我正在尝试执行矩阵乘法（动态内存分配），用户可以输入任何有效的矩阵乘法顺序（即column1 = row2）。 两个矩阵的相同阶数（2x2或3x3）的输出导致正确的计算，但是像mat1 2x3和amp; mat2 3x2..give分段错误。当我事先进行内存分配时，我无法确定如何访问任何非法内存。
请小心地告诉我，如果我犯了一些愚蠢的错误，请原谅我......以下是完整的代码：
#include<stdio.h>
#include<stdlib.h>
main(){
int **mat1, **mat2,**res,i,j,r1,c1,r2,c2;
printf("\nEnter the Order of the First matrix...\n");
scanf("%d %d",&r1,&c1);
printf("\nEnter the Order of the Second matrix...\n");
scanf("%d %d",&r2,&c2);
if(c1!=r2){
printf("Invalid Order of matrix");
exit(EXIT_SUCCESS);
}
mat1= (int**) malloc(r1*sizeof(int*));
for(i=0;i<c1;i++)
mat1[i]=(int*)malloc(c1*sizeof(int));
mat2= (int**) malloc(r2*sizeof(int*));
for(i=0;i<c2;i++)
mat2[i]=(int*)malloc(c2*sizeof(int));
res=(int**)calloc(r1,sizeof(int*));
for(i=0;i<c2;i++)
res[i]=(int*)calloc(c2,sizeof(int));
//Input Matrix1
for(i=0;i<r1;i++)
for(j=0;j<c1;j++)
scanf("%d",&mat1[i][j]);
//Input Matrix2
for(i=0;i<r2;i++)
for(j=0;j<c2;j++)
scanf("%d",&mat2[i][j]);
//Printing Input Matrix 1 and 2
printf("\n Entered Matrix 1: \n");
for(i=0;i<r1;i++){
for(j=0;j<c1;j++)
printf("%d ",mat1[i][j]);
printf("\n");
}
printf("\n Entered Matrix 2: \n");
for(i=0;i<r2;i++){
for(j=0;j<c2;j++)
printf("%d ",mat2[i][j]);
printf("\n");
}
//Computation
//Multiplication
for(i=0;i<r1;i++){
for(j=0;j<c2;j++){
res[i][j]=0;
for(k=0;k<c1;k++)
res[i][j]+= mat1[i][k]*mat2[k][j];
}
printf("\n");
}
printf("\nThe Multiplication of two matrix is\n");
for(i=0;i<r1;i++){
printf("\n");
for(j=0;j<c2;j++)
printf("%d\t",res[i][j]);
}
printf("\n");
/* Addition
for(i=0;i<r1;i++)
for(j=0;j<c2;j++)
res[i][j]=mat1[i][j]+mat2[i][j];
printf("\nThe Addition of two matrix is\n");
for(i=0;i<r1;i++){
printf("\n");
for(j=0;j<c2;j++)
printf("%d\t",res[i][j]);
}
*/
return 0;}
答案 0 :(得分：2)
mat1= (int**) malloc(r1*sizeof(int*));
for(i=0;i<c1;i++) < c1 instead of r1
mat1[i]=(int*)malloc(c1*sizeof(int));
mat2= (int**) malloc(r2*sizeof(int*));
for(i=0;i<c2;i++) < c2 instead of r2
mat2[i]=(int*)malloc(c2*sizeof(int));
您在for而不是r1 / 2中使用c1 / 2.
如果r1＆lt; c1，你最终在你分配的内存之外。
如果r1> c1，你最终得到了未初始化的指针。
与此问题无关但你应该写int main()而不是main()，第二个被接受，但第一个更容易阅读。
答案 1 :(得分：1)
这是任何有效矩阵乘法的代码.... 对“查询”感到自由....
