pointers!!!

i have to write code for swapp() using int* as an argument.code is as follows-->

#include<iostream>

void swap_(int* a, int*b){

int c=(int) a;

int m=(int) b;

int n=c;

cout<<" the swapped value of a and b are"<<m<<"\t"<<n;

}

void main()

{
cout<<" \nenter a and b";
int m,n;
cin>>m>>n;
int *l=&m;
int *o=&n;
swap_( l, o );
}


is it right or am i doing wrong??
help me out.
actually what is happening is that 'a' is storing the address of 'm' and 'b' is storing the adress value of 'n'.
like this

for example
-----------
the value in a= | 0xffcd | (this is address of m)
-----------
-----------
the value in b= | 0xffce | (this is address of n)
-----------

now when i swap it.. value in a and b are swapped but i never get the value in integer.. i mean like if
m=4,n=5. after swapping it should be m=5;n=4;and let me remind you that i am restricted to use only int* as the argument.
would be glad to see some help'S message around.

that's
--------- ---------
| 0xffcd| and | 0xffce|
--------- ---------
;-)

look at the following:

Code:

---

#include <iostream.h>

void swap(int* a, int* b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}

int main()
{
      int a = 4;
      int b = 16;

      swap(&a,&b);
      cout << a << " " << b;
      return 0;
}

---

whenever you need to do a swap, you'll need a temporary variable.

Quote:

---

Originally posted by The Dog
whenever you need to do a swap, you'll need a temporary variable.

---

Not necessarily. Here's another version of the swap function:

Code:

---

void swap(int* a, int* b)
{   
        *a^=*b^=*a^=*b;
}

---

> *a^=*b^=*a^=*b;
Very disappointing

Even without that, this is a complete disaster if a and b happen to be pointing at the same location.

> *a^=*b^=*a^=*b;
Even without that, this is a complete disaster if a and b happen to be pointing at the same location.
--------------------
Yes, but the request was for an int swapper, and :

Quote:

---

from the faq:
If the values are integers, a well-known trick using exclusive-OR could perhaps be used,....

---

The fact it doesn't work for other structural types is irrelevant as this is not required here.

And if the programmer passes pointers that are using the same end variable, then surely that's a bug in itself? I know this code doesn't highlight that fact though, so I suppose it's bad because its hiding a programmer's bug?

Or is there something else your trying to tell me here?

And next time, don't be so darn blunt. :)

>If the values are integers, a well-known trick using exclusive-OR could perhaps be used,....
XOR swapping is also well known to be scary and no self-respecting C or C++ programmer would be caught dead using it. Yes it may work for integers, but you can't be sure of this. Better to use a temporary variable. Now for my next rant...

>*a^=*b^=*a^=*b;
.
.
>The fact it doesn't work for other structural types is irrelevant as this is not required here
The types in general are irrelevant since this swap is wrong and results in undefined behavior. The modification of a variable more than once in between sequence points is undefined. The "correct" (if you can call it that) way would look something like this:

Code:

---

void swap(int* a, int* b)
{   
  *a^=*b;
  *b^=*a;
  *a^=*b;
}

---

-Prelude

>and no self-respecting C or C++ programmer would be caught dead using it.
I agree, I wouldn't and haven't, if nothing, it's ugly. I only gave it to "prove a point". But it didn't (well, sort of), so I live and learn :rolleyes: :D

>Yes it may work for integers, but you can't be sure of this.
Why can't you be sure of it (assuming you're using the "expanded" version of the code)?

>Why can't you be sure of it
What if both items you are swapping are the same item in memory? Even if the type is int, the result is something similar to putting tin foil in the microwave. And before you say that something like that shouldn't happen, remember how often impossible situations do happen. :rolleyes:

-Prelude

>same memory
I'm glad that's all the problem is, I thought I was going mad, not being able to see another problem! Yes, I see what you mean, that particular occasion was covered ealier by Salem (and eskimo).

For the purposes of the example, I was relying on the fact it wouldn't happen, but I suppose, in the real world, black magic can happen, and make things go a little strange!

Anyway, enough of this convo, it reached it's end about 5 posts ago!

>and make things go a little strange!
Strange behavior in C and C++? Never!

>Anyway, enough of this convo, it reached it's end about 5 posts ago!
That never stopped us before. ;)

-Prelude

tell me something

void swap ( int* a , int* b )
{
/*cant i just swap the memory address???or is it just not possible??*/

// and cant i print the swapped value inside this function ???

//rather than printing in main()

// what i mean to say is that.

//now swapp the address and print the value stored

//in a and b

//cant i do that???

//help me guys.

//now i know that a= 0xfff ..and b= 0xfffe...

int c = a;//c gets a's value (address of m)

a = b; // a gets b's value ( address of n)

b = c ;// b gets a's value ( address of m)

//now what i want to do is that print the swapped value of m

//and n right here using this method( not in main()).

i would love to hear..help me



void main(){
//codes....i wanna use m and instead of a and b.
int( &m, &n);}

> void main()
ROFL LMAO!!!!!

After this entire thread too..

Quote:

---

Originally posted by Shadow
> void main()
ROFL LMAO!!!!!

After this entire thread too..

---

whats that now????

This was just a long thread about undefined behavior.
void main is undefined behavior.
Undefined behavior is very bad.

HELLO SHADOW!!! :D :D :D

i ask you straightly . can you help me?? i dont think you are eligible to help!!

that was for shadow!

Well...

Your algorithm for swapping seems okay, but you are not coding it properly. I think the others already presented the correct codes for you. :) You should try them and tell the difference between what you did and how the others did it. :D You can't learn if you don't analyze the problem.

Okay how about this:

In order to swap the values, you should not dereference the variables.

int *a, *b, *c;

what you did was like this: c = a; you are swapping the addresses but not necessarily the values. You should have done this *c = *a; this way you will surely get what you want.

I'm not sure if I have explained this quite accurately, but I'm sure this is the main idea. :)

> HELLO SHADOW!!!
Oh no! They let the trolls out again!
I'm outta here..

> can you help me??
We have - study the first post from The Dog

If you want to output inside the swap function, try
cout << "new value of a is " << *a << endl;

>>Oh no! They let the trolls out again!
I'm outta here..<<


Oh cmon Shadow :D Don't go just now... You can't imagine what I have gone through...

I had to endure these past few weeks of boredom because I really missed your stupidity HWAHAHAHAHAHAHAHAHAHAHAHA!!! :D:D:D