Error when assigning value in 2D array

Question

double** Make2DDoubleArray(int arraySizeX, int arraySizeY) {
    double** theArray;
    theArray = (double**) malloc(arraySizeX*sizeof(double*));
    for (int i = 0; i < arraySizeX; i++)
    {
        theArray[i] = &(malloc(arraySizeY*sizeof(double)));
        memset(&(theArray[i]), 0, arraySizeY*sizeof(double));
    }
    return theArray;
} 

I need to make a 2D array with malloc in C. I used the above method.

When I access the returned array (let's call it Matrix ) with Matrix[1,2] = 2.0; it throws an error:

cannot convert 'double' to 'double*' in assignment

But I don't see why Matrix[1,2] is a pointer and not a value? Matrix is a pointer of pointers, and Matrix[i] is a pointer to a list of double s, so Matrix[i,j] should be a double and not a double * !

Answer 1

Matrix[i, j] is the item in Matrix at the index i, j , which is the result of the comma operator on the values i and j , which is just Matrix[j] . You probably mean Matrix[i][j] .

In addition to that:

• Don't cast the return value of malloc :

   double** theArray = malloc(arraySizeX * sizeof(double)); 

• You are using the wrong sizeof ; it's sizeof(double*) :

   double** theArray = malloc(arraySizeX * sizeof(double*)); 

• This makes no sense; drop the & :

   theArray[i] = &(malloc(arraySizeY*sizeof(double))); 

• This is wrong; drop the & again:

   memset(&(theArray[i]), 0, arraySizeY*sizeof(double)); 

Answer 2

These lines are wrong:

theArray[i] = &(malloc(arraySizeY*sizeof(double)));
memset(&(theArray[i]), 0, arraySizeY*sizeof(double));

I am surprised the compiler didn't complain about taking the address of a temporary in the first call.

You need:

theArray[i] = malloc(arraySizeY*sizeof(double));
           // ^^ Drop the &
memset(theArray[i], 0, arraySizeY*sizeof(double));
    // ^^ Drop the &