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c++ - gRPC - C++ 异步 HelloWorld 客户端示例不异步执行任何操作

转载 作者:行者123 更新时间:2023-12-04 11:28:22 28 4
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我正在尝试学习如何在 C++ 中异步使用 gRPC。在 https://github.com/grpc/grpc/blob/v1.33.1/examples/cpp/helloworld/greeter_async_client.cc 上查看客户端示例
除非我误解,否则我没有看到任何异步被证明。只有一个 RPC 调用,它会阻塞在主线程上,直到服务器处理它并将结果发回。
我需要做的是创建一个可以进行一个 RPC 调用的客户端,然后在等待第一个从服务器返回的结果的同时启动另一个。
我不知道该怎么做。
有没有人有一个工作示例,或者任何人都可以描述如何实际异步使用 gRPC?
他们的示例代码:

/*
*
* Copyright 2015 gRPC authors.
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*
*/

#include <iostream>
#include <memory>
#include <string>

#include <grpcpp/grpcpp.h>
#include <grpc/support/log.h>

#ifdef BAZEL_BUILD
#include "examples/protos/helloworld.grpc.pb.h"
#else
#include "helloworld.grpc.pb.h"
#endif

using grpc::Channel;
using grpc::ClientAsyncResponseReader;
using grpc::ClientContext;
using grpc::CompletionQueue;
using grpc::Status;
using helloworld::HelloRequest;
using helloworld::HelloReply;
using helloworld::Greeter;

class GreeterClient {
public:
explicit GreeterClient(std::shared_ptr<Channel> channel)
: stub_(Greeter::NewStub(channel)) {}

// Assembles the client's payload, sends it and presents the response back
// from the server.
std::string SayHello(const std::string& user) {
// Data we are sending to the server.
HelloRequest request;
request.set_name(user);

// Container for the data we expect from the server.
HelloReply reply;

// Context for the client. It could be used to convey extra information to
// the server and/or tweak certain RPC behaviors.
ClientContext context;

// The producer-consumer queue we use to communicate asynchronously with the
// gRPC runtime.
CompletionQueue cq;

// Storage for the status of the RPC upon completion.
Status status;

// stub_->PrepareAsyncSayHello() creates an RPC object, returning
// an instance to store in "call" but does not actually start the RPC
// Because we are using the asynchronous API, we need to hold on to
// the "call" instance in order to get updates on the ongoing RPC.
std::unique_ptr<ClientAsyncResponseReader<HelloReply> > rpc(
stub_->PrepareAsyncSayHello(&context, request, &cq));

// StartCall initiates the RPC call
rpc->StartCall();

// Request that, upon completion of the RPC, "reply" be updated with the
// server's response; "status" with the indication of whether the operation
// was successful. Tag the request with the integer 1.
rpc->Finish(&reply, &status, (void*)1);
void* got_tag;
bool ok = false;
// Block until the next result is available in the completion queue "cq".
// The return value of Next should always be checked. This return value
// tells us whether there is any kind of event or the cq_ is shutting down.
GPR_ASSERT(cq.Next(&got_tag, &ok));

// Verify that the result from "cq" corresponds, by its tag, our previous
// request.
GPR_ASSERT(got_tag == (void*)1);
// ... and that the request was completed successfully. Note that "ok"
// corresponds solely to the request for updates introduced by Finish().
GPR_ASSERT(ok);

// Act upon the status of the actual RPC.
if (status.ok()) {
return reply.message();
} else {
return "RPC failed";
}
}

private:
// Out of the passed in Channel comes the stub, stored here, our view of the
// server's exposed services.
std::unique_ptr<Greeter::Stub> stub_;
};

int main(int argc, char** argv) {
// Instantiate the client. It requires a channel, out of which the actual RPCs
// are created. This channel models a connection to an endpoint (in this case,
// localhost at port 50051). We indicate that the channel isn't authenticated
// (use of InsecureChannelCredentials()).
GreeterClient greeter(grpc::CreateChannel(
"localhost:50051", grpc::InsecureChannelCredentials()));
std::string user("world");
std::string reply = greeter.SayHello(user); // The actual RPC call!
std::cout << "Greeter received: " << reply << std::endl;

return 0;
}

最佳答案

你是对的,这是一个非常糟糕的例子,它阻塞而不是异步的。
最好看看这个例子:grpc/greeter_async_client2 .
在这里,您可以在主要内容中看到它们以异步非阻塞方式循环发送 rpc 消息:
客户端异步发送函数:

void SayHello(const std::string& user) {
// Data we are sending to the server.
HelloRequest request;
request.set_name(user);

// Call object to store rpc data
AsyncClientCall* call = new AsyncClientCall;

call->response_reader =
stub_->PrepareAsyncSayHello(&call->context, request, &cq_);

// StartCall initiates the RPC call
call->response_reader->StartCall();

call->response_reader->Finish(&call->reply, &call->status, (void*)call);
}
客户端异步接收函数:
// Loop while listening for completed responses.
// Prints out the response from the server.
void AsyncCompleteRpc() {
void* got_tag;
bool ok = false;

// Block until the next result is available in the completion queue "cq".
while (cq_.Next(&got_tag, &ok)) {
// The tag in this example is the memory location of the call object
AsyncClientCall* call = static_cast<AsyncClientCall*>(got_tag);

if (call->status.ok())
std::cout << "Greeter received: " << call->reply.message() << std::endl;
else
std::cout << "RPC failed" << std::endl;

// Once we're complete, deallocate the call object.
delete call;
}
}
主要功能:
int main(int argc, char** argv) {
GreeterClient greeter(grpc::CreateChannel(
"localhost:50051", grpc::InsecureChannelCredentials()));

// Spawn reader thread that loops indefinitely
std::thread thread_ = std::thread(&GreeterClient::AsyncCompleteRpc, &greeter);

for (int i = 0; i < 100; i++) {
std::string user("world " + std::to_string(i));
greeter.SayHello(user); // The actual RPC call!
}

std::cout << "Press control-c to quit" << std::endl << std::endl;
thread_.join(); //blocks forever

return 0;
}
添加
正如@nmgeek 所指出的,此解决方案中存在潜在的内存泄漏,请参阅 memory-leak-in-grpc-async-client .

关于c++ - gRPC - C++ 异步 HelloWorld 客户端示例不异步执行任何操作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64639004/

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