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代碼的目的是最初在grpc服務不可用時測試grpc存根的操作。然而,我所看到的行爲表明,有些事情我不明白 - 因此是一個問題。Grpc客戶端C++通道析構函數花費10秒
在此代碼:
#define IN_MILLISECONDS(x) (std::chrono::system_clock::now() + std::chrono::milliseconds(x))
string NowString()
{
char buf[128];
SYSTEMTIME timeBuf;
::GetLocalTime(&timeBuf);
sprintf(buf, "%02d:%02d:%02d.%03d - ", timeBuf.wHour, timeBuf.wMinute, timeBuf.wSecond, timeBuf.wMilliseconds);
return string(buf);
}
void testStub(std::shared_ptr<grpc::Channel> chan)
{
MessageProcessor::Stub client(chan);
Void _void;
AccumulateAmount amount;
amount.set_amount(42);
grpc::ClientContext ctx;
ctx.set_deadline(IN_MILLISECONDS(100));
cout << NowString() << " Making RPC\n";
grpc::Status st = client.Accumulate(&ctx, amount, &_void);
cout << NowString() << " Leaving testStub()\n";
}
void test()
{
auto chan = grpc::CreateChannel("localhost:54321", grpc::InsecureChannelCredentials());
cout << NowString() << " Channel Up- Testing Stub\n";
testStub(chan);
cout << NowString() << " Leaving test()\n";
}
int main()
{
cout << NowString() << "Calling test()\n";
test();
cout << NowString() << "Exiting 'main'\n";
return 1;
}
輸出是
11:42:05.400 - Calling test()
11:42:05.403 - Channel Up- Testing Stub
11:42:05.404 - Making RPC
11:42:05.506 - Leaving testStub()
11:42:05.507 - Leaving test()
11:42:15.545 - Exiting 'main'
Press any key to continue . . .
應該由Channel的析構函數正在剛剛超過10秒時戳是顯而易見的。
我的問題是這樣的:我能做些什麼來顯着減少銷燬grpc通道所需的時間?