1. ## negative to posative

How can i make -a into +a?

2. Multiply by -1.

3. Or you can use the absolute value function.
http://www.cprogramming.com/fod/abs.html

4. Originally posted by gamer4life687
Or you can use the absolute value function.
http://www.cprogramming.com/fod/abs.html
that page is 100% wrong. there is no abs() which takes and returns an int!

5. If you want to change the sign on any variable just do:
Code:
`a=-a;`
But if you only want to change negatives to positives, not the other way around, then use the absolute value function.

6. Code:
```Originally posted by Moi
that page is 100% wrong. there is no abs() which takes and returns an int```

Its funny you say it doesnt work. I tested the exact program on that site and it works 100%!. Maybe your compiler doesnt have that.

7. Originally posted by gamer4life687
Its funny you say it doesnt work. I tested the exact program on that site and it works 100%!. Maybe your compiler doesnt have that. [/B]
Right, you should mention your compiler on for example
mathematical questions

8. think before you speak

of course the program will work, ints and doubles can be implicitly converted around. that is not what i said though.

9. Simply return 0 - x.

Quzah.

10. To get the inverse of a number...

Code:
```//To get the (-) negative equiv. of 100 (-100) just reverse the bits and add 1,
//same for negative to positive.

int main(){
int i = 100;
i = ~i + 1;
cout<<i<<endl;
return 0;
}```
OR Multiply by -1.

11. ahh, i see. there is an int abs (int), but its in <stdlib.h>, not <math.h>. therefore the flaw in the tutorial program is #including the wrong header file, because without <stdlib.h> int abs (int) is not in scope, and double abs (double) is used with implicit conversions (although if <stdlib.h> had been there it would use an overloaded abs() instead). happy now?

12. Interesting what we can come up with with only one arguement. I sure hope that KrappyKoder responds to this and gets his answer. Now that we know how to do the change plz lets not argue anymore. I get the point.

13. No, clearly the best solution is the following:

Code:
```int a = -4394;

for (int i = 0; i < a*-2; i++)
a++;```

14. ## meh

The method i ended up using was from a friend, playing with unsigned and signed