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I have two .proto files (a.proto and b.proto) having the same contents in them:

syntax = "proto3";

message env {
  string broker = 1;
}

When I execute the compiler (I want to generate Java source files) and specify both files on the command line

protoc.exe --java_out=. a.proto b.proto

I get error messages:

b.proto:4:10: "env.broker" is already defined in file "a.proto".
b.proto:3:9: "env" is already defined in file "a.proto".

I'd expect the compiler to generate two Java classes (A and B), each having a nested class Env. This is how I understand the docs. But this does not happen.

What am I doing wrong?

Thank you for any hints.

fml2
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  • A somewhat similar question (without a good answer IMO) is https://stackoverflow.com/questions/62911250/why-i-am-getting-proto-class-is-is-already-defined-in-file. – fml2 Dec 16 '22 at 07:19

1 Answers1

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b.proto:4:10: "env.broker" is already defined in file "a.proto".
b.proto:3:9: "env" is already defined in file "a.proto".

Most protobuf libraries put all messages with same package name into same namespace. The Java library is special in generating the outer class, which is needed because Java does not allow multiple top-level classes per file.

Protoc is doing you a service checking the rule here. If you went ahead and used the same message name in multiple files without specifying a package, you would make it very difficult for other people to use your .proto files with different programming languages.

jpa
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  • OK, that's a sensible reasoning, even if it's questionable. This way, protoc ensures that the interoperability is easily possible. This should be covered in the docs IMO. As of now, it can be confusing. – fml2 Dec 16 '22 at 09:31
  • @fml2 I agree. The protoc developers seem to often look at things from a C++ perspective, which is visible in e.g. enum scoping rules also. – jpa Dec 16 '22 at 09:39