Swapping objects using pointers - c

Question

I'm trying to swap objects for a homework problem that uses void pointers to swap objects. The declaration of my function has to be:
void swap(void *a, void *b, size_t size);
I'm not looking for the exact code how to do it so I can figure it out by myself, but I'm not sure if I understand it correctly. I found that one problem is by doing:
void *temp;
temp = a;
a = b;
b = temp;
only changes what the pointers point to. Is that correct? If it is correct, why doesn't swapping pointers actually change the contents between *a and *b. Because if your pointer points to something different, couldn't you dereference it and the objects would now be different?
Similarly, just switching the values like:
void *temp;
*temp = *a;
*a = *b;
*b = *temp;
Is not correct either, which I'm not sure why. Because again, it seems to me that the content is switched.
Does swapping objects mean complete swapping of memory and value of what a pointer points to?
So it seems like I have to use malloc to allocate enough space for my swap. If I allocate enough memory for one object, assuming they are the same size, I don't really see how it is different than the other two methods above.
void *temp = malloc(sizeof(pa));
// check for null pointer
temp = a;
// do something I'm not sure of since I don't quite get how allocating space is any
// different than the two above methods???
Thanks!

Answer 1

Swapping pointers does not change the pointed-to values. If it did, that would be like swapping address labels on envelopes moving me into your house and you into mine.
You were nearly there:
void swap(void *a, void *b, size_t size) {
char temp[size]; // C99, use malloc otherwise
// char serves as the type for "generic" byte arrays
memcpy(temp, b, size);
memcpy(b, a, size);
memcpy(a, temp, size);
}
The memcpy function copies memory, which is the definition of objects in C. (Called POD or plain ol' data in C++, to compare.) In this way, memcpy is how you do assignment without caring about the type of the object, and you could even write other assignments as memcpy instead:
int a = 42, b = 3, temp;
temp = b;
b = a;
a = temp;
// same as:
memcpy(&temp, &b, sizeof a);
memcpy(&b, &a, sizeof a);
memcpy(&a, &temp, sizeof a);
This is exactly what the above function does, since you cannot use assignment when you do not know the type of the object, and void is the type that stands in for "unknown". (It also means "nothing" when used as function return type.)
As a curiosity, another version which avoids malloc in common cases and doesn't use C99's VLAs:
void swap(void *a, void *b, size_t size) {
enum { threshold = 100 };
if (size <= threshold) {
char temp[threshold];
memcpy(temp, b, size);
memcpy(b, a, size);
memcpy(a, temp, size);
}
else {
void* temp = malloc(size);
assert(temp); // better error checking desired in non-example code
memcpy(temp, b, size);
memcpy(b, a, size);
memcpy(a, temp, size);
free(temp);
}
}

Answer 2

To answer your first question, let's fill in some values to see what is happening:
void* a = 0x00001000; // some memory address
void* b = 0x00002000; // another memory address
/* Now we'll put in your code */
void* temp; // temp is garbage
temp = a; // temp is now 0x00001000
a = b; // a is now 0x00002000
b = temp; // b is now 0x00001000
So at the end of those statements, the pointer's values have been swapped, that is, whatever a was pointing to is now pointed to by b, and vice versa. The values of what those pointers are pointing to are unmodified, it's just that now their memory addresses are being held by different pointers.
To answer your second question, you cannot dereference a void*. The reason for this is that void has no size, so to try and dereference or assign to something that has no size is nonsensical. Thus, void* is a way of guaranteeing you can point to something, but you will never know what that something is without more information (hence the size parameter to your routine).
From there, knowing the pointer and the size of the data the pointer points to, you can use a routine like memcpy to move the data pointed to by one pointer into the location pointed to by another.

Answer 3

Parameters are like local variables, with values copied into them before the function starts executing. This prototype:
void swap(void *a, void *b, size_t size);
Means that the two addresses are copied into new variables called a and b. So if you change what is stored in a and b, nothing you do will have an any effect after swap returns.

Answer 4

I had a question similar to this for my C course. I think memcopy is probably best but you can also try this:
typedef unsigned char * ucp;
void swap(void *a, void *b, int size){
ucp c=(ucp)a;
ucp d=(ucp)b;
for(int i=0; i<size; i++){
int temp=(int)c[i];
c[i]=(int)d[i];
d[i]=temp;
}
}
Basically what this does is cast both pointers to an unsigned char pointer type. Then you increment the pointer, which in the case of an unsigned char, increments one BYTE at a time. Then what you're doing is basically copying the contents of each byte at a time in memory. If anyone wants to correct or clarify on this I would appreciate it too.

Answer 5

First of all, note that any changes to the pointers inside the function won't be propagated to outside the function. So you're going to have to move memory around.
The easiest way to do that is with memcpy - allocate a buffer on the stack, memcpy the appropriate size from a into it, memcpy from b to a, and one last memcpy from temp into b.

Answer 6

If you were writing a function to swap two integers, given pointers to them, your solution of swapping the values pointed to would work. However, consider the situation with
struct {
int a;
int b;
} a, b;
swap(&a, &b, sizeof(a));
You need to figure out a way to swap the contents of each value passed without any knowledge of what they actually consist of.

Answer 7

You're close.
The problem is: you are "swapping" only pointers a and b which are local variables in the function.
I assume outside of the function you have some variables, let's call them:
void *x = ...;
void *y = ...;
When you call:
swap(x, y, some_size);
a and b points to the same objects as x and y respectively. Now, when you swap what a and b points too, x and y are still pointing to what they where pointing before.
To change what memory x and y points you would have to pass a pointer to the x variable, so a pointer to a pointer :)
Because you can't change function declaration you can only swap the content of the memory where x (and a) and y (and b) points to. Some solutions are in other answers :) Generally memcpy is what you want.

Answer 8

To have any real effect, you need to do the equivalent of the second block you mentioned:
void *temp;
*temp = *a;
*a = *b;
*b = *temp;
The problem here is that 'void' doesn't have a size, so you can't assign 'void's as they stand. You need to allocate space for temp to point at, then copy the values using something like memcpy().

Answer 9

To change the pointer inside and keep it outside, you have to pass the pointer by reference or a double pointer.
If your function has to be like:
void swap(void *a, void *b, size_t size);
I suppose that you have to implement something like:
void * temp;
temp = malloc(size);
memcpy(temp,a,size);
memcpy(a,b,size);
memcpy(b,temp,size);
free(temp);

Answer 10

We need not use memcpy for swapping two pointers, following code works well(tested for swapping int* and char* strings):
void swap(void **p, void **q)
{
void *t = *p;
*p = *q;
*q = t;
}