I have a cross platform c++ program where I’m using the boost libraries to create an asynchronous timer.
I have a global variable:
bool receivedInput = false;
One thread waits for and processes input
string argStr;
while (1)
{
getline(cin, argStr);
processArguments(argStr);
receivedInput = true;
}
The other thread runs a timer where a callback gets called every 10 seconds. In that callback, I check to see if I’ve received a message
if (receivedInput)
{
//set up timer to fire again in 10 seconds
receivedInput = false;
}
else
exit(1);
So is this safe? For the read in thread 2, I think it wouldn’t matter since the condition will evaluate to either true or false. But I’m unsure what would happen if both threads try to set receivedInput at the same time. I also made my timer 3x longer than the period I expect to receive input so I’m not worried about a race condition.
Edit:
To solve this I used boost::unique_lock when I set receivedInput and boost::shared_lock when I read receivedInput. I used an example from here
This is fundamentally unsafe. After thread 1 has written
truetoreceivedInputit isn’t guaranteed that thread 2 will see the new value. For example, the compiler may optimize your code making certain assumptions about the value ofreceivedInputat the time it is used as the if condition or caching it in a register, so you are not guaranteed that main memory will actually be read at the time the if condition is evaluated. Also, both compiler and CPU may change the order of reads and writes for optimization, for exampletruemay be written toreceivedInputbeforegetLine()andprocessArguments().Moreover, relying on timing for synchronization is a very bad idea since often you have no guarantees as to the amount of CPU time each thread will get in a given time interval or whether it will be scheduled in a given time interval at all.
A common mistake is to think that making
receivedInputvolatilemay help here. In fact,volatileguarantees that values are actually read/written to the main memory (instead of for example being cached in a register) and that reads and writes of the variable are ordered with respect to each other. However, it does not guarantee that the reads and writes of thevolatilevariable are ordered with respect to other instructions.You need memory barriers or a proper synchronization mechanism for this to work as you expect.