How do I access the memory using pointers for a multidimensional array?

Question

I have been learning memory allocation and pointers lately, and I made this program which will ask the user to input the dimensions of a matrix and enter it's elements, after which it displays the elements in a matrix format. Here's the code that I've typed.

#include"stdio.h"
#include"stdlib.h"

int *minput();

int *minput(int x,int y)
{
    int *M;
    M=(int*)malloc(x*y*sizeof(int));
    for(int i=0;i<=(x-1);i++)
    {
        for(int j=0;j<=(y-1);j++)
        {
            printf("A(%d,%d)=",i+1,j+1);
            scanf("%d",(M+i+j));
        }
    }
    return M;
}

int main()
{
    int *A,a,b;
    printf("Matrix is (m*n)\n\n");
    printf("m=");
    scanf("%d",&a);
    printf("n=");
    scanf("%d",&b);
    A=minput(a,b);
    printf("\n");

    for(int k=0;k<=(a-1);k++)
    {
        for(int l=0;l<=(b-1);l++)
        {
            printf("%d ",*(A+k+l));
        }
        printf("\n");
    }
    free(A);
    return 0;
}

However when I gave my inputs, I got this:

Matrix is (m*n)

m=3
n=3
A(1,1)=1
A(1,2)=2
A(1,3)=3
A(2,1)=4
A(2,2)=5
A(2,3)=6
A(3,1)=7
A(3,2)=8
A(3,3)=9

1 4 7
4 7 8
7 8 9

What's wrong? Amn't I supposed to get

1 2 3
4 5 6
7 8 9

Is there anything I had made wrong in my code?

Answer 1

You get incorrect output because *(A+k+l) is not the right way of accessing matrix element at matrix[k][l] .

For addressing a matrix stored as a "flat" array you need to multiply the value of one of the indexes by the size of the opposite dimension. Depending on which index you multiply you get either a row-major order or a column-major order.

You need to apply the same fix to (M+i+j) inside minput function.

// Input
scanf("%d",(M+y*i+j));
...
// Output
printf("%d ",*(A+b*k+l));

The idea behind multiplying k by b is to make k "count faster". For each increment of k by 1 you need to skip an entire row of b elements. In your three-column matrix example, if you would like to access elements at A[k][0] (the initial column) of your matrix for each row, your index would count by three: 0 , 3 , 6 . This is accomplished by multiplying k by b . The rest is the same as the usual pointer arithmetic: *(A+b*k+l) is equivalent to A[b*k+l] .

Demo.

Answer 2

Lets take a close look at one of your loops (I use the input loop, but the error is in the output loop as well):

for(int i=0;i<=(x-1);i++)
{
    for(int j=0;j<=(y-1);j++)
    {
        printf("A(%d,%d)=",i+1,j+1);
        scanf("%d",(M+i+j));
    }
}

The first iteration of the outer loop, i is zero. Then the inner loop runs, and we read into (in turn) M+0+0 , M+0+1 and M+0+2 .

Then we run the second iteration of the outer loop, where the inner loop will read into M+1+0 , M+1+1 and M+1+2 .

In these two iterations of the outer loop you will read into M+1 and M+2 twice . That's because M+0+1 and M+1+0 are the same element .

---

To fix this, lets take a look at your "matrix" as it is in memory

+---+---+---+---+---+---+---+---+---+
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
+---+---+---+---+---+---+---+---+---+

The numbers inside are the indexes. Now, 0 , 1 and 2 is the first row. 3 , 4 and 5 is the second. And 6 , 7 and 8 is the last.

From this we can see that to get to the next line, we have to add the number of columns.

From this we get the formula line * number_of_columns + column to get the index.

Putting it in place, your input loop could look like

for(int i=0;i<x;i++)
{
    for(int j=0;j<y;j++)
    {
        printf("A(%d,%d)=",i+1,j+1);
        scanf("%d",M+i*y+j);
    }
}

Answer 3

To choose the address in which to write a value, you use:

M+i+j

Let's try that for a few values of (i,j) :

0,0  -> M + 0
0,1  -> M + 1
0,2  -> M + 2  // So far so good.
1,0  -> M + 1  // not right
1,1  -> M + 2  // not right
... etc.

You want M + (i * x) + j . For x == 3 :

0,0  -> M + 0
0,1  -> M + 1
0,2  -> M + 2  
1,0  -> M + 3  
1,1  -> M + 4  
... etc. 

The same goes for the pointer arithmetic when reading from the same memory.

---

Furthermore, since the pointer just goes up by one each time, you could get the same behaviour with:

int *m = M;
for(int i=0;i<x;i++)
{
    for(int j=0;j<=(y-1);j++)
    {
        printf("A(%d,%d)=",i+1,j+1);
        scanf("%d",m);
        m++;
    }
}

Or even:

for(int i=0; i<x*y; i++) {
   printf("A(%d,%d)=", i/3, i%3);
   scanf("%d", M + i);
}

---

Other points:

In one method you use variables x,y,i,j , and in another you use a,b,k,l . I assume you've done this because you don't want to overwrite one with the other. But because of scope , that's not a factor. x is local to the function minput() -- you can have another x in main() and they will be completely independent of one another. Use x,y,i,j in both places, because they are the "sensible" names for dimensions and loop counters.

for(i=0; i<x; i++) is the conventional way of looping x times. Your i<=(x-1) is equivalent but messy and confusing.

Casting the result of malloc() is discouraged nowadays. Do I cast the result of malloc?

int *M = malloc(x*y*sizeof(int));