On 2017年03月28日 11:19, Zhang, Jerry (Junwei) wrote:
On 03/27/2017 01:53 PM, Chunming Zhou wrote:
Change-Id: I17b40aec68404e46961a9fda22dfadd1ae9d6f2c
Signed-off-by: Chunming Zhou <[email protected]>
---
drivers/gpu/drm/amd/amdgpu/gmc_v9_0.c | 6 ++++++
1 file changed, 6 insertions(+)
diff --git a/drivers/gpu/drm/amd/amdgpu/gmc_v9_0.c
b/drivers/gpu/drm/amd/amdgpu/gmc_v9_0.c
index 6625a2f..613c8f6 100644
--- a/drivers/gpu/drm/amd/amdgpu/gmc_v9_0.c
+++ b/drivers/gpu/drm/amd/amdgpu/gmc_v9_0.c
@@ -508,6 +508,12 @@ static int gmc_v9_0_vm_init(struct amdgpu_device
*adev)
* amdkfd will use VMIDs 8-15
*/
adev->vm_manager.num_ids = AMDGPU_NUM_OF_VMIDS;
+ /* Because of four level VMPTs, vm size at least is 256GB.
+ 256TB is OK as well */
If each PT size is 9, and contains 3 levels PT + 1 PD.
9 * 3 + 12(page size) + 1(at least 1 bit for PD) = 40 bits here.
block size is 9 means 512 entries each PDB/PTB. then one PDE of 4 levels
is 512* 512 *512*4KB = 512GB.
will send another patch to correct this typo.
Thanks,
David Zhou
Then it looks 1T is minimum size.
Please confirm it.
Jerry.
+ if (amdgpu_vm_size < 256) {
+ DRM_WARN("vm size at least is 256GB!\n");
+ amdgpu_vm_size = 256;
+ }
adev->vm_manager.num_level = 1;
amdgpu_vm_manager_init(adev);
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