C Grid using pointers and Malloc

I am Trying to work through some class examples and have gotten stuck on the following:

The array grid should have length width with each entry representing a column of cells. Columns that have some occupied cells should be a malloc'ed character array of length height.

with the given header:

void grid(char **grid, int width, int height)

grid is defined in another file as:

char **grid;

As I have said I have gotten stuck on using malloc, I currently have:

int x;
*grid = malloc(width * sizeof(char));

        for(x = 0; x < width; x++){
            grid[x] = malloc(height * sizeof(char));
        }

Can any one take a look at give me some pointers on the correct way to accomplish "Columns that have some occupied cells should be a malloc'ed character array of length height.", As I dont understand how the line:

grid[x] = malloc(height * sizeof(char));

is equivalent to an开发者_如何学编程 array of char's

Thanks

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char** grid; is a pointer to pointer.

grid = malloc( width* sizeof( char* ) ) ;  // Statement 1

for( int i=0; i<height; ++i )
{
    grid[i] = malloc( height ) ; // Statement 2
}

Understand char** -> char* -> char. So first need to reserve for holding addresses amounting to width. By Statement 1, it is achieved. Now each of these index should point to a memory location holding height characters and is achieved by Statement 2.

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Diagramatic representation sounds more clear than description. Hope that helps !

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In C, an array is a pointer to the first element of the array. So an array is just a memory block, with the array variable pointing onto the first element in this memory block.

malloc() reserves a new memory block of the specified size. To know the size of a type (i.e. number of bytes needed to store one variable of that type), one uses the sizeof operator. Hence a character needs sizeof(char) bytes, and hence height characters need height * sizeof(char).

So with the malloc() call you allocate a memory block to store all the elements of the array, and malloc() returns a pointer onto the first of them.

With C's definition for an array variable (a pointer onto the first element), you can assign the results of malloc(...) to your array variable.

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Use this:

grid = malloc(width * sizeof(char *));

You want to allocate space for width pointers to char - And then you correctly allocate the individual pointers to height chars in the loop.

Using a typedef makes this more visible:

typedef char * charpointer;
charpointer * grid = malloc(width * sizeof(charpointer));

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1. First you allocate space for array of [width] char pointers. So instead of *grid = malloc(width * sizeof(char)); which allocates space for [width] chars, you should use *grid = malloc(width * sizeof(* char)); (the difference is that char is one byte, and char pointer is int (usually 32 bit, but architecture dependent)
2. In your loop, each time you allocate (an array of) [hight] chars and store the pointer to it in one of the pointers that you allocated in (1). so grid[x], actually points to an allocated buffer of chars (which is your array)

hope I made it clear.