resolvers implement both, Resolver and Builder.
r.InitialState(newState) will set the new list of peers the next time that Build is called
| } | ||
| } | ||
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| func TestResolverAlreadyExists(t *testing.T) { |
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Test creating a new resolver for the same service with a different list of peers
| rb := resolver.Get(scheme) | ||
| if rb != nil { | ||
| if r, ok := rb.(*manual.Resolver); ok { | ||
| r.InitialState(newState) |
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InitialState doesn't use mutex inside, so it's a potential data race.
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resolver.UnregisterForTestinghas been removed from grpc-go grpc/grpc-go#6471, this change is needed in order to update to grpc 1.59.0