此程序case 5中存在分段错误（转置）此分段错误仅在输入行大于输入列的情况下发生。希望这是由于我没有相应分配内存的原因。
b= (int**)malloc(r*sizeof(int*));
for(i=0; i<c; i++)
{
b[i] = (int*)malloc(sizeof(int));
}
如果我这样做的话就是这样：
b= (int**)malloc(c*sizeof(int*));
for(i=0; i<r; i++)
{
b[i] = (int*)malloc(sizeof(int));
}
还有seg。错误，这次它没有为任何订单的矩阵产生正确的输出。
以下是我的完整代码：
#include<stdio.h>
#include<stdlib.h>
int** inputmatrix(int **a,int r, int c)
{
int i, j;
a = (int**)malloc(r*sizeof(int));
for(i=0; i<c; i++)
{
a[i] = (int*)malloc(sizeof(int));
}
printf("\n Input the Elements of the Matrix :");
for(i=0; i<r; i++)
{
for(j=0; j<c; j++)
{
scanf("%d",&a[i][j]);
}
}
return a;
}
void showmatrix(int** a, int r, int c)
{
int i, j;
for(i=0; i<r; i++)
{
printf("\n");
for(j=0; j<c; j++)
{
printf(" %d",a[i][j]);
}
}
}
int** add(int **a, int **b, int r, int c)
{
int i,j;
for(i=0; i<r; i++)
{
for(j=0; j<c; j++)
{
a[i][j] = a[i][j]+b[i][j];
}
}
return a;
}
int** multiplication(int** a, int **b, int r1, int c1, int c2)
{
int **c,i,j,k;
c = (int**)malloc(r1*sizeof(int*));
for(i=0; i<c2; i++)
{
c[i] = (int*)malloc(sizeof(int));
}
for(i=0; i<r1; i++)
{
for(j=0; j<c2; j++)
{
c[i][j] = 0;
for(k=0; k<c1; k++)
{
c[i][j] = c[i][j] + a[i][k]*b[k][j];
}
}
}
return c;
}
int minval(int **a, int r, int c)
{
int i, min;
min = a[r][0];
for(i=0; i<c; ++i)
{
if(a[r][i]<min)
{
min = a[r][i];
}
}
return min;
}
int maxval(int **a, int r, int c)
{
int i, max;
max = a[0][c];
for(i=0; i<r; ++i)
{
if(a[i][c] > max )
{
max = a[i][c];
}
}
return max;
}
void saddlepoint(int **a, int r, int c)
{
int i, j, rpos, cpos, flag = 0,sp;
for(i=0; i<r; ++i)
{
for(j=0; j<c; ++j)
{
if(a[i][j] == minval(a, i, c) && a[i][j] == maxval(a, r, j))
{
sp = a[i][j];
flag = 1;
rpos = i;
cpos = j;
}
}
}
if(flag == 1)
{
printf("\n The Saddle point of the Matrix is found at position (%d,%d) value is %d ", rpos, cpos,sp);
}
else
{
printf("\n There is no saddle point in the Matrix ");
}
}
int magicsquare(int **a, int r, int c)
{
int i, j, row_sum, col_sum, d1, d2, flag = 0;
if(r==c)
{
for(i =0 ;i<r; i++) // for digonals
{
d1 = d1 + a[i][i];
d2 = d2 + a[i][r-i-1];
}
for(i=0; i<r; i++)
{
row_sum = 0;
for(j=0; j<c; j++)
{
row_sum = row_sum + a[i][j];
}
if(row_sum == d1)
flag = 1;
else
break;
}
for(i=0; i<r; i++)
{
col_sum = 0;
for(j=0; j<c; j++)
{
col_sum = col_sum + a[j][i];
}
if(col_sum == d1)
flag =1;
else
break;
}
}
else
{
printf("\n This Matrix is not a Magic Square ");
}
return flag;
}
int** transpose(int **a, int r, int c)
{
int i, j, **b;
b= (int**)malloc(c*sizeof(int*));
for(i=0; i<r; i++)
{
b[i] = (int*)malloc(sizeof(int));
}
for(i =0; i<r; i++)
{
for(j=0; j<c; j++)
{
b[j][i] = a[i][j];
}
}
return b;
}
int main()
{
int **a, **b,r1,c1,r2,c2, i,j,ch,f;
int **c;
printf("\n enter your choice : \n1.Addition \n2.Multiplication \n3.Saddle Point \n4. Magic Square \n5.Transpose\n");
