Where we are setting flag[j] = true and turn = i ?
Example explained with single process i.
Unable to perform dry run with 2 parallel processes
Peterson’s Algorithm is used to synchronize two processes. It uses two variables, a bool array flag of size 2 and an int variable turn to accomplish it.
In the solution i represents the Consumer and j represents the Producer. Initially the flags are false. When a process wants to execute it’s critical section, it sets it’s flag to true and turn as the index of the other process.
This means that the process wants to execute but it will allow the other process to run first. The process performs busy waiting until the other process has finished it’s own critical section.
After this the current process enters it’s critical section and adds or removes a random number from the shared buffer. After completing the critical section, it sets it’s own flag to false, indication it does not wish to execute anymore.