【发布时间】:2017-05-29 18:45:17
【问题描述】:
代码的目的最初是在 grpc 服务不可用时测试 grpc 存根的操作。但是,我看到的行为表明发生了一些我不明白的事情——因此提出了这个问题。
在这段代码中:
#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;
}
输出是
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 . . .
从时间戳可以看出,Channel 的析构函数只用了 10 多秒。
我的问题是这样的:我可以做些什么来显着减少销毁 grpc 通道所需的时间?
【问题讨论】: