Test ISA11

Illustration of a cornercase relative to the atomicity axiom in the total order formulation. Although the store g lies between the read b and the store c in all valid global memory order, the test is allowed. This is due to store a lying between g and c.

Let “<” be a valid global memory order, we show that we must have b < g < a < c.

  1. We first have a < c, as the final value of x is written by c.
  2. Furthermore, b < d (b is a read-acquire), d < e (the final value of y is written by e) and e < g (g is a write-release). Thus, we have b < g by transitivity.
  3. Now, let us consider the relative positions of the writes g and a. We cannot have a < g: if so, as b < g (see above), we would contradict the atomicity axiom. Hence we must have g < a, as “<” is a total order.
  4. We conclude b < g < a < c.



  uint64_t x;
  uint64_t y;
  0:s0=x; 0:s1=y;
  1:s0=x; 1:s1=y;

 P0                     |  P1                     ;
 li t1,2                | li t3,2                 ;
 li t2,1                | li t4,1                 ;
 sd t1,0(s0)            | sd t3,0(s1)             ;
 amoor.d.aq a0,t2,(s0)  | amoswap.d.rl x0,t4,(s0) ;
 sd t2,0(s1)            |                         ;

exists 0:a0=2 /\ x=3 /\ y=2