在 C 中创建访问共享内存

发布于 2024-08-09 23:00:52 字数 845 浏览 1 评论 0原文

所以我有一个问题，我真的不知道如何解决。我希望你能让我知道如何处理它。

我需要在共享内存中分配 N 个缓冲区。每个缓冲区应该初始化为 0。然后我必须分叉 N/2 个子进程。

然后每个子进程（i）将值（i）写入缓冲区（i），休眠一秒钟。然后读取当前缓冲区中的值，如果该值同时发生变化，则显示一条消息。然后孩子移动 i 位置 N/2 次。

So I can have 
N Buffers = 11. (must be prime)
N/2 children = 5.

So child 2 would:
write into buffer 2, sleep, read from buffer 2. (now move 2 positions)
write into buffer 4, sleep, read from buffer 4. (now move 2 positions)
**remmeber we only have five buffers so we go around again*****
write into buffer 1, sleep, read from buffer 1. (now move 2 positions)
write into buffer 3, sleep, read from buffer 3. (now move 2 positions)
write into buffer 5, sleep, read from buffer 5. ***we stop moving here. N/2 moves already performed.

有什么想法应该如何做到这一点？

我可以创建一个包含 N 个整数的数组吗？由于所有孩子都将访问相同的元素，这会成为问题吗？

原文

So I have a problem that I don't really know how to go about. I was hoping maybe you could let me know how to deal with it.

I need to allocate N number of buffers in shared memory. Each buffer should be initialized to 0. Then I must fork N/2 number of child processes.

Each child(i) then writes value (i) into buffer (i), sleeps for one second. Then reads the value at the current buffer, if the value changed in the mean time then it displays a message. The child then moves i-positions N/2 number of times.

So I can have 
N Buffers = 11. (must be prime)
N/2 children = 5.

So child 2 would:
write into buffer 2, sleep, read from buffer 2. (now move 2 positions)
write into buffer 4, sleep, read from buffer 4. (now move 2 positions)
**remmeber we only have five buffers so we go around again*****
write into buffer 1, sleep, read from buffer 1. (now move 2 positions)
write into buffer 3, sleep, read from buffer 3. (now move 2 positions)
write into buffer 5, sleep, read from buffer 5. ***we stop moving here. N/2 moves already performed.

Any ideas how this should be done?