Basic Question about passing pointer by reference

how can I pass a pointer by reference? I want this function to duplicate the string. I know only a copy of the pointer is passed, but how do I pass the pointer by reference? I tried having the method accept char*&, but it doesnt enter the method at all (made sure with breakpoints) [beforehand it was just char*].
Thanks in advanced

Code:

void strdup (char*& str)
	{		
		int len=strlen(str);		
		char* strDup = new char[2*len+1];
		for (int i=0;i<len;i++)
		{
			strDup[i]=str[i];
			strDup[i+len]=str[i];
		}
		str= strDup;
	}

the whole code:

Code:

#include <iostream>
using namespace std; 

namespace myStr {
	
	int strlen (char* str)
	{
		int counter=0;
		char* ch = &str[0];
		while (*ch!=0)
		{
			ch++;
			counter++;
		}
		return counter;
	}

	void strdup (char*& str)
	{		
		int len=strlen(str);		
		char* strDup = new char[2*len+1];
		for (int i=0;i<len;i++)
		{
			strDup[i]=str[i];
			strDup[i+len]=str[i];
		}
		str= strDup;
	}
}

using namespace myStr;

int main () {	
	char str1[] = "Hello";
	cout << str1 << endl;	
	cout << strlen(str1) << endl;
	strdup(str1);
	cout << str1 << endl;
	return 0;
}

There are two problems. First of all, you forgot to null terminate the string after the copy. Second, str1 is an array, not a pointer. And if it were, you would need to delete the old memory before reassigning it. And anyway, it's just a bad design. Provide the reference as a second parameter if anything and leave the original string as is. Or better yet, just return the new string from the function. Or if you want to go all out and do the right thing, return a RAII-compliant container, such as std::string. Saves you a lot of headaches, and much easier, too.

I suppose this is not going to work. You are passing an array to the function, and expect the array variable to turn magically into a pointer to dynamically allocated block of chars.

Code:

int main () {	
	char str1[] = "Hello";
        char* p = str1;
	cout << str1 << endl;	
	cout << strlen(str1) << endl;
	strdup(p);
	cout << p<< endl;
	return 0;
}

Still it all looks rather problematic. You expect the pointer that is passed in not to point to dynamically allocated buffer (or you'll leak it), but the pointer that the user gets back does point to dynamically allocated memory.

A better idea:

Code:

//returns pointer to dynamically allocated char array, containing twice the contents of source
char* strDup(const char* source);

thanks for your comments

edit: I read the requirement as having the given string changed, and not returning anything. I now understand this is not accepted? I must have understood it wrongly

As you have it, it is extremely bad. In goes a pointer to the stack, out comes a pointer to the heap. How, for example, would you imagine calling the same function twice without leaking memory.

If you really want to modify the same pointer, make it a requirement that it points to dynamically allocated memory to begin with and handle memory accordingly.

The problem with the reference parameter approach is that you are essentially promising to manage the caller's memory, which generally just leads to very nasty bugs. By returning the string directly, you are basically telling the caller "I'll let you figure out what to do with this", which is an infinitely better idea altogether.