Question

I have a global shared variable and that is being updated 5 times by each of the 5 threads spawned. As per my understanding the increment operation is consisting of 3 instructions

load  reg, M
inc reg
store reg, M

因此,我要问,在这种情形下,在5条深线中任意搭桥会带来最大和最低价值。

So according to me the maximum value will be 25 ( I am 100% sure that it can be more than 25) and the minimum value is 5. But I am not so sure on minimum value. Can it be less that 5 in some arbitrary interleaving ? Any inputs will be greatly appreciated.

/* Global Variable */
int var = 0;

/* Thread  function */
void thread_func()
{
     for(int c = 0; c < 5; c++)
             var++;
}

Answer 1

Given your definition of increment, I agree with your max of 25.

然而,我认为,在以下情况下,这名人可能是2人。我点名的是A、B、C、D和E。

A loads 0
C, D, E run to completion
B runs through 4 of its 5 iterations.
A increments 0 to 1 and stores the result (1).
B loads 1
A runs to completion
B increments 1 to 2 and stores 2.

Answer 2

如果我采用jtdubs的相同逻辑,那么在以下情况下,最低价值应为1。

让我们使用与A、B、C、D和E一样的5条镜头。

A loads 0
B, C, D, E run to completion and incremented to maximum value 20 (5 increments by each of 4 threads).
A increments 0 to 1 and store the result 1.

Answer 3

我同意至少2项(而不是1项)。

The minimum equals 1 solution ignores the fact that A still hasn t run to completion after it stores 1 in the shared memory. With no other thread left to "interfere", thread A must still run through the remaining 4 iterations ending with the result 5.

What the minimum of 2 solution enables is an end-game between the two remaining threads A and B, after all other threads have finished running, leading to the minimum possible outcome. B "wastes" 4 iterations only to load 1 again, increment it and store 2 after A has run to completion.

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