a still has a negative value! Why?Code:int a= INT_MIN; a = -a; printf("%d", a);
It also mentions an overflow... how come?
a still has a negative value! Why?Code:int a= INT_MIN; a = -a; printf("%d", a);
It also mentions an overflow... how come?
Check the value of INT_MAX. Chances are you are dealing with a two's complement representation, so -INT_MIN cannot be stored in an int.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
Let me explain it with a 16bit arithmetic. The value of INT_MIN for it will be -32768 and INT_MAX will be 32767. Thus when you negate INT_MIN you get 32768 but this can't be represented in it coz the maximum value of INT is 32767, thus it'll go to the other side and print the corresponding value which in this case is -32768(the same as INT_MIN). Would it had been 32769 it would have printed -32767 and so on.
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
Whats a two's complement representation?
Yes i can print INT_MIN value. If i try printing -(INT_MIN) it's same as INT_MIN instead of being positive.
Oh! Thanks, this made it preety clear now.
For 2's complement, look here.
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
In 2's-complement, INT_MIN == -INT_MAX-1. For instance, if INT_MIN is -128, then INT_MAX is 127. So the negative of INT_MIN is 128, which is out of range.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
And rolls over to -128
Which of course is really just pure luck, because the behavior is undefined, since you are talking about signed values. It is not portable behavior.
Quzah.
Hope is the first step on the road to disappointment.
Yes, but that's no reason not to discuss what is actually happening, since we of course know that.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
Sure, I'm just being pedantic.
Quzah.
Hope is the first step on the road to disappointment.
