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?
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+1
和x=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。)
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)
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.