swap addresses ?

[pras.vaidya]

hi, please help me with this problem : -

how to swap two addresses .For eg variable i is located at 2000 and j
at 3000 . Now i call swap function . Result should be that i should be
now having address 3000 and j should be having 2000.

Is it the normal swapping of two number ? if no please help me out.

Thanks in advance

 

[osmium]

how to swap two addresses .For eg variable i is located at 2000 and j
at 3000 . Now i call swap function . Result should be that i should be
now having address 3000 and j should be having 2000.

Is it the normal swapping of two number ? if no please help me out.

Something like this:

int i, j,temp;
/* give values to i and j*/
temp = i;
i = j;
j = temp;

That is the *right* way to do it. A question such as this may invoke some
so called "cute" digressions.

Note that we didn't swap the address, we swapped the variables. I followed
your description, not your message title. The numbers 2000 and 3000 have
nothing to do with it - high level programmers are not supposed to know the
numbers of addresses. They deal in variable *names*.

 

[Malcolm]

how to swap two addresses .For eg variable i is located at 2000 and j
at 3000 . Now i call swap function . Result should be that i should be
now having address 3000 and j should be having 2000.

Is it the normal swapping of two number ? if no please help me out.

Thanks in advance

Nearly.
If you use the identifiers "i" and "j" for your variables in C you have no
control over the address.

temp = i;
i = j;
j = temp;

will swap them

If you have pointers to the variables, such that ptr2k holds the address
"2000" and ptr3k holds the address "3000", in your platform-dependent
machine representation, then you can do this.

int temp = *ptr2k;
*ptr2k = *ptr3k;
*ptr3k = temp;

That will swap the contents of those addresss

 

[Joe Wright]

Nearly.
If you use the identifiers "i" and "j" for your variables in C you have no
control over the address.

temp = i;
i = j;
j = temp;

will swap them

If you have pointers to the variables, such that ptr2k holds the address
"2000" and ptr3k holds the address "3000", in your platform-dependent
machine representation, then you can do this.

int temp = *ptr2k;
*ptr2k = *ptr3k;
*ptr3k = temp;

That will swap the contents of those addresss

And if you really want to swap addresses..

void *temp = ptr2k;
ptr2k = ptr3k;
ptr3k = temp;

 

[Sorav Bansal]

Something like this:

int i, j,temp;
/* give values to i and j*/
temp = i;
i = j;
j = temp;

That is the *right* way to do it. A question such as this may invoke some
so called "cute" digressions.

Perhaps, this will qualify as a "cute" digression. But nonetheless, the
following code is more efficient in terms of both space and time:

void
swap (int i, int j)
{
i = i ^ j ;
j = i ^ j ;
i = i ^ j ;
}

 

[Sorav Bansal]

Perhaps, this will qualify as a "cute" digression. But nonetheless, the

following code is more efficient in terms of both space and time:

void
swap (int i, int j)
{
i = i ^ j ;
j = i ^ j ;
i = i ^ j ;
}

Oops, the function should be replaced by a macro (since all parameters
are call-by-value in C):
something like:

#define swap(i,j) do { i=i^j; j=i^j; i=i^j; } while (0)

 

[Walter Roberson]

how to swap two addresses .For eg variable i is located at 2000 and j
at 3000 . Now i call swap function . Result should be that i should be
now having address 3000 and j should be having 2000.

Are i and j the same type? Are they the same size? Is it certain
that the location of i is suitably aligned for a value of j's type
(and vice versas) ?

 

[Ben Pfaff]

Perhaps, this will qualify as a "cute" digression. But nonetheless,
the following code is more efficient in terms of both space and time:

void
swap (int i, int j)
{
i = i ^ j ;
j = i ^ j ;
i = i ^ j ;
}

Please read the FAQ.

20.15c: How can I swap two values without using a temporary?

A: The standard hoary old assembly language programmer's trick is:

a ^= b;
b ^= a;
a ^= b;

But this sort of code has little place in modern, HLL
programming. Temporary variables are essentially free,
and the idiomatic code using three assignments, namely

int t = a;
a = b;
b = t;

is not only clearer to the human reader, it is more likely to be
recognized by the compiler and turned into the most-efficient
code (e.g. using a swap instruction, if available). The latter
code is obviously also amenable to use with pointers and
floating-point values, unlike the XOR trick. See also questions
3.3b and 10.3.