scanf("%d",&ch);
printf("\n enter the oder of matrix A :");
scanf("%d%d",&r1,&c1);
a = inputmatrix(a,r1,c1);
switch(ch)
{
case 1:
printf("\n enter the oder of matrix B :");
scanf("%d%d",&r2,&c2);
if(r1==r2 && c1==c2)
{
b = inputmatrix(b,r2,c2);
a = add(a,b,r1,c1);
printf("\n the result of the addition of matrices is :");
for(i=0; i<r1; i++)
{
printf("\n");
for(j=0;j<c1; j++)
{
printf("%d\t",a[i][j]);
}
}
}
else
{
printf("\n these matrices can't be added ");
}
break;
case 2 :
printf("\n Enter the Order of Matrix B :");
scanf("%d%d",&r2,&c2);
b = inputmatrix(b,r2,c2);
if(c1 == r2)
{
c = multiplication(a, b, r1, c1, r2);
for(i=0; i<r1; i++)
{
printf("\n");
for(j=0; j<c2; j++)
{
printf("%d\t",c[i][j]);
}
}
}
else
{
printf("\n Sorry, These Matrices Can't be Multiplied ");
}
break;
case 3 :
printf("\n The Matrix you Entered is :");
for(i=0; i<r1; i++)
{
printf("\n");
for(j=0; j<c1; j++)
{
printf(" %d",a[i][j]);
}
}
saddlepoint(a,r1,c1);
break;
case 4 :
printf("\n The Matrix you Entered is :");
for(i=0; i<r1; i++)
{
printf("\n");
for(j=0; j<c1; j++)
{
printf(" %d",a[i][j]);
}
}
int f = magicsquare(a, r1, c1);
if(f==1)
printf("\n This Matrix is a Magic Square ");
else
printf("\n This Matrix is not a Magic Square ");
break;
case 5 :
printf("\n The Matrix you enter is :");
showmatrix(a,r1,c1);
b = transpose(a,r1,c1);
printf("\n the transpose of the entered matrix is :");
for(i=0; i<c1; i++)
{
printf("\n");
for(j=0; j<r1; j++)
{
printf(" %d",b[i][j]);
}
}
break;
default : printf("\n Sorry, This is a Wrong Choice ");
}
return 0;
}
一些输出案例也在下面：
案例1：
enter your choice :
1.Addition
2.Multiplication
3.Saddle Point
4. Magic Square
5.Transpose
5
enter the oder of matrix A :3
2
Input the Elements of the Matrix :1
2
3
4
5
Segmentation fault (core dumped)
案例2：
enter your choice:
1.Addition
2.Multiplication
3.Saddle Point
4. Magic Square
5.Transpose
5
enter the oder of matrix A :2
3
Input the Elements of the Matrix :1
2
3
4
5
6
The Matrix you enter is :
1 2 3
4 5 6
the transpose of the entered matrix is :
1 4
2 5
3 6
在乘法函数中也存在一些问题，也没有显示右矩阵。
答案 0 :(得分：0)
b= (int**)malloc(r*sizeof(int*));
这会分配一个r int*元素数组。
for(i=0; i<c; i++)
这会覆盖此数组的第一个c元素。如果是c>r，那么
b[i] = (int*)malloc(sizeof(int));
上述行的行为未定义。如果c<=r，它定义明确但不是很有用，因为它为每个矩阵行分配一个元素。当您尝试访问第一列之后的元素时，它可能会在以后崩溃。
要分配包含r行和c列的矩阵，您可能希望这样做：
b = malloc(r * sizeof(int*)); /* allocate `r` rows */
for(i = 0; i < r; i++) /* for each of the `r` rows */
b[i] = malloc (c * sizeof(int)); /* allocate `c` columns */
答案 1 :(得分：0)
使用malloc声明数组时出错。 当您希望声明“r”行和“c”列的二维数组时，语句应该类似于
int **arr=malloc(r*sizeof(int *));
for(i=0;i<r;i++)
arr[i]=malloc(c*sizeof(int));
这将消除分段错误。
第一个语句声明了一个'r'整数指针数组。
for循环执行'r'次，因为你必须初始化前一个语句声明的每个'r'指针。
for循环中的语句声明了一个长度为'c'的整数数组，用于存储每行的'c'元素。 此外，它使用这些行的地址（按正确的顺序）初始化数组'arr'的每个指针。
因此我们得到'r'行的'c'元素，因此得到（r x c）矩阵。