Why is it not work?

Question

#include <iostream>
using namespace std;
void swap(int x,int &y)
int temp;
temp=x;
x=y;
y=temp;
int main() {
int a=100;
int b=200;
cout <<"before swap , value of a "<<a<<endl;

swap(a,b);
cout <<"after swap , value of a "<<a<<endl;
cout <<"after swap , value of b "<<b<<endl;
return 0;
}

Answer 1

Yes by my smallmistake

Answer 2

#include <iostream>
using namespace std;
void wap(int x,int &y)
{int temp;
temp=x;
x=y;
y=temp;
}
int main() {
int a=100;
int b=200;
cout <<"before swap , value of a "<<a<<endl;

wap(a,b);
cout <<"after swap , value of a "<<a<<endl;
cout <<"after swap , value of b "<<b<<endl;
return 0;
}
function name is swap in that code
now see function name is wap

Answer 3

Close the swap function..
edit:
Sangeetha Santhiralingam
oh. you don't have opening brace also...

void swap( int x, int &y) { //add it

//function body..

} //add this
int main() {
..
}

Answer 4

Yes Jayakrishna🇮🇳

Answer 5

😑 okay

Answer 6

why is there a need to define swap?

"The function std::swap() is a built-in function in the C++ Standard Template Library (STL) which swaps the value of two variables."

you can use swap without having to define it yourself.😁

#include <iostream>
using namespace std;

int main()
{
int a = 5;
int b = 50;
cout << "Before swap:\n";
cout << "a = " << a << endl;
cout << "b = " << b << endl;
swap(a, b);
cout << "After swap:\n";
cout << "a = " << a << endl;
cout << "b = " << b << endl;

return 0;
}

Answer 7

Pass both <x> and <y> by reference

void swap( int& x, int& y )

Answer 8

What is your expected outcome by your code?
Your code only swaps y value with x. x remains same so x, y have same values.
edit: code is fine now.Sangeetha Santhiralingam

Answer 9

Sangeetha Santhiralingam
I guess it is for practice or an assignment. Make your own tools instead of using tools in the toolbox.

Answer 10

me too i have that problems

Answer 11

Hello dear

Answer 12

Gekjcx