Accessing outside the dimensions in array created by a double pointer

Question

Is this the correct method to define an 5*3 matrix using double pointers?`

 int **M1;
 M1 = (int **)malloc(5 * sizeof(int *));
 for (i=0;i<5;i++)
 {
    M1[i] = (int *)malloc(3 * sizeof(int));
 }`

If so, how can I assign M1[3][15] = 9 in the code and still get no error? And why am I getting a segmentation error in assigning M1[6][3]=2?

I understood after few such initializations that I created a 5*xx array, i.e. I couldn't go above 5th row but I could assign any value to the number of columns. How should I create just a 5*3 array?

Please [see why not to cast](http://stackoverflow.com/q/605845/2173917) the return value of `malloc()` and family in `C`. — Sourav Ghosh, Aug 13 '15 at 10:57
If you go out of bounds you will have *undefined behavior*. The C language has no bounds checking, it allows you to [shoot yourself in the foots](http://idioms.thefreedictionary.com/shoot+in+the+foot) as much as you like. Just don't do it. — Some programmer dude, Aug 13 '15 at 10:58
@JoachimPileborg: I was able to go out of bounds by extending rows not columns. I was confused as to how did it happen. I don't intend to go out of bounds. — re3el, Aug 13 '15 at 11:01

Sourav Ghosh · Answer 1 · 2015-08-13T11:08:06.990

1

In your code, you're allocating memory for 5 pointers

M1 = (int **)malloc(5 * sizeof(int *));

and later, you're trying to access beyond that, based on an unrelated value of m

for (i=0;i<m;i++)

when m goes beyond 4, you're essentially accessing out of bound memory.

A better way to allocate will be

int m = 5;
M1 = malloc(m * sizeof(*M1));
if (M1) 
{
 for (i=0;i<5;i++)
 {
    M1[i] = malloc(3 * sizeof(*M1[i]));
 }
}

couldn't go above 5th row but I could assign any value to the number of columns.

NO, you can not. In any way possible, accessing out of bound memory invokes undefined behaviour.

edited Aug 13 '15 at 11:08

answered Aug 13 '15 at 10:58

Sourav Ghosh

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sorry, I forgot to edit m to 5. Please recheck the question. – re3el Aug 13 '15 at 11:03

score 0 · Answer 2 · answered Aug 13 '15 at 11:23

Since Sourav tackled the UB case, I'll answer

How should I create just a 5*3 array?

Why not rely on automatic variables? Unless you've a compelling reason not to, use them

int matrix[5][3];

If you don't know the dimensions in advance, and don't prefer doing the double pointer manipulation, flatten it like this:

int *m = malloc(sizeof(int) * rows * cols);
// accessing anything from 0 to (rows * cols) - 1 is permitted

// helper to make usage easier
int get_element(int *m, int i, int j, int cols) {
    return m[i * cols + j];
}

OTOH, if you only don't know the first dimension at compile-time, then you may do:

typedef int (Int5) [5];   // cols known at complie-time
int rows = 3;
Int5 *r = malloc(rows * sizeof(Int5));
r[0][0] = 1;   // OK
r[0][5] = 2;   // warning: out of bounds access

With this method you get a bit more type safety due to the compiler knowing the size in advice.

Accessing outside the dimensions in array created by a double pointer

2 Answers2