Passing Pointers by reference

I know how to simulate passing by reference by using pointers. How does one do this for something that already IS a pointer? For example a C-style string?

Code:

---

int main() {
  char *mystring;
  int myint = 4;

  make_my_string(mystring, &myint);
}

void make_my_string(char *some_string, int *some_int) {
  some_string="hello";
  some_int = 0;

}

---

In this example, mystring is still null, and myint is now 0. I want mystring = "hello"

Why would it change? Pass a pointer to the thing you want to be "by reference".

As I explained in an earlier post, a copy of the variable you pass is made, so if you want to change the original, you must pass the address.
The type of the pointer is always the type it points to first, and then a *.
So the type becomes: char**.

And string literals should be const, because they are non-modifiable.
And your code is bugged, since it will change the pointer to 0 (which will not even raise a warning, sadly).

And it's impossible to pass anything by reference in C (I just had to add that one!). It's only possible to pass by address or pointer!

#1) Never do this:

Code:

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void make_my_string(char *some_string, int *some_int) {
  some_string="hello";

---

altho you could do this:

Code:

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void make_my_string(char *some_string, int *some_int) {
  strcpy(some_string,"hello");

---

However, in your example mystring does not have any memory space allocated to it! To do that you must use malloc, in main():

Code:

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int main() {
  char mystring=malloc(6);

---

Be sure you understand that this means some_string now points to 6 bytes of memory into which you can store data. If you redirect this pointer, that is, if you use it on the left side of an = equal sign, then you have reassigned it, and those 6 bytes of storage DO NOT come with it!!!!

That's pretty much already summed up: http://apps.sourceforge.net/mediawik...llocating_them
And if you're going to strcpy, you had better make sure you know the size of the buffer to avoid buffer overruns.

On second thought I don't have anything more to say.

But you're assigning tmp to the local copy of some_string, so that's kinda useless.
That means you'll have to return a pointer, making the whole char* some_string argument useless.
And this also means you have to free the pointer, so you can't simply pass in a buffer. And allocating a buffer on the stack vs malloc, especially for small amounts, is much faster, so it's advantages vs disadvantages, which is usually why functions take the buffer as argument, along with its size.

Quote:

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Originally Posted by Elysia

But you're assigning tmp to the local copy of some_string, so that's kinda useless.
That means you'll have to return a pointer, making the whole char* some_string argument useless.

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Sorry, you're right, I don't know how many times I have to get this wrong ;)

Well you've been very helpful in telling me what was wrong with my little "off the top of my head" example. However you've not helped me understand what I need to do to pass a pointer into a function so that when I modify its value inside the function, the value outside changes as well... I went to Elysia's post on another thread and used her examples and I get about 3 errors/warnings about suspicious pointer conversions.

I did mention it.

Code:

---

int main()
{
    const char* mystring;
    int myint = 4;
    make_my_string(&mystring, &myint);
}

void make_my_string(const char** some_string, int* some_int)
{
    *some_string = "hello";
    *some_int = 0;
}

---

And I have no idea what warnings and things you got.
The examples were 100% correct, but they were missing including the header, string.h, and a prototype for the called function. Things every C programmer should know about.

Ah... My bad. I forgot to add the '*' in the body of the function...

"Thanks Elysia. You're a programming master! How the hell do you know every thing?"

:P