Lines Matching refs:r2
124 int r2 = 0;
128 r2 = READ_ONCE(buf);
136 data from buf into a second private variable r2 for copying to
154 from flag and buf, or equivalently, what values r1 and r2 might end up
158 predict that r1 = 42 or r2 = -7, because neither of those values ever
163 r2 will both be 0 at the end. Or P0 might run entirely before P1
164 begins, in which case r1 and r2 will both be 1.
168 store to buf but before the store to flag. In this case, r1 and r2
170 unconditionally then we would instead have r1 = 0 and r2 = 1.)
172 However, the most interesting possibility is where r1 = 1 and r2 = 0.
193 that the undesired result r1 = 1, r2 = 0 cannot occur. The reasoning
202 P1 must load 0 from buf before P0 stores 1 to it; otherwise r2
216 r2 = 0.
459 r2 = READ_ONCE(a[r1]);
651 int r1, r2;
654 r2 = READ_ONCE(x);
657 If r1 = 5 (reading from P0's store) and r2 = 0 (reading from the
660 the r2 load in program order, so it must not read from a store that
665 and r2 = 0! This results from parallel execution of the operations
1042 r2 = READ_ONCE(*r1);
1092 int r2;
1095 r2 = READ_ONCE(*r1);
1100 and r2 = 0 at the end, in spite of the address dependency.
1107 second load executed. So why doesn't r2 end up equal to 1?
1133 int r2;
1136 r2 = READ_ONCE(*r1);
1139 then we would never get r1 = &x and r2 = 0. By the time P1 executed
1231 int r1, r2;
1234 r2 = READ_ONCE(x);
1242 If r1 = 0 and r2 = 9 at the end then P0's accesses must have executed
1266 int r2;
1269 r2 = READ_ONCE(buf);
1273 stores. If r1 = 1 and r2 = 0 at the end then there is a prop link
1280 The prop link says that in order to obtain the r1 = 1, r2 = 0 result,
1283 have propagated to P1 by the time P1's load from buf executed, so r2
1294 conclude that if an smp_rmb() fence is added, the r1 = 1, r2 = 0
1324 int r2;
1326 r2 = READ_ONCE(y);
1330 If x = 2, r0 = 1, and r2 = 1 after this code runs then there is a prop
1420 to P0's. This means that if both r1 and r2 were 0 there would be a
1474 int r1, r2;
1478 r2 = READ_ONCE(y);
1482 never end with r1 = 1 and r2 = 0. The reasoning is as follows. r1 = 1
1486 other hand, r2 = 0 means that P0's store to y, which occurs before the
1684 int r1, r2;
1688 W: r2 = READ_ONCE(y);
1692 If r2 = 0 at the end then P0's store at Y overwrites the value that
1731 int r2;
1734 r2 = READ_ONCE(z);
1739 If r0 = r1 = r2 = 1 at the end, then similar reasoning to before shows
1759 r2 = READ_ONCE(z)
1828 int r1, r2;
1834 r2 = READ_ONCE(y);
1847 cannot have r1 = 1 and r2 = 0 at the end (this is an instance of the
1856 int r1, r2, r3;
1861 r2 = READ_ONCE(y);
1869 r2 = READ_ONCE(y);
1873 and thus it could load y before x, obtaining r2 = 0 and r1 = 1.
1902 int r2, r3;
1904 r2 = READ_ONCE(y);
1911 before the store to y does, so we cannot have r2 = 1 and r3 = 0.
1944 int r2 = 0;
1948 r2 = READ_ONCE(*r1);
1957 possible final values for r2 are 6 and 0, depending on whether or not
1967 int r2 = 0;
1970 r2 = READ_ONCE(*x);
2089 int r2 = 0;
2093 Y: r2 = READ_ONCE(buf);
2170 int r2 = 0;
2175 V: r2 = buf;