returning a multidimensional pointer in c - c

Question

error lvalue returned as left operand of assignment
What is the error in this program;
int(*matrix)[row][col];
int i = 0;
if(n == 1)
{
for (int i = 0; i < no_matrices; i++)
{
printf("Matrix %d", i + 1);
(matrix + i) = GetMatrixU(row, col); // (error here)
}
}
int* GetMatrixU(int row, int col)
srand(time(NULL));
int(* matrix)[col] = malloc(col*sizeof(int));
for (int i = 0; i < row; i ++)
{
for (int j = 0; j < col; j ++)
{
matrix[i][j] = rand()%100;
}
}
return matrix[];
}

Answer 1

The problem is this line:
(matrix + i) = GetMatrixU(row, col);
This tries to make an assignment. The expression on the right is evaluated; this is the "r-value" ("r" for "right"). Then the result should be assigned to the expression on the left, which is the "l-value" ("l" for "left").
Well, (matrix + i) is not a valid l-value. matrix by itself is a single pointer to a two-dimensional array. In C, you can't just specify a pointer value and assign to it; you must use the * operator to dereference the array, and assign through that.
Here is a short code sample that shows how to dereference a pointer, and then repeats your error.
main()
{
int a[10];
*(a + 1) = 0; // proper use of * to dereference
a + 1 = 0; // will not compile; error is "l-value required"
}
But in C, there is a shortcut way to add to a pointer and then dereference it. It is the use of square brackets and an index value:
a[1] = 0; // index the array
In C, the expressions *(a + i) and a[i] mean exactly the same thing, by definition.
http://en.wikipedia.org/wiki/C_syntax#Accessing_elements
It looks like you are trying to create a random matrix, one row at a time. And you are allocating memory for each row. Your basic matrix declaration should be an array of pointers to rows.
int *matrix[rows];
for (i = 0; i < rows; ++i)
{
// either of these lines will work; pick just one
*(matrix + i) = AllocateRandomRow(cols); // explicit add-and-dereference
matrix[i] = AllocateRandomRow(cols); // simpler: just index array
}
I favor the second, simpler syntax for solving this problem. You are indexing an array; just use the square brackets.
Here is a free online text that describes how to dynamically allocate a matrix.
http://www.eskimo.com/~scs/cclass/int/sx9b.html

Answer 2

Here's an example of simpler case to understand this more complex case:
int b = 4;
int *a = &b; // Allowed and works because a is a pointer to an integer
a+1 = 5; // Not allowed no idea about the type of a+1 and it could be used!!
Why this relates to your problem ?
int(*matrix)[5][5];
printf("%p\n",matrix);
printf("%p\n",matrix+1);
Output
0xb7753000
0xb7753064
Difference between the above pointers is of 6*16 + 4 = 100. This number is nothing but row*col*sizeOf(int) = 5*5*4 = 100
Therefore, what you want to do was accessing the cells in matrix, however what you are actually doing is trying to access a memory that your not allowed to use as in the first simple example.
Example of what you are actually doing vs what you wanted the program to do:
int(*matrix)[5][5];
matrix + 0 => address of matrix
matrix + 1 => address of matrix + 100 bytes (you expected + 4 bytes instead)
matrix + 2 => address of matrix + 200 bytes (you expected + 8 bytes instead)
etc...