janetsc commented on code in PR #13181:
URL: https://github.com/apache/tvm/pull/13181#discussion_r1005850161
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src/runtime/hexagon/hexagon_hvx.cc:
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@@ -31,25 +31,28 @@ namespace tvm {
namespace runtime {
namespace hexagon {
-HexagonHvx::HexagonHvx() {
- // Reserve HVX.
- int res = qurt_hvx_reserve(QURT_HVX_RESERVE_ALL_AVAILABLE);
- CHECK((res != QURT_HVX_RESERVE_NOT_SUPPORTED) && (res !=
QURT_HVX_RESERVE_NOT_SUCCESSFUL))
- << "error reserving HVX: " << res;
+HexagonHvx::HexagonHvx() { Acquire(); }
- // Lock HVX.
+HexagonHvx::~HexagonHvx() { Release(); }
+
+void HexagonHvx::Acquire() {
+ reserved_count_ = qurt_hvx_reserve(QURT_HVX_RESERVE_ALL);
+ CHECK(reserved_count_ == QURT_HVX_RESERVE_ALL) << "error reserving HVX: " <<
reserved_count_;
+}
+
+void HexagonHvx::Release() {
+ int rel = qurt_hvx_cancel_reserve();
+ CHECK(rel == 0) << "error releasing HVX: " << rel;
+}
+
+void HexagonHvx::Lock() {
int lck = qurt_hvx_lock(QURT_HVX_MODE_128B);
CHECK(lck == 0) << "error locking HVX: " << lck;
}
-HexagonHvx::~HexagonHvx() {
Review Comment:
This is by design. Reserve/release happen in the constructor/destructors,
and lock/unlock must happen on the thread that is dedicated to that resource.
In the case of HVX, it will be called 4 times.
You bring up a good point, which is what happens if one of the threads
crashes before it can unlock? I tested offline removing the unlock. (There's
not really a good way to incorporate that test, because other tests that run in
parallel depend on the resources being acquired.)
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