Thread: Allocating memory for multidimensional arrays - how to?

First, I don't remember if we can leave one dimension of the array without a value and the other with a value.

No; at least, not in this context.

Second, are this malloc and this type cast correct? buffercodes = (char **) malloc(numofflights* sizeof(char));

There is no need for a cast. And the malloc() call is wrong. What you want is something like:

Code:

char (*buffercodes)[7];
buffercodes = malloc(numofflights * sizeof *buffercodes);

Here “buffercodes” is a pointer to an array of char. So buffercodes[0], [1], [2]... are each a char[7]. The size of each element is thus 7 bytes, or sizeof(char[7]). However, an often-recommended technique is to use the variable name with sizeof, not the type, because there's less of a chance of flubbing the size.

So, if buffercodes is a pointer to char[7], then *buffercodes (which is the same as buffercodes[0]) is a char[7], and (sizeof *buffercodes) is the same as taking the size of a char[7].

Code:

char **mat = malloc(num * sizeof(char*));
for (i = 0; i < num; i++) {
	mat[i] = malloc(STRLEN);
}

Then for any multidimensional array we start allocating from right to left ? In this example, we fill collumns and then rows, so I can acces any of the strings using mat[row].

Am I right?

Originally Posted by MK27

Well, that looks better doesn't it? Just when I thot I knew everything

Yes I prefeer the solution proposed by Salem and cas, but still you helped me remembering something about dynamic allocation


I think I understood how to do it, I'll rearrange my code now

Thanks everyone