habanalabs: share a single ctx-mutex between all MMUs
Multiple locks are usually a source of problems, which in the MMU case can be avoided since it is relatively rare that both MMU tables are updated at the same time. Therefore, use a single shared lock instead of two separate ones. Signed-off-by: Moti Haimovski <mhaimovski@habana.ai> Reviewed-by: Oded Gabbay <ogabbay@kernel.org> Signed-off-by: Oded Gabbay <ogabbay@kernel.org>
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@ -81,6 +81,8 @@ int hl_mmu_ctx_init(struct hl_ctx *ctx)
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if (!hdev->mmu_enable)
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if (!hdev->mmu_enable)
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return 0;
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return 0;
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mutex_init(&ctx->mmu_lock);
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if (hdev->mmu_func[MMU_DR_PGT].ctx_init != NULL) {
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if (hdev->mmu_func[MMU_DR_PGT].ctx_init != NULL) {
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rc = hdev->mmu_func[MMU_DR_PGT].ctx_init(ctx);
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rc = hdev->mmu_func[MMU_DR_PGT].ctx_init(ctx);
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if (rc)
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if (rc)
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@ -115,6 +117,8 @@ void hl_mmu_ctx_fini(struct hl_ctx *ctx)
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if (hdev->mmu_func[MMU_HR_PGT].ctx_fini != NULL)
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if (hdev->mmu_func[MMU_HR_PGT].ctx_fini != NULL)
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hdev->mmu_func[MMU_HR_PGT].ctx_fini(ctx);
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hdev->mmu_func[MMU_HR_PGT].ctx_fini(ctx);
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mutex_destroy(&ctx->mmu_lock);
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}
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}
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/*
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/*
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@ -481,9 +481,7 @@ static void hl_mmu_v1_fini(struct hl_device *hdev)
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*/
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*/
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static int hl_mmu_v1_ctx_init(struct hl_ctx *ctx)
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static int hl_mmu_v1_ctx_init(struct hl_ctx *ctx)
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{
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{
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mutex_init(&ctx->mmu_lock);
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hash_init(ctx->mmu_shadow_hash);
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hash_init(ctx->mmu_shadow_hash);
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return dram_default_mapping_init(ctx);
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return dram_default_mapping_init(ctx);
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}
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}
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@ -516,8 +514,6 @@ static void hl_mmu_v1_ctx_fini(struct hl_ctx *ctx)
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pgt_info->phys_addr, ctx->asid, pgt_info->num_of_ptes);
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pgt_info->phys_addr, ctx->asid, pgt_info->num_of_ptes);
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_free_hop(ctx, pgt_info);
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_free_hop(ctx, pgt_info);
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}
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}
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mutex_destroy(&ctx->mmu_lock);
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}
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}
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static int _hl_mmu_v1_unmap(struct hl_ctx *ctx,
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static int _hl_mmu_v1_unmap(struct hl_ctx *ctx,
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