adding two two-dimensional arrays in C

Question

i start with:

 typedef struct
{
    int rows, cols;
    int **element;
}Matrix;

and i create two matrices:

void matrixInit (Matrix *m, int r, int c )
{
    m->element = malloc(sizeof(int*)*r);
    for (int j=0;j<3;++j)
    {
        m->element[j] = malloc(sizeof(int)*c);
    }
    m->rows=r;
    m->cols=c;
 }
Matrix m1,m2,m3;
matrixInit(&m1,3,3);
    for (k=0; k<m1.cols*m1.rows; k++)
        {
            m1.element[k]=q;
            q++;
        }

then i do matrix 2 using copy function created

void matrixCopy (Matrix *m, Matrix *n )
{
    int r=n->rows,c=n->cols,k;
    for (k=0; k<r*c; k++)
    {
        m->element[k]=n->element[k];
    }
}

matrixInit(&m2, 3, 3);
matrixCopy(&m2, &m1);

and then i create a third to be the result of the addition

matrixInit(&m3, 3, 3);

then i do the addition. this is where my problem lies. I cannot seem to get this to work.

my code for the function is below: (the prototype must stay the same)

Matrix* matrixAdd (Matrix *m, Matrix *n )
{
    int q;
    for (q=0; q<m->rows*m->cols; q++)
    {
        m->element[q]=*m->element[q]+*n->element[q];
    }
    return m;
}
m3=*matrixAdd(&m1, &m2);

Add the programming language, where you perform this implementation. — Praveen
– Praveen, Commented Dec 5, 2013 at 4:10
Is the addition function supposed to allocate a new matrix? Because your's currently modifies the first-parameter in-place, then returns its address. The result in your code usage is leaked memory from m3, and two-matrices (m3 and m1) that are pointing to the same buffer. In summary, that isn't good. — WhozCraig
– WhozCraig, Commented Dec 5, 2013 at 4:41
(a) The for (int j=0; j<3; ++j) loop should be for (int j=0; j<r; ++j) (r instead of 3); (b) the function creates but does not initialize the matrix; the values in the cells is indeterminate. For pity's sake, please use spaces: the loop for (q=0; q<m->rows*m->cols; q++) is damned hard to read, compared with: for (q = 0; q < m->rows * m->cols; q++). — Jonathan Leffler
– Jonathan Leffler, Commented Dec 5, 2013 at 4:56
Also, since m3 == &m1 after the call, returning the matrix as a pointer is a little pointless. Either you need to allocate a result matrix in the function or you don't need to return a pointer to the result. Since you can't change the prototype, you need to allocate the matrix in the addition function. — Jonathan Leffler
– Jonathan Leffler, Commented Dec 5, 2013 at 5:55

doptimusprime · Accepted Answer · 2013-12-05 04:37:49Z

1

matrixAdd function should be as follows: * on element will refer to first value only.

Matrix* matrixAdd (Matrix *m, Matrix *n )
{
    int q;
    int r;
    for(r = 0; r<m->rows; ++r)
       for (q=0; q<m->cols; q++)
       {
           m->element[r][q]=m->element[r][q]+n->element[r][q];
       }

    return m;
}

This will be more readable can be understood.

In your case,

   *m->element[q] = *m->element[q]+*n->element[q];

will lead you into problem as you have allocated r pointers to int * (therefore r continuous element) and at each pointer you have allocated c integers.

Therefore, if q is beyond r, behaviour is not defined.

edited Dec 5, 2013 at 4:37

answered Dec 5, 2013 at 4:25

doptimusprime

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4 Comments

runnercto Over a year ago

*m->element[q] = *m->element[q]+*n->element[q]; gets me "Thread 1: EXC_BAD_ACCESS (code=1,address=0x1)"

doptimusprime Over a year ago

sorry for inconvenience. I edited the answer by stating the reason.

Stack Exchange Broke The Law Over a year ago

Is q<m->rows*m->cols correct? Looks like it should be q<m->cols

Jonathan Leffler Over a year ago

This is one of the times when m->element[r][q] += n->element[r][q]; really helps; you don't have to look hard to see if two of the three expressions are the same.

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