问题
Based on the following code, I need to find the minimum and maximum final values of x
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
I figured that the minimum value x can have is 8, if the the loops are executed in order. And the maximum value x can have is 16, if the program enters one of the loops first, switches to the other loop and execute it until x=8 and i=4, and finishes the first loop, then x=16 and i=5. Is this correct? Am I missing any case where x could be either greater or lower?
回答1:
The answers you came up with are correct!
回答2:
It depends on whether i=i+1 and x=x*2 are atomic operations (meaning nothing can happen between the value of i is got and before it's set).
If not atomic:
Maximum: 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)
Minimum: 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)
If atomic:
Maximum: 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)
Minimum: 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)
回答3:
It turns out the minimum was 2 and maximum 512 in the non-atomic case.
For 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.
For 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.
来源：https://stackoverflow.com/questions/14988055/maximum-and-minimum-value-of-cobegin-coend-block