swap_floats function

[alielsaadi]

Im supposed to create a function that swaps the floats...
the outcome of this code gives me: 2,4... i want it to give me 4,2...
how would i do that?

#include <iostream>
#include <math.h>
using namespace std;

void swap_floats (float nb1, float nb2 );

void main ()
{
	float nbb1= 2;
	float nbb2= 4;

	swap_floats (nbb1, nbb2);
	cout << nbb1 << " " << nbb2; 

	system ( " Pause " );
}

void swap_floats ( float nb1, float nb2 )
{
	int x;
	x = nb1;
	nb1 = nb2;
	nb2 = x;



}

[kemotoe]

Your main function must return an int value, so swap it from void to int.

To swap values in a void returning function you must pass the values by reference.
Your logic in the function is right otherwise.
Look at this tutorial it should answer your question.
http://www.cplusplus.com/doc/tutorial/functions/

[alielsaadi]

that wasnt the problem. I figured it out.... in the function i had to set " float x = nb1; " instead of int x; and then saying x=nb1.

Thanks anyways

[Zhuge]

kemotoe is right; your call on line 12 is a fancy no-op. It does nothing as is, and will still do nothing if you make x a float.

[alielsaadi]

this worked fine for me

#include <iostream>
#include <math.h>
using namespace std;

void swap_floats (float & a, float & b);
void main ()
{
	float nb_1;
	float nb_2;

 cin >> nb_1;
 cin >> nb_2;

 swap_floats ( nb_1, nb_2 );
 cout << nb_1 << " " << nb_2;
		
 system ( " pause " );
}
void swap_floats (float & a, float & b)
{
 float temp = a;
 a = b;
 b = temp;
}

[Zhuge]

You made the changes kemotoe suggested, including the reference parameters. That's why. When you said "that wasn't the problem", it sounded like you weren't going to do that.