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Generic Swap without using template

The idea behind this was, I wanted to write a function that will take two parameters of the same type as a parameter and then it will swap them. It is a kind of generic swap but without the use of a C++ template. So the best way of doing it using "void *" as a parameter. As we know "void *" represents any arbitrary type that actually eases my job of writing a generic swap function. So, the function signature can be like below: 
void swap(void *arg1, void *arg2);
"void *" points to the starting address of the arbitrary location in the memory, irrespective of the bit pattern. Try to write the function like the below: 
void swap(void *arg1, void *arg2)
{
    void temp = *arg1;
    arg1 = *arg2;
    *arg = temp;
}
Oops, this is full of errors. 
1. We can't declare a variable of type "void".
2. "void *" can't be dereferenced. 
3. We also interested in swapping values. 
   So, number of bytes making up the values to 
   be passed as parameter.
Here is the functional version of the swap function. 
void swap(void *arg1, void *arg2, int size)
{
    char *buffer = new char[size];
    memcpy(buffer, arg1, size);
    memcpy(arg1, arg2, size);
    memcpy(arg2, buffer, size);
    delete []buffer;
}
How do we going to use this function from our client code? We can do it like below: (I have used vc++, 2005) 
int _tmain(int argc, _TCHAR* argv[])
{
 int one = 1;
 int two = 2;
 swap(&one, &two, sizeof(int *));

 cout << one << ":" << two << endl;

 char *n = _strdup("World");
 char *t = _strdup("Hello");
 swap(&n, &t, sizeof(char *));

 cout << n << ":" << t << endl;
 return 0;
}
and the output is like below: Please remember to pass the same type of parameter always as above. Avoid trying to pass arguments like: 
    int one = 1;
    short two = 2;
    swap(&one, &two, sizeof(short *));
The advantage of this over the template is the same assembly code being used for multiple calls as opposed to a template, which creates a separate copy for each call.

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