I'm getting error what changes should I make to my code? #swapping no. By using function

Question

1.00/5 (1 vote)

See more:

C++

#include<iostream>
using namespace std;
int swap(int *, int *);
int main()
{
	int a, b;
	cout <<"Enter two number:\nA.";
	cin >> a;
	cout <<"B.";
	cin >> b;
	cout<<swap(&a, &b);
	return 0;
}
int swap(int x, int y)
{
	x = x + y;
	y = x - y;
	x = x - y; 
	return(x, y);

}

What I have tried:

#include<iostream>
using namespace std;
int swap(int *, int *);
int main()
{
int a, b;
cout <<"Enter two number:\nA.";
cin >> a;
cout <<"B.";
cin >> b;
cout<

Posted 8-Feb-20 16:54pm

thund3r cod3

Updated 10-Feb-20 2:07am

Add a Solution

4 solutions

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

phil.o · Answer 1 · 2020-02-08T17:32:00

- The implementation of the swap method does not match its declaration. Arguments should be pointers to integers, so that variables can be swapped. The mismatch should prevent compilation anyway, and compiler should issue an explicit error.

C++

int swap(int *x, int *y) // ...

- The return type of the swap function (an integer) does not match what you are actually trying to return (some kind of tuple). That should also be signaled as an error by the compiler. If you define x and y parameters as pointers, you won't need to return anything, anyway.

KarstenK · Answer 2 · 2020-02-08T22:41:00

Solution 2

You should use the reference syntax for your swap function:

C++

int swap(int &x, int &y)

Than no new instances of the variables are created on the call and you direct access the variables in the call.

Posted 8-Feb-20 22:41pm

KarstenK

Comments

CPallini 10-Feb-20 7:59am

Indeed. My 5.
Note the return value could possibly be void.

KarstenK 10-Feb-20 11:41am

You are right, but the result is used for output. ;-)

Shao Voon Wong · Answer 3 · 2020-02-09T15:40:00

Solution 3

The corrected code with pointers should be below.

C++

#include <iostream>
using namespace std;
void swap(int*, int*);

int main()
{
    int a, b;
    cout << "Enter two number:\nA.";
    cin >> a;
    cout << "B.";
    cin >> b;
    swap(&a, &b);
    cout << "a,b is now " << a << "," << b << endl;
    return 0;
}

// xor swap
void swap(int* x, int* y)
{
    *x = (*x) + (*y);
    *y = (*x) - (*y);
    *x = (*x) - (*y);
}

The sample output is below.

Enter two number:
A.56
B.89
a,b is now 89,56

Posted 9-Feb-20 15:40pm

Shao Voon Wong

Updated 10-Feb-20 2:35am

v3

Comments

Stefan_Lang 10-Feb-20 7:15am

Why did you post that solution? There is no indication another swap() implementation is needed, and the actual problem was aleady addressed.

Shao Voon Wong 10-Feb-20 7:49am

Indeed!

CPallini 10-Feb-20 8:55am

Have my 5.

KarstenK 10-Feb-20 11:42am

The code is buggy, when some overflow conditions occur.

CPallini · Answer 4 · 2020-02-10T02:07:00

To elaborate KarstenK solution (that is, by the way, the cleanest one), since you apparently don't master pointer (as well reference) syntax:

C

#include<iostream>
using namespace std;
void swap(int &, int &);

int main()
{
  int a, b;
  cout <<"Enter two number:\nA.";
  cin >> a;
  cout <<"B.";
  cin >> b;

  swap( a, b);
  cout << "After swap call\n";

  cout << "A = " << a << "\nB = " << b << "\n";
}
void swap(int & x, int & y)
{
  x = x + y;
  y = x - y;
  x = x - y;
}