Defected value when passing a matrix by pointer

Question

I am using an algorithm where I need to copy a 2D matrix in a more sized 2D matrix. When I am working in the same main stack it works well but when I pass the pointer of the copied matrix to another function, the values became corrupted.

The steps to generate the problem: 1- Copy a 2D square matrix of size x to another square matrix of size y (y>x). 2- Pass the pointer of copied matrix to a function. 3- Print the matrix inside the function.

The result will be corrupted values.

int main(void) {
    int tab[3][3] = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
    int tab2[5][5];
    int size = 3;

    // Coppying tab to tab2
    for(int i = 0; i<size; i++)
    {
        for(int j = 0; j<size; j++)
        {
            tab2[i][j] = tab[i][j];
        }
    }


    // printing tab2 using myPring Function
    myPrint(tab2, size);
    return 0;
}

void myPrint (int* tab, int size)
{
    for(int i = 0; i<size; i++)
    {
        printf("|");
        for(int j = 0; j<size; j++)
        {
           printf("tab[%d][%d] => %d |", i, j,  *((tab+i*size) + j)  );
        }
        printf("\n");
    }
    printf("____________________________________________\n");
}

Turn up your compiler warnings, and pay attention to what they tell you. — Mark Benningfield, Apr 26 '19 at 01:21
I am sorry I still didn't get, do I need to cast tab2 to (int*) ? — Ahmed Garssallaoui, Apr 26 '19 at 01:29
Possible duplicate of [How to pass 2D array (matrix) in a function in C?](https://stackoverflow.com/questions/3911400/how-to-pass-2d-array-matrix-in-a-function-in-c) — Mark Benningfield, Apr 26 '19 at 01:33

score 0 · Accepted Answer · answered Apr 26 '19 at 01:31

0

*((tab+i*size) + j) computes a pointer using size, which is three. However, although only three values were stored in each row of tab2, it actually has five elements per row.

answered Apr 26 '19 at 01:31

Eric Postpischil

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Defected value when passing a matrix by pointer

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