forked from pi7mcrg2k/operator_optimization
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70 lines
2.1 KiB
70 lines
2.1 KiB
#include <stdio.h>
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#include <time.h>
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#include <stdlib.h>
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#define SIZE 1024
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void sparce_matmul_coo(float*, int*, int*, int,
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float*, int*, int*, int, float*, int*, int*, int*);
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int main() {
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// 矩阵 A 的 COO 格式
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float A_values[] = {1, 2, 3, 4, 5};
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int A_rowIndex[] = {0, 0, 1, 2, 2};
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int A_colIndex[] = {0, 2, 1, 0, 2};
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int A_nonZeroCount = 5;
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// 矩阵 B 的 Coo 格式
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float B_values[] = {6, 8, 7, 9};
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int B_rowIndex[] = {0, 2, 1, 2};
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int B_colIndex[] = {0, 0, 1, 2};
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int B_nonZeroCount = 4;// 结果矩阵 C 的 Coo 格式
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float C_values[SIZE];
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int C_rowIndex[SIZE];
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int C_colIndex[SIZE];
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int C_nonZeroCount = 0;
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clock_t start = clock();
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sparce_matmul_coo(A_values, A_rowIndex, A_colIndex, A_nonZeroCount,
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B_values, B_rowIndex, B_colIndex, B_nonZeroCount,
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C_values, C_rowIndex, C_colIndex, &C_nonZeroCount);
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clock_t end = clock();
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printf("基础的稀疏矩阵乘法时间:%lf\n", (double)(end-start) / CLOCKS_PER_SEC);
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}
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void sparce_matmul_coo(float* A_values, int* A_rowIndex, int* A_colIndex, int A_nonZeroCount,
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float* B_values, int* B_rowIndex, int* B_colIndex, int B_nonZeroCount,
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float* C_values, int* C_rowIndex, int* C_colIndex, int* C_nonZeroCount) {
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int currentIndex = 0;
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int i, j, k;
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int rowA, colA, rowB, colB;
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float valueA, valueB, product;
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// 遍历 A 的非零元素
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for(i=0; i<A_nonZeroCount; i++) {
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rowA = A_rowIndex[i];
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colA = A_colIndex[i];
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valueA = A_values[i];
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// 遍历 B 的非零元素
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for(j=0; j<B_nonZeroCount; j++) {
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rowB = B_rowIndex[j];
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colB = B_colIndex[j];
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valueB = B_values[j];
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// 如果 A 的列和 B 的行匹配,则计算乘积并存储结果
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if (colA == rowB) {
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product = valueA * valueB;
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// 检查是否已有此(rowA, colB) 项
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int found = 0;
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for(k=0; k<currentIndex; k++) {
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if(C_rowIndex[k] == rowA && C_colIndex[k] == colB) {
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C_values[k] += product;
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break;
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}
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}
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if (!found) {
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C_values[currentIndex] = product;
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C_rowIndex[currentIndex] = rowA;
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C_colIndex[currentIndex] = colB;
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currentIndex++;
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}
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}
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}
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}
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*C_nonZeroCount = currentIndex;
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} |