How can i make -a into +a?
How can i make -a into +a?
Multiply by -1.
Or you can use the absolute value function.
http://www.cprogramming.com/fod/abs.html
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that page is 100% wrong. there is no abs() which takes and returns an int!Originally posted by gamer4life687
Or you can use the absolute value function.
http://www.cprogramming.com/fod/abs.html
hello, internet!
If you want to change the sign on any variable just do:
But if you only want to change negatives to positives, not the other way around, then use the absolute value function.Code:a=-a;
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.
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Right, you should mention your compiler on for exampleOriginally 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]
mathematical questions
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.
hello, internet!
Simply return 0 - x.
Quzah.
Hope is the first step on the road to disappointment.
To get the inverse of a number...
OR Multiply by -1.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; }
Last edited by OneStiffRod; 12-27-2002 at 08:15 PM.
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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?
hello, internet!
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.
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No, clearly the best solution is the following:
Code:int a = -4394; for (int i = 0; i < a*-2; i++) a++;
The method i ended up using was from a friend, playing with unsigned and signed