如何在google测试中测试多线程？

Question

The following testcode creates a server- and a clientsocket. Then client sends a message to server and server replies. Thats all. But I can't compile. All ASSERT_EQ in the threadfunctions raise the error "error: void value not ignored as it ought to be" . I have no clue what this should tell me. What is the problem in here? Type is irrelevant as ASSERT_EQ(1, 1); raises the errors too.

EDIT Found this in FAQ from google:

Q:My compiler complains "void value not ignored as it ought to be." What does this mean?

A: You're probably using an ASSERT_XY() in a function that doesn't return void. ASSERT_XY() can only be used in void functions.

How shall I understand this?

void * serverfunc(void * ptr);  
void * clientfunc(void * ptr);    

TEST(netTest, insert)
{
  pthread_t mThreadID1, mThreadID2;
  ::pthread_create(&mThreadID1, nullptr, serverfunc, nullptr);
  ::sleep(1);
  ::pthread_create(&mThreadID1, nullptr, clientfunc, nullptr);
  ::pthread_join(mThreadID1, nullptr);
  ::pthread_join(mThreadID2, nullptr);        
}

void * serverfunc(void * ptr)
{
  net::ServerSocket serv(IPV4, TCP, 55555,5);
  net::ServerSocket * conn = serv.accept();
  net::Message msg;

  conn->recvmsg(&msg);

  ASSERT_EQ(msg.size(),5);
  ASSERT_EQ(msg[0],1);
  ASSERT_EQ(msg[1],2);
  ASSERT_EQ(msg[2],3);
  ASSERT_EQ(msg[3],4);
  ASSERT_EQ(msg[4],5);

  msg = {9,8,6};
  ASSERT_EQ(msg.size(),3);
  ASSERT_EQ(msg[0],9);
  ASSERT_EQ(msg[1],8);
  ASSERT_EQ(msg[2],6);

  conn->sendmsg(msg);

  ::sleep(1);

  delete conn;
  return 0;
}

void * clientfunc(void * ptr)
{
  net::ClientSocket clie(IPV4, TCP, "localhost",55555);
  net::Message msg;

  msg = {1,2,3,4,5};
  ASSERT_EQ(msg.size(),5);
  ASSERT_EQ(msg[0],1);
  ASSERT_EQ(msg[1],2);
  ASSERT_EQ(msg[2],3);
  ASSERT_EQ(msg[3],4);
  ASSERT_EQ(msg[4],5);

  clie.sendmsg(msg);

  clie.recvmsg(&msg);

  ASSERT_EQ(msg.size(),3);
  ASSERT_EQ(msg[0],9);
  ASSERT_EQ(msg[1],8);
  ASSERT_EQ(msg[2],6);

  return 0;
}

Answer 1

Q:My compiler complains "void value not ignored as it ought to be." What does this mean?

A: You're probably using an ASSERT_XY() in a function that doesn't return void. ASSERT_XY() can only be used in void functions.

How shall I understand this?

Your functions don't return void , they return void* - ie they return something ( void* is a pointer-to-anything) while they should return nothing. The FAQ says it is required for the functions which use ASSERT_EQ() to have the void return type.

Answer 2

I have the same problem, too, and I found an "ugly" way to solve it:

void* your_func(void* ptr)
{
    _your_func(ptr);
    reutrn NULL;
}

void _your_func(void* ptr)
{
    ...
    ASSERT_EQ(1, 1);
    ...
}

Answer 3

It looks like the ASSERT_EQ can only be called in a function with the right return type (which appears to be void while yours return void *)

Personally I dislike macro over-use but that's the way it is. The issue with the macro is it obfuscates the code, so you can't see what it's doing wrong.

So just write such a function and get serverfunc and clientfunc to call it.

Answer 4

至于建议，应更换void*类型，请法void ，并return 0与return 。