swapping two number using pointer in C

Question

I tried to swap two integer using pointers...

#include<stdio.h>
int main()
{
int a,b,*i,*j;
printf("Enter two integer:");
scanf("%d%d",&a,&b);
i=&a;
j=&b;
a=*j;
b=*i;
printf("\n %d \t %d",a,b);
return 0;
}

The input is

12 45

The output is

45 45

After some trials, I found that If I assigned the b=*i first and then assigned a=*j, the first integer i.e 12 is repeating..

Why this happens ? In my understanding of pointer, this is what I've done. I've assigned the *j (i.e value of variable stored in address of a) to b and *i(i.e value of variable stored in address of b) to a..

Please explains what really happens in this program..

It is unclear whether you are looking for a C or C++ solution. Because several answers already gave C++ solutions, I wouldn't remove the C++ tag. Please next time be specific on C++ XOR C, don't mix those languages up. — Sebastian Mach
– Sebastian Mach, Commented Oct 26, 2013 at 18:22

Timothy Shields · Accepted Answer · 2013-10-26 18:12:15Z

4

Conceptually, this is what you want to do:

int temp = a; //temp <- 12
a = b;        //a <- 45
b = temp;     //b <- 12

Conceptually, this is what you're doing:

a = b; //a <- 45
b = a; //b <- 45

You can do this "elegantly" if you're using C++11:

std::tie(b, a) = std::make_tuple(a, b);

answered Oct 26, 2013 at 18:12

Timothy Shields

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1 Comment

sehe Over a year ago

+1 I like the C++11 variation. Could use an explanation why tie(b,a) = tie(a,b) wouldn't work though. And possibly a demo like tie(d,a,b,c) = make_tuple(a,b,c,d) to to show how it's more versatile. (For bonus points, a comparison of generated assembly code to show that it compiles to the same instructions)

VoidStar · Accepted Answer · 2013-10-26 18:15:14Z

3

This is what happens:

i=&a; //i points to a
j=&b; //j points to b
a=*j; //assign the value of b to a
b=*i; //assign the value of a, which has been assigned to the value of b in the previous step

This is a workaround:

int temp = a;
a = b;
b = temp;

answered Oct 26, 2013 at 18:15

VoidStar

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2 Comments

Wazir Ahmed Over a year ago

Thanks.. I don't know how I forgot this simple logic. :D

VoidStar Over a year ago

Please chose an answer which most suitably answers your question and accept it as correct, so that it is shown at the very top.

sehe · Accepted Answer · 2013-10-26 18:17:22Z

3

Simples:

#include<iterator>
std::iter_swap(i,j);

Or indeed

#include<algorithm>
std::swap(a,b);

Or for purists

using std::swap;
swap(a,b); // allow for ADL of user-defined `swap` implementations

answered Oct 26, 2013 at 18:17

sehe

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Comments

Gangadhar · Accepted Answer · 2013-10-26 18:18:02Z

2

a=*j;
b=*i;

i points address of a after first statement a value become 45 and next assigning a value to b so b also becomes 45

      addressof `a` 
      ^
i-----|

      addressof 'b' 
      ^ 
j-----|

Now when you make change to a then value dereferenced by i also changes.

simply use a temporary variable

int temp;
temp=*i;
*i=*j;
*j=temp;

edited Oct 26, 2013 at 18:18

answered Oct 26, 2013 at 18:12

Gangadhar

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Comments

niko · Accepted Answer · 2013-10-29 10:51:41Z

2

i=&a; //i points to a
j=&b; //j points to b
a=*j; // this statement is equvivalent to a = b; (since *j and b both are same)
so a = 45;
now
b=*i; // this statement is equvivalent to b = a; (since *i and a both are same) 
but a value is changed to 45 in previous statement, so the same 45 is assigned to b variable also

You could use temp variable

edited Oct 29, 2013 at 10:51

answered Oct 26, 2013 at 19:13

niko

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1 Comment

Wazir Ahmed Over a year ago

Yeah.. I forgot this simple logic.. Thanks :D

Ilya Kobelevskiy · Accepted Answer · 2013-10-26 18:22:19Z

0

This is because i points to a, so as soon as you assign something to a, original value stored in a is lost. You can either introduce third variable temp to store value between assignments (like in one of the answers posted), or do some trick like

a=a+b;
b=a-b;
a=a-b;

to avoid using third variable.

EDIT This technique would work only for unsigned integer type.

edited Oct 26, 2013 at 18:22

answered Oct 26, 2013 at 18:14

Ilya Kobelevskiy

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6 Comments

sehe Over a year ago

integer overflow lurking

Ilya Kobelevskiy Over a year ago

@sehe Wouldn't it still work despite of overflow - since it is circularly overflowing, so substructions would take care of overflow?

Andrei Over a year ago

@IlyaKobelevskiy except signed integer overflow is undefined

Sebastian Mach Over a year ago

@IlyaKobelevskiy: C++ defines only unsigned overflow.

michael_wu Over a year ago

Overflowing does reliably wrap-around only for unsigned data types. For signeds, it is undefined behavior.

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