Question

double *f(int n, double v)
{
  double *a, *p;
  a = malloc(n * sizeof(double));
  if (a != NULL)
    for (p = a; p < a + n; p++)
      *p = v;
  return a;
}

Can you explain me what this function is needed for? Does it copy the content of v in n? If yes, why does it return a? I really don’t get it… Thanks in advance.

Answer 1

double *f(int n, double v) 
{
  double *a, *p;
  a = malloc(n * sizeof(double));  // allocate memory enough for "n" doubles (an array)
  if (a != NULL)                   // if the allocation was successful
    for (p = a; p < a + n; p++)    // loop from the beginning of the array to the end
      *p = v;                // fill every element of the array with the value "v"
  return a;                  // return the new array
}

So if I called this function:

double * myarray;
myarray = f(3, 1.3);

Now I have:

myarray[0] = 1.3
myarray[1] = 1.3
myarray[2] = 1.3

So to answers your questions:

Can you explain me what this function is needed for?

• allocates and initializes an array of doubles.

Does it copy the content of v in n?

• No. Considering v is a double and n is an int, that doesn’t even make sense. It makes an array n large and initializes it with the value v.

If yes, why does it return a?

• It returns a so you have a reference to the newly created array. (see example above on how it could be used)