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copied!<p>It is no matter on which type of system this code will run. Switching between threads goes in the same way. Worst case for all cases is: 1000 + 1000 * 999 * 998 * 997 * ... * 2 * 1 times. (incorrect!!! correct one is in update)</p> <p>When first thread tries to increase a variable (it already read a value, but not written yet), second thread can make all the loop from the start value of i, but when second thread finishing it's last loop, first thread increases a value of i, and second thread starts it's long job again :)</p> <p><em>*</em> Updated (A little more details)</p> <p>Sorry, real formula is:</p> <p>1000 + 1000 + 999 + 998 + ... + 2 + 1 times, or 500999</p> <p>Each iteration of the loop looks like this:</p> <ol> <li>Check condition.</li> <li>Make a work.</li> <li>Read value from i</li> <li>Increase read value</li> <li>Write value to i</li> </ol> <p>Nobody said that step 2 has constant time. So I suppose that it has a varying and suitable for my worst case time.</p> <p>Here is this worst case:</p> <p>Iteration 1:</p> <p>[1st thread] Makes steps 1-4 of the first loop iteration (very long work time)</p> <p>[2nd thread] Makes all the loop (1000 times), but doesn't check a condition last time</p> <p>Iteration 2:</p> <p>[1] Maskes step 5, so now i == 1, and makes steps 1-4 of next loop iteration</p> <p>[2] Makes all the loop from current i (999 times)</p> <p>Iteration 3: the same as before, but i == 2</p> <p>...</p> <p>Iteration 1000: the same as before , but i == 999</p> <p>In the end we will have 1000 Iterations and each iteration will have 1 execution of loop code from first thread and (1000 - Iteration number) executions from second thread.</p>

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