 

[Emmanuel Delahaye]

Sorav Bansal wrote on 23/07/05 :

Oops, the function should be replaced by a macro (since all parameters are
call-by-value in C):
something like:

#define swap(i,j) do { i=i^j; j=i^j; i=i^j; } while (0)

Lack of parenthesis around the parameters, multiple evaluation of the
parameters, non portable code, you are cooked ! (^ works portably on
unsigned integer types)

--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html

"Mal nommer les choses c'est ajouter du malheur au
monde." -- Albert Camus.

 

[CBFalconer]

Oops, the function should be replaced by a macro (since all parameters
are call-by-value in C):
something like:

#define swap(i,j) do { i=i^j; j=i^j; i=i^j; } while (0)

It is a herd of errors waiting to trap you. Try calling it with:

swap(a, a[j]);

oh, you are going to make the passed items pointers. Alright, then
try the same swap when i == j and see what you get. Read the FAQ.

 

[Keith Thompson]

hi, please help me with this problem : -

how to swap two addresses .For eg variable i is located at 2000 and j
at 3000 . Now i call swap function . Result should be that i should be
now having address 3000 and j should be having 2000.

Is it the normal swapping of two number ? if no please help me out.

Several people have offered possible solutions, but I'm not sure that
anybody understands what you're asking. I know I don't.

A variable's address is determined when the variable is created, and
doesn't change over the lifetime of the variable. Changing the
address of a variable doesn't even make sense; if it has a different
address, it's a different variable.

Just to add to the confusion, you tell us that the addresses are 2000
and 3000. It's important to keep in mind that addresses are not
integers (though they might sometimes be represented that way).

The closest thing I can think of to what you're asking for is to
declare two variables, then declare two pointers that point to them.
You can then swap the values of the pointer variables. Swapping two
variables is (or should be) very straightforward, whether they're
pointers or anything else; others have posted examples using a
temporary variable.

If you can clarify what you're asking for, perhaps we can help.

 

[Kenny McCormack]

Nice Cajun-speak there. Well done, sir.

Several people have offered possible solutions, but I'm not sure that
anybody understands what you're asking. I know I don't.

I think it is perfectly clear what he is asking for.

And that the answer is, as it is to most questions asked here, that it
can't be done in standard C.

 

[Clark S. Cox III]

Perhaps, this will qualify as a "cute" digression. But nonetheless, the
following code is more efficient in terms of both space and time:

On almost every single processor that I've ever worked on, that is
simply false. Additionally, it will give the wrong result when both
variables have the same value.

Put simply, don't do it.

 

[Antonio]

<added code we're talking about>

i ^= j;
j ^= i;
i ^= j;

On almost every single processor that I've ever worked on, that is
simply false. Additionally, it will give the wrong result when both
variables have the same value.

Wrong. It still works if both variables have the same value. See the
breakdown just below:

int a = SOME_VALUE_WE_DONT_CARE_ABOUT;
int i = a;
int j = a;

i ^= j; // Effectively i = i ^ j = a ^a = 0. So after this step i = 0,
j = a;
j ^= i; // j = j ^ i = a ^ 0 = a; After this step i = 0, j = a;
i ^= j; // i = i ^ j = 0 ^ a = a; Finally i = a, j = a;

As you can see it still works if both variables have the same value.

The problem arises when the two variables are the same variable, as in
the following code:

void swap (int *i, int *j) {
*i ^= *j;
*j ^= *i;
*i ^= *j;
}

int main (int argc, char **argv) {
int a = SOME_VALUE_WE_DONT_CARE_ABOUT;

swap (&a, &a);
printf("%d\n", a);
}

In this case in the swap function we have:

*i ^= *j; // ==> a ^= a ==> a = a ^ a = 0;
*j ^= *i; // ==> a ^= a ==> a = a ^ a = 0 ^ 0 = 0;
*i ^= *j; // ==> a ^= a ==> a = a ^ a = 0 ^ 0 = 0;

So we end up with a 0 in a, not the initial value as was intended.

Put simply, don't do it.

Agreed.

 

[Chris Dollin]

Perhaps, this will qualify as a "cute" digression. But nonetheless, the
following code is more efficient in terms of both space and time:

void
swap (int i, int j)
{
i = i ^ j ;
j = i ^ j ;
i = i ^ j ;
}

It certainly is, since it can be optimised to

void swap( int i, int j ) {}

and this is futher subject to cheap inlining.

However, it's semantics are a tad unlike any kind of swap function.