While replacing linalg.copy with the more desired memref.copy
I found a bug in the support library for rank 0 memref copying.
The code would loop for something like the following, since there
is code for no-rank and rank > 0, but rank == 0 was unexpected.
memref.copy %0, %1: memref<f32> to memref<f32>
Note that a "regression test" for this will follow using the
sparse compiler migration to memref.copy which exercises this
case many times.
Reviewed By: herhut
Differential Revision: https://reviews.llvm.org/D106036
DynamicMemRefType<char> dst(*dstArg);
int64_t rank = src.rank;
+ char *srcPtr = src.data + src.offset * elemSize;
+ char *dstPtr = dst.data + dst.offset * elemSize;
+
+ if (rank == 0) {
+ memcpy(dstPtr, srcPtr, elemSize);
+ return;
+ }
+
int64_t *indices = static_cast<int64_t *>(alloca(sizeof(int64_t) * rank));
int64_t *srcStrides = static_cast<int64_t *>(alloca(sizeof(int64_t) * rank));
int64_t *dstStrides = static_cast<int64_t *>(alloca(sizeof(int64_t) * rank));
- char *srcPtr = src.data + src.offset * elemSize;
- char *dstPtr = dst.data + dst.offset * elemSize;
-
// Initialize index and scale strides.
for (int rankp = 0; rankp < rank; ++rankp) {
indices[rankp] = 0;