如何交错执行两个共享变量进程以输出"x is 7"？

First, let's restate the identical code for both processes clearly:

shared int x;
x = 7;
while (true) {
    x = x - 1;
    x = x + 1;
    if (x != 7) printf("x is %d", x);
}

Normally, if a process runs its loop body sequentially, x will always stay at 7 after the x-1 and x+1 pair—so the if condition never triggers, and nothing gets printed. To get the output "x is 7", we need to exploit the gap between the if (x !=7) check and the subsequent printf call: these are two separate memory operations, so another process can modify x in between them.

Here's a step-by-step interleaving that makes this happen (let's call the processes P1 and P2):

1. Both processes finish their initial x =7 assignment (shared memory ensures x is 7 here).
2. P1 runs x = x -1; → x becomes 6.
3. P2 runs x = x -1; → x becomes 5.
4. P1 runs x = x +1; → x becomes 6.
5. P1 hits the if (x !=7) check and reads x → it reads 6, so the condition is true (6≠7), and proceeds to the printf statement.
6. Before P1 reads x for printf, P2 runs x = x +1; → x becomes 6.
7. P2 runs x = x +1; again → x becomes 7.
8. P1 finally reads x for the printf → it reads 7, so the output is "x is 7".

Why This Works #

The key is the time between the if condition's read of x and the printf's read of x. Since these are distinct operations, there's a window where the other process can tweak the shared x variable. The if triggers because x wasn't 7 when checked, but by the time printf fetches the value, the other process has restored x to 7.

内容的提问来源于stack exchange，提问作者user9648060