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
int main()
{
int *ans,*first,*second;
int *A,*B,*C;
int i,j,k=0;
int rowA,colA,sizeA,sizeB,sizeC;
int rowB,colB;
printf("Enter the row's and column of 1st matrix\n");
scanf("%d%d",&rowA,&colA);
printf("Enter the row's and column of 2nd matrix\n");
scanf("%d%d",&rowB,&colB);
if(colA!=rowB)
{
printf("Error => colA must be equal to rowB\n");
getch();
exit(EXIT_SUCCESS);
}
sizeC = rowA*colB;
sizeA = rowA*colA;
sizeB = rowB*colB;
A = (int *)malloc(sizeA*sizeof(int *));
first = A;
B = (int *)malloc(sizeB*sizeof(int *));
second = B;
C = (int *)malloc(sizeC*sizeof(int *));
ans = C;
printf("Enter the elements of the first matrix A\n");
for(i=0;i<sizeA;i++,first++)
scanf("%d",first);
printf("Enter the elements of the second matrix B\n");
for(i=0;i<sizeB;i++,second++)
scanf("%d",second);
first=A;
second= B;
if(rowA==1 && colB==1)
{
for(i=0;i<rowA;i++)
{
for(j=0;j<colB;j++)
{
*ans=0;
for(k=0;k<rowB;k++)
*ans = *ans + (*(first + (k + i*colA))) * (*(second + (j+k*colB)));
ans++;
}//j
}//i
}//if
else
{
for(i=0;i<rowA;i++)
{
for(j=0;j<colB;j++)
{
*ans=0;
for(k=0;k<rowB;k++)
*ans = *ans + (*(first + (k + i*colA))) * (*(second + (j+k*rowB)));
ans++;
}//j
}//i
}
printf("\nThe value of matrix 'C' = \n");
ans = C;
for(i=0;i<rowA;i++)
{
printf("\n");
for(j=0;j<colB;j++,ans++)
printf("%d\t",*ans);
}
free(A);
free(B);
free(C);
getch();
}
答案 2 :(得分：0)
我附上了Matrix Multiplication的代码，用于动态内存分配的任何正确顺序
为了完整性，我使用了3种不同的矩阵乘法方法：一个函数double** multMatrixpf（参见等效函数Fortran / Pascal）和两个子程序/过程（Fortran / Pascal之类），其中首先是void multMatrixp你需要allocate_mem(&c,ro1,co2)外部和第二个子程序void multMatrixpp，矩阵c1被分配在子程序中。所有这三种方法都给出了相同的结果。
我也使用不同的方法初始化数组。
#include <stdio.h>
#include <stdlib.h>
void allocate_mem(double*** arr, int rows, int cols);
void deallocate_mem(double*** arr, int n);
double** readMatrixf(int rows, int cols);
void readMatrix(double ***a, int rows,int cols);
void printMatrix(double** a, int rows, int cols);
void printMatrixE(double** a, int rows, int cols);
void multMatrixp(double **A, double **B, double **C,int r1,int c1,int r2,int c2);
void multMatrixpp(double **A, double **B, double ***C,int ro1,int co1,int ro2,int co2);
double** multMatrixpf(double **A, double **B, int ro1,int co1,int ro2,int co2);
//______________________________________________________________________________
int main()
{
int ro1, co1, ro2, co2;
double **a1, **b1, **c1;
ro1=2; co1=3;
ro2=3; co2=4;
printf("Ex1:__________________________________________________"
"__________________________ \n");
double (*(a[])) = {
(double[]) { 1, 3, 5},
(double[]) {2, 4, 0}};
double (*(b[])) = {
(double[]) {6, 2, 4, 8},
(double[]) {1, 7, 0, 9},
(double[]) {0, 3, 5, 1}};
printMatrix(a,ro1,co1);
printMatrix(b,ro2,co2);
printf("MatMult \n");
double **c;
allocate_mem(&c,ro1,co2);
multMatrixp(a, b, c, ro1, co1, ro2, co2);
printMatrix(c,ro1,co2);
printMatrixE(c,ro1,co2);
deallocate_mem(&c,ro1);
printf("Ex2:__________________________________________________"
"__________________________ \n");
scanf("%d%d", &ro1, &co1);
readMatrix(&a1,ro1,co1);
printMatrix(a1,ro1,co1);
//deallocate_mem(&a1,ro1);
//printMatrix(a1,ro1,co1);
scanf("%d%d", &ro2, &co2);
readMatrix(&b1,ro2,co2);
printMatrix(b1,ro2,co2);
printf("MatMult \n");
multMatrixpp(a1, b1, &c1, ro1, co1, ro2, co2);
printMatrix(c1,ro1,co2);
printMatrixE(c1,ro1,co2);
deallocate_mem(&a1,ro1);
deallocate_mem(&b1,ro2);
deallocate_mem(&c1,ro1);
printf("Ex3:__________________________________________________"
"__________________________ \n");
scanf("%d%d", &ro1, &co1);
a1=readMatrixf(ro1,co1);
printMatrix(a1,ro1,co1);
//deallocate_mem(&a1,ro1);
//printMatrix(a1,ro1,co1);
scanf("%d%d", &ro2, &co2);
b1=readMatrixf(ro2,co2);
printMatrix(b1,ro2,co2);
printf("MatMult \n");
c1=multMatrixpf(a1, b1, ro1, co1, ro2, co2);
printMatrix(c1,ro1,co2);
printMatrixE(c1,ro1,co2);
deallocate_mem(&a1,ro1);
deallocate_mem(&b1,ro2);
deallocate_mem(&c1,ro1);
return 0;
}
//______________________________________________________________________________
void allocate_mem(double*** arr, int rows, int cols)
{
int i;
*arr = (double**)malloc(rows*sizeof(double*));
for( i=0; i<rows; i++)
(*arr)[i] = (double*)malloc(cols*sizeof(double));
}
