Help creating a 2D array

**livin4th** · 02-28-2017

Hello,

I am working on a problem statement which typedefs an array like

Code:

typedef signed char Type[6]

the task is to create a function that will take input parameters as rows and cols and create a 2-dimensional pointer array of data type 'Type' having the number of rows and cols passed in as inputs. There is an additional requirement that all memory allocations should happen in only one call to malloc()

I tried this but don't think it is correct.

Can someone provide some pointers?

Code:

Type **testArray(size_t rows, size_t cols)
{
    Type **temp = NULL;

    temp = (Type **)malloc(rows * cols * sizeof(Type));
    if (temp == NULL)
    {
        //error out
    }
    
    return temp;
}

**Salem** · 02-28-2017

The amount of memory you need to allocate is
rows * sizeof(Type*) + rows * cols * sizeof(Type)

The first part will be your array of Type* pointers, the beginning of which is your actual return value.

You then need to initialise 'rows' of pointers to point to 'cols' sized blocks of Type in the second part of the memory block.

Also, this is C, so there is no need to cast the return result of malloc (see the FAQ).

**Malcolm McLean** · 03-01-2017

2D arrays don't work well in C when the dimensions are not known at compile time.

So what most C programmers will do is to allocate a large 1D array, then access with

Code:

      array[y*width+x] = value;

but it's also possible to create an array of pointers

Code:

        Type **array = malloc(height * sizeof(Type *));
        for(i=0;i<height;i++)
           array[i] = malloc(width * sizeof(Type));
        array[y][x] = value;

The access syntax is the same as for a built-in 2D array, but the underlying machine operations are different.

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