The section "Control Dependencies" in memory-barriers.txt has the following text:
662 In addition, you need to be careful what you do with the local variable 'q', 663 otherwise the compiler might be able to guess the value and again remove 664 the needed conditional. For example: 665 666 q = ACCESS_ONCE(a); 667 if (q % MAX) { 668 barrier(); 669 ACCESS_ONCE(b) = p; 670 do_something(); 671 } else { 672 barrier(); 673 ACCESS_ONCE(b) = p; 674 do_something_else(); 675 } 676 677 If MAX is defined to be 1, then the compiler knows that (q % MAX) is 678 equal to zero, in which case the compiler is within its rights to 679 transform the above code into the following: 680 681 q = ACCESS_ONCE(a); 682 ACCESS_ONCE(b) = p; 683 do_something_else(); Given that there is an explicit barrier() in both the branches of if/else statement, how can the above transformation happen? The compiler cannot just remove the barrier(), right? I think it will transform to the following if MAX is defined to 1: q = ACCESS_ONCE(a); barrier(); ACCESS_ONCE(b) = p; do_something_else(); and hence the ordering will be preserved. What am I missing here? -- Pranith -- To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majord...@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/