//______________________________________________________________________________
void deallocate_mem(double*** arr, int rows){
int i;
for (i = 0; i < rows; i++)
free((*arr)[i]);
free(*arr);
}
//______________________________________________________________________________
double** readMatrixf(int rows, int cols)
{
double **a; // Define a local pointer to keep rest of the code intact
int i, j;
a= (double**) malloc(rows*sizeof(double*));
for(i=0;i<rows;i++)
a[i]=(double*)malloc(cols*sizeof(double));
for(i=0;i<rows;i++)
for(j=0;j<cols;j++)
scanf("%lf",&a[i][j]);
return a;
}
//______________________________________________________________________________
void readMatrix(double ***a, int rows,int cols)
{
int i, j;
*a= (double**) malloc(rows*sizeof(double*));
for(i=0;i<rows;i++)
(*a)[i]=(double*)malloc(cols*sizeof(double));
for(i=0;i<rows;i++)
for(j=0;j<cols;j++)
scanf("%lf",&(*a)[i][j]);
}
//______________________________________________________________________________
void printMatrix(double** a, int rows, int cols)
{
int i, j;
printf("Matrix[%d][%d]\n",rows,cols);
for(i=0;i<rows;i++){
for(j=0;j<cols;j++)
printf("%8.3lf ",a[i][j]);
printf("\n");
}
printf("\n");
}
//______________________________________________________________________________
void printMatrixE(double** a, int rows, int cols)
{
int i, j;
printf("Matrix[%d][%d]\n",rows,cols);
for(i=0;i<rows;i++){
for(j=0;j<cols;j++)
printf("%9.2e ",a[i][j]);
printf("\n");
}
printf("\n");
}
//______________________________________________________________________________
void multMatrixp(double **A, double **B, double **C,int ro1,int co1,int ro2,int co2)
{
int i, j, k;
for(i = 0; i < ro1; i++) {
for(j = 0; j < co2; j++) {
C[i][j] = 0;
for(k = 0; k < co1; k++) {
C[i][j] += A[i][k] * B[k][j];
}
}
}
}
//______________________________________________________________________________
void multMatrixpp(double **A, double **B, double ***C,int ro1,int co1,int ro2,int co2)
{
int i, j, k;
*C= (double**) malloc(ro1*sizeof(double*));
for(i=0;i<ro1;i++)
(*C)[i]=(double*)malloc(co2*sizeof(double));
for(i = 0; i < ro1; i++) {
for(j = 0; j < co2; j++) {
(*C)[i][j] = 0.0;
for(k = 0; k < co1; k++) {
(*C)[i][j] += A[i][k] * B[k][j];
}
}
}
}
//______________________________________________________________________________
double** multMatrixpf(double **A, double **B, int ro1,int co1,int ro2,int co2)
{
int i, j, k;
double **C;
C= (double**) malloc(ro1*sizeof(double*));
for(i=0;i<ro1;i++)
C[i]=(double*)malloc(co2*sizeof(double));
for(i = 0; i < ro1; i++) {
for(j = 0; j < co2; j++) {
C[i][j] = 0.0;
for(k = 0; k < co1; k++) {
C[i][j] += A[i][k] * B[k][j];
}
}
}
return C;
}
作为输入矩阵我们有in.txt
4 4
1 1 1 1
2 4 8 16
3 9 27 81
4 16 64 256
4 3
4.0 -3.0 4.0
-13.0 19.0 -7.0
3.0 -2.0 7.0
-1.0 1.0 -1.0
3 4
1 2 -2 0
-3 4 7 2
6 0 3 1
4 2
-1 3
0 9
1 -11
4 -5
在unix中，如cmmd行执行命令：
$ time ./Matmult＆lt; in.txt＆gt; out.txt
你得到了输出
out.txt
Ex1:____________________________________________________________________________
Matrix[2][3]
1.000 3.000 5.000
2.000 4.000 0.000
Matrix[3][4]
6.000 2.000 4.000 8.000
1.000 7.000 0.000 9.000
0.000 3.000 5.000 1.000
MatMult
Matrix[2][4]
9.000 38.000 29.000 40.000
16.000 32.000 8.000 52.000
Matrix[2][4]
9.00e+00 3.80e+01 2.90e+01 4.00e+01
1.60e+01 3.20e+01 8.00e+00 5.20e+01
Ex2:____________________________________________________________________________
Matrix[4][4]
1.000 1.000 1.000 1.000
2.000 4.000 8.000 16.000
3.000 9.000 27.000 81.000
4.000 16.000 64.000 256.000
Matrix[4][3]
4.000 -3.000 4.000
-13.000 19.000 -7.000
3.000 -2.000 7.000
-1.000 1.000 -1.000
MatMult
Matrix[4][3]
-7.000 15.000 3.000
-36.000 70.000 20.000
-105.000 189.000 57.000
-256.000 420.000 96.000
Matrix[4][3]
-7.00e+00 1.50e+01 3.00e+00
-3.60e+01 7.00e+01 2.00e+01
-1.05e+02 1.89e+02 5.70e+01
-2.56e+02 4.20e+02 9.60e+01
Ex3:____________________________________________________________________________
Matrix[3][4]
1.000 2.000 -2.000 0.000
-3.000 4.000 7.000 2.000
6.000 0.000 3.000 1.000
Matrix[4][2]
-1.000 3.000
0.000 9.000
1.000 -11.000
4.000 -5.000
MatMult
Matrix[3][2]
-3.000 43.000
18.000 -60.000
1.000 -20.000
Matrix[3][2]
-3.00e+00 4.30e+01
1.80e+01 -6.00e+01
1.00e+00 -2.00e+01
答案 3 :(得分：0)
在计算矩阵乘法时，您必须运行从0到c2的k循环，而不是从0到c1。