使用Apache Thrift通过exec执行二进制文件

Question

I am using Apache Thrift to set up a c++ server. On the server side I want to execute

execl("a.out","",(char *)NULL);

but either the above line of code is executed first or server.serve() is executed first. Is there any way for the server to start and have the line of code run on top of the server?

I tried putting it in the message_test function but I only want it to execute once not every time when I start up a client.

a.cpp:

 int main(){
     cout<<"I have started up"<<endl;
     string message;
     while (cin >> message){
         cout<<"your message is: "<<message<<endl;
     }
     cout<<"I have now exited"<<endl;
  }

Server.cpp

class ThforkServiceHandler : virtual public ThforkServiceIf {
     public:
     ThforkServiceHandler() {
       // Your initialization goes here

     }
     void message_test(const std::string& message) {
       // Your implementation goes here
        printf("message_test\n");
     }
 };

int main(int argc, char **argv) {
    int port = 9090;
    shared_ptr<ThforkServiceHandler> handler(new ThforkServiceHandler());
    shared_ptr<TProcessor> processor(new ThforkServiceProcessor(handler));
    shared_ptr<TServerTransport> serverTransport(new TServerSocket(port));
    shared_ptr<TTransportFactory> transportFactory(new TBufferedTransportFactory());
    shared_ptr<TProtocolFactory> protocolFactory(new TBinaryProtocolFactory()); 
    TSimpleServer server(processor, serverTransport, transportFactory, protocolFactory);
    server.serve();
    return 0;
}

Answer 1

According to the docs, execl() replaces the current process with the program called. On the other hand, server.serve() will block until someone forces the server to exit the server loop inside.

That explains why you get only one of the two. The one replaces the server and the other call blocks. In essence, either one of these two calls will be (more or less) the last thing that is executed in the server code.

I'd suggest to use a system() call or something like that instead.