malloc inside a function

Hi,

I'm trying to pass a pointer to a function, and then have the function allocate some memory for that pointer. Is this possible?

I tried the following code, but doesn't seem to work. Once I return from the function, the pointer seems to be void. The program crashes when I try to access it in main with memcpy() or some other function.

Code:

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int main()
{
  unsigned char *ptr;
  int memSize;
  unsigned char temp[1024]; // 1024 is larger than memSize

  foo(ptr, &memSize);
  memcpy(temp, ptr, memSize);
}

void foo(unsigned char *ptr, int memSize)
{
  *memSize = someotherfunction();
  ptr = (unsigned char *)malloc(*memSize)
}

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Quote:

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Originally Posted by galapogos

Hi,

I'm trying to pass a pointer to a function, and then have the function allocate some memory for that pointer. Is this possible?

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Yes it is possible. The problem you're having is that you are changing the value of a temporary within a function.

When you pass a parameter by value, the function makes a temporary copy. When that function returns, that temporary is gone. A pointer is nothing but a value you're passing, and the same rules apply.

To change a parameter value and have it reflect back to the caller, you either pass a pointer to the variable, or a reference to the variable. So you either pass a pointer to the pointer, or a reference to the pointer.

Quote:

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I tried the following code, but doesn't seem to work.

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Just like this code won't work:

Code:

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void foo(int x)
{
    x = 10;
}

int main()
{
  int p = 0;
  foo( p );
  // p is still 0.  Why isn't it 10?
}

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Same principle.

Regards,

Paul McKenzie