arrays of pointers to pointers 2d arrays

Question

I'm trying to create a continuous block of memory in one function call that has the first part of the memory as pointer arrays to the other blocks. The function works with an index notation though I'd like to use compact pointer

Type **TwoD(size_t rows, size_t cols) 
{
    Type **p1, **prows, *pcol;

    p1 = (Type **)malloc(rows * sizeof(Type *) + rows * cols * sizeof(Type));


//  ??? something wrong here ??? I'd rather use this style if possible
//  
    //index notation works

    return prows;
}

user1500049 · Accepted Answer · 2012-10-19 06:01:36Z

2

The value of prows returned is different.

prows is changed in this case:

//  for (; prows < (Type **)pcol; ++prows, pcol += cols)
//      *prows = pcol;
return prows;

while the other case, no change of prows (which is still p1):

for (unsigned ii = 0; ii < rows; ++ii)
{
   prows[ii] = pcol;
   pcol += cols;
}
return prows;

So, what you can do is at the end return p1 instead of prows.

edited Oct 19, 2012 at 6:01

answered Oct 19, 2012 at 5:42

user1500049

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3 Comments

blitzeus Over a year ago

I still get a Segmentation fault: 11 when I use p1 as a return value.

blitzeus Over a year ago

I figured it out, every time I incremented pcol+= cols, I was changing the termination part of the for loop so for (end = p1 + rows; prows < end; ++prows, pcol += cols)

user1500049 Over a year ago

Personally I prefer the index notation way, seems easier to follow.

zetsin · Accepted Answer · 2012-10-19 05:58:35Z

0

try this one:

// ======i make change here=========================
// =====================V===========================
for (; prows < (Type **)&pcol; ++prows, pcol += cols)
    *prows = pcol;

return p1;

answered Oct 19, 2012 at 5:58

zetsin

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Comments

Hai Vu · Accepted Answer · 2012-10-19 06:00:13Z

0

I think two-dimensional array is a little complicated. I always use a single-dimensional array to represent a 2D one, for example:

typedef int Type; // for simplicity, assume int

Type *TwoD(size_t rows, size_t cols) {
    Type *array = (Type*)malloc(rows * cols * sizeof(Type));
    Type *p = array;
    Type counter = 0;
    for (size_t row = 0; row < rows; row++) {
        for (size_t col = 0; col < cols; col++) {
            *p++ = counter++;
        }
    }
    return array;
}

/* The array will look like this, for rows=3, cols=4:
0  1  2  3
4  5  6  7
8  9 10 11
*/

In the above example, I use a counter to initialize the array, but you can do differently.

answered Oct 19, 2012 at 6:00

Hai Vu

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