cobegin / coend块的最大值和最小值

时间:2013-02-20 19:12:43

标签: concurrency operating-system

根据以下代码,我需要找到 x的最小和最大最终值

x=1
i=1

cobegin

while (i<4)    while (i<4)
  begin          begin
    x=x*2          x=x*2
    i=i+1          i=i+1 
  end            end

coend

如果循环按顺序执行,我认为最小值x可以是8 。 并且最大值x可以是16 ,如果程序首先进入其中一个循环,切换到另一个循环并执行它直到x = 8且i = 4,并完成第一个循环,然后x = 16,i = 5。它是否正确?我错过了x可能更大还是更低的任何情况?

3 个答案:

答案 0 :(得分:1)

你提出的答案是正确的!

答案 1 :(得分:1)

这取决于i=i+1x=x*2是否是原子操作(意味着在i的值之前和设置之前不会发生任何事情。)

如果不是原子的:

最高: x = 64

x = 1
i = 1
x = 2 (from 1)
x = 4 (from 2)
i = 2 (from 1+2) // get i=1 for both
x = 8 (from 1)
x = 16 (from 2)
i = 3 (from 1+2) // get i=2 for both
x = 32 (from 1)
x = 64 (from 2)

最低限度: x = 4

x = 1
i = 1
x = 2 (from 1+2) // get x=1 for both
i = 2 (from 1)
i = 3 (from 2)
x = 4 (from 1)
i = 4 (from 2)

如果是原子:

最高: x = 16

x = 1
i = 1
x = 2 (from 1)
x = 4 (from 2)
i = 2 (from 1)
i = 3 (from 2)
x = 8 (from 1)
x = 16 (from 2)
i = 4 (from 1)
i = 5 (from 2)

最低限度: x = 8

x = 1
i = 1
x = 2 (from 1)
x = 4 (from 2)
i = 2 (from 1)
i = 3 (from 2)
x = 8 (from 1)
i = 4 (from 2)

答案 2 :(得分:1)

事实证明,在非原子情况下,最小值为2,最大值为512。

对于x = 2:

Process 2 (right loop) executes the [MOV r1, x] assembly instruction of line x=x*2, then switches to Process 1 (left loop). 
Process 1 loops until x=16 and i=4, then it exits.
Back to process 2, which executes [MUL r1, r1], [MOV x,r1], completing the line x=x*2. It then executes i++, yielding i=5, and exits the loop. 
The final value of x is 2.

对于x = 512:

Process 2 executes x=x*2 (x=2) and [MOV r1,i], then switches.
Process 1 loops, yielding (x=4,i=2), (x=8,i=3), (x=16,i=4), then switches.
Process 2 executes [inc r1] and [MOV i,r1]. Now i=2. Process 2 loops and executes x=x*2 (x=32), then [mov r1,i], and switches with i=2.
Process 1 loops, yielding (x=64,i=3), (x=128,i=4), then switches.
Process 2 executes [inc r1] and [MOV i,r1]. Now i=3. Process 2 loops and executes x=x*2 (x=256), then [mov r1,i], and switches.
Process 1 loops, yielding (x=512,i=4), then switches.
Process 2 executes [inc r1] and [MOV i,r1]. Now i=4.
Process 1 and 2 exit. x=512 and i=4.