pointer to preallocated memory as an input parameter and have the function fill it

Question

Test code:

void modify_it(char * mystuff)
{
   //last element is null i presume for c style strings here.
   char test[7] = "123456";

   //when i do this i thought i should be able to gain access to this
   //bit of memory when the function is destroyed but that does not
   //seem to be the case.
   //static char test[] = "123123";

   //this is also creating memory on stack and not the heap i reckon
   //and gets destroyed once the function is done with.
   //char * test = new char[7];

   //this does the job as long as memory for mystuff has been
   //allocated outside the function.
   strcpy_s(mystuff,7,test); 

   //this does not work. I know with c style strings you can't just do
   //string assignments they have to be actually copied. in this case
   //I was using this in conjunction with static char test thinking
   //by having it as static the memory would not get destroyed and i can
   //then simply point mystuff to test and be done with it. i would later
   //have address the memory cleanup in the main function.
   //but anyway this never worked.
   mystuff = test;
}

int main(void)
{
  //allocate memory on heap where the pointer will point 
  char * mystuff = new char [7];
  modify_it(mystuff);
  std::string test_case(mystuff);

  //this is the only way i know how to use cout by making it into a c++ string.
  std::cout<<test_case.c_str();

  delete [] mystuff;
  return 0;
}

1. In the case of a static array in the function why would it not work?
2. In the case when I allocated memory using new in the function does it get created on the stack or heap?
3. In the case when I have a string which needs to be copied into a char * form. everything I see usually requires const char* instead of just char*.

I know I could use a reference to take care of this easily. Or char ** to send in the pointer and do it that way. But I just wanted to know if I could do it with just char *. Anyway your thoughts and comments plus any examples would be very helpful.

Answer 1

char * mystuff = new char [7];
delete mystuff;

delete mystuff is causing undefined behavior. You must delete[] what you new[].

Answer 2

The line mystuff = test; causes the variable mystuff to contain the address of the test array. However, this assignment is local to the function. The caller never sees the modified value of mystuff. This is generally true for C/C++: function parameters are passed by value, and local modifications to that value are invisible outside of the function. The only exception to this is if you use the & operator in the parameter list in C++, which causes the parameter to be passed by reference. Like so:

void modify_it(char* &str) { /* ... */ }

However, if you do this, your program still won't work correctly, and will probably crash. That's because the address of test is stack memory, and that memory will be overwritten when modify_it returns. You'll be giving the caller the address of invalid stack memory, which can only lead to bad things. The correct thing to do is one of the following:

/* function allocates, caller frees */
void modify_it(char* &str) {
    str = new char[7]; // allocate enough memory for string
    memcpy(str, 7, test);
}

Or this:

/* caller allocates and frees */
void modify_it(char* str, size_t str_len) {
    if (str_len < 7) { /* report an error. caller didn't allocate enough space. */ }
    memcpy(str, 7, test);
}