C 使用指针进行矩阵乘法时出错

我正在尝试进行矩阵乘法优化，如下所示：

期望输出矩阵为

  matrix C 

  2    2 

  2    2

但对于如上所示的使用指针的矩阵乘法优化，输出是不同的

   matrix C

   0    0 

   7    9

也应该是

A= (int*)malloc(sizeof(int)*(n * n));//not sizeof(int*)
B= (int*)malloc(sizeof(int)*(n * n));
C= (int*)calloc(n*n, sizeof(int));

---

矩阵可以看作是2D数组，这样您就可以使用数组寻址，这有助于读取代码

我会像这样实现动态数组乘法

这可能是一种可能的实现方法，可以将2个矩阵相乘并动态存储结果，我假设您可以通过命令行将矩阵输入到程序中

#include<stdio.h>
//this function returns a pointer to the new array. 
int **makeMatrix(int,int);

int **makeMatrix(ROWS,COLS){
int i=0;
//your code u use sizeof(*int) this is wrong suppose you use a 64bit machine 
//then sizeof(*int) is going to be 8bytes(64bits)!, whereas an integer can be 32bits, 
//the size of the integer is completely compiler dependent, does not matter if you 
//have 64bit hardware.


int **array1 = malloc(sizeof(int *)*ROWS);
for(i=0;i<Columns;i++)
array1[i] = malloc(sizeof(int)*COlUMNS);
return array1;
//we are making a 2D array here so we can use array notation to address the memory
//A[row][column] is the same as *(*(A + row) + column), this makes the code neat unlike
//your method which is also very correct but i prefer the old array notation.
}

int main(void){

int set_row1,set_row2,set_row3,set_column1,set_column2,set_column3;
int row1,column1,column;

int **array1 = makeMatrix(set_row1,set_column1);
int **array2 = makeMatrix(set_row2,set_column2);
int **result = makeMatrix(set_row3,set_column3);
//now we have 3 matrices that we can use//i assume you initialise them somehow 
 //with numbers

//now the multiplication part.we can use normal array notation since we are 
// using   essentialy a 2D array, your method is also corret where you allocate the 
// entire array on a boxed piece of memory.  

//also need to check if the 2 matrix can be multiplied.
 int temp =0;
if(Column1 == Row2) //they can be multiplied together
{

for(row1=0;row1<set_row1;row1++){
   for(column2=0;column2<set_column2;column2++){
       for(column1=0;column1<set_column1;column1++){
        temp += array1[row1][column1]*array2[column1][column2]
        }
   result[row1][column2] = temp;
   }
}
else
//do something

#包括
//此函数返回指向新数组的指针。
int**makeMatrix（int，int）；
int**makeMatrix（行、列）{
int i=0；
//您的代码u使用sizeof（*int）假设您使用64位机器，这是错误的
//那么sizeof（*int）将是8字节（64位）！，而整数可以是32位，
//整数的大小完全取决于编译器，如果
//拥有64位硬件。
int**array1=malloc（sizeof（int*）*行）；
对于（i=0；i如何显示您的预期输出以及您得到了什么？将
int*
的倍数分配到
int
应该在的位置，然后尝试访问
int*
，就好像它是
double*
？在退出最内部的循环并执行
C[i*n+j]之后，考虑
j
的值=sum
您应该看到逻辑错误。将矩阵大小都取为A=2*2，B=2*2，并且A和B中的每个元素值都取为1，答案应该是C[0]=2，C[1]=2，C[2]=2，C[3]=2。但答案不同
   matrix C

   0    0 

   7    9

sum=0;
for ( j = 0 ;  j< n ; j++ )
{
    double c =A[i*n + k];
    double d =B[k*n + j];
    sum+=c*d;
}
C[i*n + j] =sum;

for ( i = 0 ; i < n ; i++ ){
    for ( k = 0 ; k < n ; k++ ){
        for ( j = 0 ;  j< n ; j++ ){
            C[i*n + j] +=A[i*n + k]*B[k*n + j];//C initialized by 0
        }
    }
}

A= (int*)malloc(sizeof(int)*(n * n));//not sizeof(int*)
B= (int*)malloc(sizeof(int)*(n * n));
C= (int*)calloc(n*n, sizeof(int));

#include<stdio.h>
//this function returns a pointer to the new array. 
int **makeMatrix(int,int);

int **makeMatrix(ROWS,COLS){
int i=0;
//your code u use sizeof(*int) this is wrong suppose you use a 64bit machine 
//then sizeof(*int) is going to be 8bytes(64bits)!, whereas an integer can be 32bits, 
//the size of the integer is completely compiler dependent, does not matter if you 
//have 64bit hardware.


int **array1 = malloc(sizeof(int *)*ROWS);
for(i=0;i<Columns;i++)
array1[i] = malloc(sizeof(int)*COlUMNS);
return array1;
//we are making a 2D array here so we can use array notation to address the memory
//A[row][column] is the same as *(*(A + row) + column), this makes the code neat unlike
//your method which is also very correct but i prefer the old array notation.
}

int main(void){

int set_row1,set_row2,set_row3,set_column1,set_column2,set_column3;
int row1,column1,column;

int **array1 = makeMatrix(set_row1,set_column1);
int **array2 = makeMatrix(set_row2,set_column2);
int **result = makeMatrix(set_row3,set_column3);
//now we have 3 matrices that we can use//i assume you initialise them somehow 
 //with numbers

//now the multiplication part.we can use normal array notation since we are 
// using   essentialy a 2D array, your method is also corret where you allocate the 
// entire array on a boxed piece of memory.  

//also need to check if the 2 matrix can be multiplied.
 int temp =0;
if(Column1 == Row2) //they can be multiplied together
{

for(row1=0;row1<set_row1;row1++){
   for(column2=0;column2<set_column2;column2++){
       for(column1=0;column1<set_column1;column1++){
        temp += array1[row1][column1]*array2[column1][column2]
        }
   result[row1][column2] = temp;
   }
}
else
//do something