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PanZezhong1725
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Apr 7, 2025
infiniop/ops/embed/README.md
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| $$ Y = \alpha_1 \cdot table_1[i_1] + \alpha_2 \cdot table_2[i_2] $$ | ||
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| - 通常 $α$ 为 1； | ||
| - 某一项的 $α$ 为 0、$table$ 为空或 $i$ 为空表示不使用； |
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table显示不正常。
建议只有table2可以为空
infiniop/ops/embed/README.md
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| `infiniopCreateEmbedDescriptor` 指针，指向将被初始化的算子描述符地址； | ||
| - `y_desc` - $\{ dT | (N, D) | (..., 1) \}$: | ||
| 算子输出 `y` 的张量描述； | ||
| - `table1_desc` - $\{ dT | (..., D) | (..., 1) \}$: |
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如果是二维你可以随便写个字母，比如 (B,D)
infiniop/ops/embed/README.md
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| 算子输入 `table1` 的张量描述； | ||
| - `table2_desc` - $\{ dT | (..., D) | (..., 1) \}$: | ||
| 算子输入 `table2` 的张量描述； | ||
| - `i1_desc` - $\{ dT_i | (N) | (1) \}$: |
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类型叫dI吧，下标不好看。把可以为nullptr这事说一下，要求alpha2必须为0
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那就是要求为空的时候几项必须对应同时表示空？也行
infiniop/ops/embed/README.md
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| 参数限制： | ||
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| - $dT$: 任意浮点类型； |
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只要支持乘加，可能是排除 bool 都行，但是不知道怎么表示
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任意浮点数、整数、无符号整数类型？我觉得是不是bool也可以有啊
| 计算输出结果； | ||
| - `table1`: | ||
| 第 1 个嵌入表； | ||
| - `table2`: |
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| <div style="background-color: lightblue; padding: 1px;"> 返回值：</div> | ||
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| - [`INFINI_STATUS_SUCCESS`], [`INFINI_STATUS_BAD_PARAM`], [`INFINI_STATUS_BAD_TENSOR_SHAPE`], [`INFINI_STATUS_BAD_TENSOR_DTYPE`], [`INFINI_STATUS_BAD_TENSOR_STRIDES`], [`INFINI_STATUS_BAD_DEVICE`]. |
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加个 INFINI_STATUS_NULL_POINTER
PanZezhong1725
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PanZezhong1725
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Apr 15, 2025
infiniop/ops/embedding/README.md
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| $$ Y = \alpha_1 \cdot table_1[i_1] + \alpha_2 \cdot table_2[i_2] $$ | ||
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| - 通常 $α$ 为 1； | ||
| - $ table2 $ 可以不使用，则公式变为 $ Y = \alpha_1 \cdot table_1[i_1] $； |
infiniop/ops/embedding/README.md
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| `infiniopHandle_t` 类型的硬件控柄。详情请看：[`InfiniopHandle_t`] | ||
| - `desc_ptr`: | ||
| `infiniopCreateEmbeddingDescriptor` 指针，指向将被初始化的算子描述符地址； | ||
| - `y_desc` - $\{ dT | (N, D) | (..., 1) \}$: |
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这个张量格式不需要用数学符号了，正常写就行
infiniop/ops/embed/README.md
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| `infiniopCreateEmbedDescriptor` 指针，指向将被初始化的算子描述符地址； | ||
| - `y_desc` - $\{ dT | (N, D) | (..., 1) \}$: | ||
| 算子输出 `y` 的张量描述； | ||
| - `table1_desc` - $\{ dT | (..., D) | (..., 1) \}$: |
Collaborator
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如果是二维你可以随便写个字母，比如 (B,D)
infiniop/ops/embed/README.md
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| 参数限制： | ||
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| - $dT$: 任意浮点类型； |
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任意浮点数、整数、无符号整数类型？我觉得是不是bool也可以有啊
Signed-off-by: YdrMaster <ydrml@hotmail.com>
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Signed-off-by: YdrMaster <ydrml@hotmail.com>
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