使用多线程矩阵乘法？ [英] Matrix multiplication using multiple threads?
问题描述
我应该乘以使用线程2矩阵。两件事情：我一直得到0，当我运行的程序。我也得到的信息错误（每，它说：警告：传递从兼容的指针类型'printMatrix'的参数1的粗线（这里我尝试打印输出上），另外需要注意的，那就是粗体的第一个块，我那是我在解决问题的尝试。我想我接近，但我可能不是。谁能帮助？谢谢:)
输出看起来是这样的：
A =
1 4
2 5
3 6
B =
8 7 6
5 4 3
A * B =
0 0 0
0 0 0
0 0 0
的#include＆LT; pthreads.h中＆GT;
＃包括LT＆;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;＃定义M 3
＃定义K 2
＃定义N 3结构v
{
INT I; //行
诠释J; //柱
};int类型的[M] [K] = {{1,4}，{2,5}，{3,6}};
INT B〔K] [N] = {{8,7,6}，{5,4,3}};
INT C [M] [N];无效*的WorkerThread（void *的数据）
{
INT I =（（结构V *）数据） - I标记;
INT J =（（结构V *）数据） - GT;焦耳;
INT累加器= 0;/ *这是你应该计算的已分配单元。您将需要使用的行（I）
A和B柱[J]厚积薄发的结果在累加器* /
**时int k;
对于（K = 0; K＆LT; k; k ++）
{
累加器= A [I] [K] * B [k]的[J]。
}
C [I] [J] =蓄能器;
了pthread_exit（NULL）; **
} 无效printMatrix为（int * matrixIn，诠释行，诠释列）
{
为int *矩阵= matrixIn;
INT I，J;
对于（i = 0; I＆LT;行;我++）
{
}INT主（INT ARGC，CHAR *的argv []）
{ 的pthread_t线程[M * N];
INT I，J;
INT计数器= 0;
INT numThreadsCreated = 0; / *下面的5行演示如何创建线程1来计算C [0] [0]，你
将需要创建一个循环为所有的C的细胞* /
结构V *数据=（结构V *）malloc的（的sizeof（结构V））;
DATA- I标记= 0; //指定C的排线程计算
DATA-＆GT; J = 0; //分配的C柱线程计算
在pthread_create（安培;线程[0]，NULL，的WorkerThread，数据）;
numThreadsCreated ++; / *等待所有线程打印出矩阵前完成* /
为（J = 0; J＆LT; numThreadsCreated; J ++）
{
在pthread_join（线程[J]，NULL）;
} 输出（A = \\ n）;
** printMatrix（A，3,2）; **
的printf（B = \\ n）;
** printMatrix（B，2,3）; **
的printf（A * B = \\ n）;
** printMatrix（C，M，N）**
了pthread_exit（NULL）;
}
您计划似乎已经实施了错误的编码算法矩阵乘法。
以下code的一件似乎很荒谬： -
的（K = 0; K＆LT; k; k ++）
{
累加器= A [I] [K] * B [k]的[J]。
}
C [I] [J] =蓄能器; //不是code矩阵相乘...
您应该实现这样的： -
为（i = 0; I＆LT;米;我++）{
为（J = 0; J＆LT; N; J ++）{
累加器= 0;
对于（INT东西= 0;＆东西LT; k;东西++）{
累加器累加器= + A [I] [东西] * B [东西] [J]。
}
C [I] [J] =蓄能器;
累加器= 0;
}
}
I am supposed to multiply 2 matrices using threads. Two things: I keep getting 0's when I run the program. I also get message errors(for each, it says "warning: passing argument 1 of 'printMatrix' from incompatible pointer type" on the bolded lines(where I try to print the output). Also to note, the first block that is bolded, I that was my attempt at solving the problem. I think I am close, but I may not be. Can anyone help? Thanks :) Output looks like this: A= 1 4 2 5 3 6 B= 8 7 6 5 4 3 A*B= 0 0 0 0 0 0 0 0 0
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
#define M 3
#define K 2
#define N 3
struct v
{
int i; //row
int j; //column
};
int A[M][K] = {{1,4},{2,5},{3,6}};
int B[K][N] = {{8,7,6},{5,4,3}};
int C[M][N];
void *workerThread(void *data)
{
int i=((struct v*)data)->i;
int j=((struct v*)data)->j;
int accumulator = 0;
/*this is where you should calculate the assigned Cell. You will need to use the row(i) of
A and column[j] of B. Accumulate the result in accumulator */
**int k;
for(k=0; k<k; k++)
{
accumulator = A[i][k]*B[k][j];
}
C[i][j]=accumulator;
pthread_exit(NULL);**
}
void printMatrix(int *matrixIn, int rows, int columns)
{
int *matrix = matrixIn;
int i,j;
for (i=0;i<rows;i++)
{
}
int main (int argc, char *argv[])
{
pthread_t threads[M*N];
int i,j;
int counter = 0;
int numThreadsCreated = 0;
/*the following 5 lines demonstrates how to create 1 thread to calculate C[0][0], you
will need to create a loop for all of C's cells*/
struct v *data = (struct v *)malloc(sizeof(struct v));
data->i = 0; //assign the row of C for thread to calculate
data->j = 0; //assign the column of C for thread to calculate
pthread_create(&threads[0], NULL, workerThread, data);
numThreadsCreated++;
/*wait for all the threads to finish before printing out the matrices*/
for(j=0; j < numThreadsCreated; j++)
{
pthread_join( threads[j], NULL);
}
printf("A=\n");
**printMatrix(A,3,2);**
printf("B=\n");
**printMatrix(B,2,3);**
printf("A*B=\n");
**printMatrix(C,M,N);**
pthread_exit(NULL);
}
Your program seems to have implemented a wrong coding algorithm for matrix multiplication.
The following piece of code seems absurd :-
for(k=0; k<k; k++)
{
accumulator = A[i][k]*B[k][j];
}
C[i][j]=accumulator; // Not a code for matrix multiplication...
You should implement something like :-
for(i=0;i<M;i++){
for(j=0;j<N;j++){
accumulator=0;
for(int something=0;something<K;something++){
accumulator=accumulator+A[i][something]*B[something][j];
}
C[i][j]=accumulator;
accumulator=0;
}
}
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