How can I return to a string address and then assign it to a new string?

Question

I have 1 function that I want to return the address of an assigned string to the main function and assign an new string pointer with the same address so that the new string will have the contents of the old string.

For example:

 unknown_datatype function()
 {
      char *old = "THE STRING";
      return old;
 }

 int main()
 {
      char *snew = "";
      snew = function();
      return 0;
 }

*unknown_datatype means I don't know that to put there...

*How can I approach this without changing anything in the main() method

Answer 1

You can simply return a char*. Since you are returning a pointer to a string literal, it's probably best to return a const char*, since you aren't able to modify the string literal:

const char* function()

And likewise you would want to assign the return value to a const char*:

const char* snew = 0;
snew = function();

Answer 2

Typically you will pass the address of the first element in an array of chars in, as well as the length, and have the function fill it.

int fillMyString(char *str, int buffer_size)
{
  if(buffer_size > strlen("test"))
  {
    strncpy(str, "test", buffer_size-1);
    str[buffer_size-1] = '\0';
    return strlen(str);
  }

  return 0;
}


//In some function or main
char buffer[1024];
fillMyString(buffer, 1024);
ASSERT(!strcmp(buffer, "test"));

Edit: You mentioned that for some reason you need to return a char*. I would suggest in that case to use malloc to allocate the string inside the function, but make sure whenever you call the function you free the return value eventually.

Answer 3

The unknown data type would have to be char* but in this case it would not do what you expect since the variables in function() are local variables that will disappear once the function exits and they get popped off the stack. You could use a malloc() call inside the function and return a pointer to the new string. Just don't forget to free it when you're done with it.

Answer 4

OK here is the answer: Since I cannot EDIT the main (MEANING I CANNOT PASS ANY PARAMETER) the following solution is possible:

char *function()
{

    char *old;
    old = (char*)malloc(9999);
    strcpy(old,"THE STRING");
    return old;
}

int main()
{
    char *snew = "";
    snew = function();
    return 0;
}

Thanks for all who replied.

Answer 5

char * function()
{
    char *old = "THE STRING";
    return old;
}

int main()
{
     char *snew = "";
     snew = function();
     return 0;
}

Answer 6

Your question and example are contradicting each other.

Is this what your looking for ? given below are two simple solutions.

char* func(char *);

int main()

{

char *newstring="Hey";

newstring = func(newstring);

printf("New string : %s\n", newstring);

}

char* func(char *s)

{

printf("Old string : %s\n", s);

s="Hello";

return(s);

}

or else the solution given by Y_Y is also fine, assuming your string is created in the function call. Just in case you do not want to use malloc, the solution below works.

char* func();

int main()

{

char *newstring="";

newstring = func();

printf("New string : %s\n", newstring);

}

char* func()

{

char *s="Hello";

printf("Old string : %s\n", s);

return(s);

}