Google Interview Question SDE1s
Country: United States
Interview Type: Phone Interview
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This results in undefined behavior in c++ if v is not an atomic variable (I am assuming that we have two threads here one calling inc and another calling disp). Thread 1 will set v to 1 and then signal s. Thread 2 then can get to line D to print v, but at the same time Thread 1 goes back to line A and tries to increment v. Since we are reading and writing v at the same time, the printed value can be any number (since the read might happen before, after or in the middle of writing).
- Mona February 24, 2014If v is an atomic variable, then the read and write can happen in any order. So the first printed value can be either 1 or 2, the next will be 2 or 3 and so on.