Suppose m101-m106 = 0.

When it is executed for the first time, the result is m100 =1m102-m106 = 0.

When M 100=0 is executed for the second time, the result is m10/= 0m102 =1m/03-m106 = 0.

The result of the third execution is m100 =10/m102 = 0m103 =1m103.

The result of the fourth execution is m100 =1=1m102 =1m103 = 0m104 = 65438+.

That is to say, the status of M 100 will be queued and written into m1kloc-0/-m106.