arrays in functions where the size is a formal argument

HI there,

I'm new to both this forum and to C programming so apologies in advance if my question is answered in the FAQ or elsewhere but I could not find it.

I would like to do the following:

within a function (other than main) definition I would like to declare an array of size equal to some number that is a formal argument of the function itself. So if "n" is a formal argument of FUNCT, say, I would like to locally (within FUNCT) declare an array of size n.

This seems a fairly reasonable thing to do and I am sure there must be a trick - but I can't get it to work. I have tried dynamic memory allocation using malloc to no avail.

Thanks in advance for any comments anyone may have.

Boggle

I am not a C programmer (I'm a C++ programmer) but I think you can do this

Code:

---

void f(int n)
{
    int* ptr = (int*)malloc(sizeof(int) * n);
}

---

However I think I have read against casting the result of malloc... Oh and don't forget free() ing memory.

You should not cast malloc in C as FAQ suggests, and you HAVE to check the return value

Code:

---

void f(int n)
{
  int* array = malloc(n* sizeof *array);
  if(array)
  {
      /* do your stuff here */
      ...
      free(array);
  }
}

---

Thanks guys.

One question about the sample code you provided vart - why the use of the "if" statement?

Also, I am not sure what you mean by the return value.

I am very grateful for the help.

Cheers.

@boggle, malloc returns a NULL pointer if it fails. You don't want to continue on thinking the memory has been allocated and really it hasn't.

However, if malloc succeeds it returns a pointer to the start of the newly allocated block of memory (ie not NULL). Just remember anything non-zero in C is true, and thing 0 is false.

A more explicit way of putting it is:

Code:

---

/* ... */
if(array != NULL)
{
/* ... */

---

you guys rock!! Thanks.