1

那些@things 是EmPy

C++ 

  const char *
  publish__@(spec.base_type.type)(void * untyped_data_writer, const void * untyped_message)
  {
    DataWriter * topic_writer = static_cast<DataWriter *>(untyped_data_writer);
    const __ros_msg_type & ros_message = *(const __ros_msg_type *)untyped_message;
    __dds_msg_type dds_message;
    conversion_cpp(ros_message, dds_message);

    @(__dds_msg_type_prefix)DataWriter * data_writer =
      @(__dds_msg_type_prefix)DataWriter::_narrow(topic_writer);
    DDS::ReturnCode_t status = data_writer->write(dds_message, DDS::HANDLE_NIL);

    // some common switch statements in C and C++

    }
  }

C

static const char *
publish(void * data_writer, const void * cool_message)
{
  if (!data_writer) {return "data writer handle is null";}
  if (!cool_message) {return "ros message handle is null";}

  DDS::DataWriter * topic_writer = static_cast<DDS::DataWriter *>(data_writer);
  __dds_msg_type dds_message;
  const char * err_msg = conversion_c(cool_message, &dds_message);
  if (err_msg != 0) {return err_msg;}

  @(__dds_msg_type_prefix)DataWriter * data_writer =
    @(__dds_msg_type_prefix)DataWriter::_narrow(topic_writer);
  DDS::ReturnCode_t status = data_writer->write(dds_message, DDS::HANDLE_NIL);
@[for field in spec.fields]@
@[if field.type.type == 'string']@
@[if field.type.is_array]@
  {
@[if field.type.array_size]@
    size_t size = @(field.type.array_size);
@[else]@
    size_t size = dds_message.@(field.name)_.length();
@[end if]@
    for (DDS::ULong i = 0; i < size; ++i) {
      // This causes the DDS::String_mgr to release the given c string without freeing it.
      dds_message.@(field.name)_[i]._retn();
    }
  }
@[else]@
  // This causes the DDS::String_mgr to release the given c string without freeing it.
  dds_message.@(field.name)_._retn();
@[end if]@
@[end if]@
@[end for]@

  // some common switch statements in C and C++
  }
}

这个问题有点特定于我正在尝试贡献的一个开源项目，所以我会指出我猜想的确切功能。这是原始的C 方法 ，这是C++ 方法

我需要使用函数指针吗？
这里发生的另一件事是 C 包依赖于 C++ 包。

（也许这不是一个好问题或者是一个模糊的问题，但我不知道该怎么做，因为我是这个代码库的新手）

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0 回答 0