为什么人们会使用像char *和buf这样的东西？

Question

I am reading a post on Stack Overflow and I saw this function:

    advance_buf( const char*& buf, const char* removed_chars, int size );

What does char*& buf mean here and why do people use it?

Answer 1

It means buf is a reference to a pointer, so its value can be changed (as well as the value of the area it's pointing to).

I'm rather stale in C, but AFAIK there are no references in C and this code is C++ _{(note the question was originally tagged c )} .

For example:

void advance(char*& p, int i) 
{       
    p += i;  // change p
    *p = toupper(*p); // change *p
}

int main() {
    char arr[] = "hello world";
    char* p = arr; // p -> "hello world";
    advance(p, 6);
    // p is now "World"
}

---

Edit: In the comments @brett asked if you can assign NULL to buff and if so where is the advantage of using a reference over a pointer. I'm putting the answer here for better visibility

You can assign NULL to buff . It isn't an error. What everyone is saying is that if you used char **pBuff then pBuff could be NULL (of type char** ) or *pBuff could be NULL (of type char* ). When using char*& rBuff then rBuff can still be NULL (of type char* ), but there is no entity with type char** which can be NULL .

Answer 2

buf 'sa (C++) reference to a pointer. You could have a const char *foo in the function calling advance_buf and now advance_buf can change the foo pointer, changes which will also be seen in the calling function.