Hi,
i'm writing a program that adds 2 4bit numbers and check if there is an overflow.
let's say i have A and B and each of them is 4 bits
i know if they have different signs, i wouldn't have overflow
my question is that how can i check for overflow after adding the A and B
Thanks
If your processor supports larger number of bits than 4 - which it's bound to do if you have a C-compiler working on it. So calculate using 8, 16 or 32-bit signed numbers (sign-extending the number to make it the longer size). Now check if the next bit up is the same sign as the sign-bit. If it isn't, you got an overflow.
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What about this:
Code:if ( ((nibble >> 4) << 4) == 0 ) { ... }
How are you doing the addition? Are you doing it yourself, bit-by-bit, or are you just using the '+' operator?Originally Posted by Lind
If you are adding by bit, all you need to do is see if the carry bit is set at the end. If you are using '+' to simulate 4-bit addition inside a larger integer type, just look to see if the value is greater than 7. If it is, there was overflow.
Which is also equivalent to:
Which is also equivalent to:Code:if((nibble & ~0x07) == 0)
Code:if(nibble > 7)
There will be an overflow only if both "most significant bits" are 1.
(in your case, that's the 3rd bit from the right)
http://en.wikipedia.org/wiki/Most_significant_bit
EDIT: Sorry, that was wrong.
Last edited by Govalant; 09-11-2007 at 09:29 PM.
No. If the leftmost bits are 0 and 1, or 1 and 0, you could still overflow if there is a carry. For instance, 1111 + 0111. The MSB's are not both 1, but it overflows.
Right, i was thinking kind of the opposite. If both MSB are 0 there will be no overflow.
INT_MAX - A < B
if this is true, then A + B would be > INT_MAX, hence overflow.
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How do you come to have 4-bit numbers? Are you using bitfields?
If so, you can simply check if the answer is less than one of the operands:Edit: Oh you said unsigned, which most of us seem to have missed...Code:struct foo { unsigned int bar:4; }; foo x, y, z; x.bar = 12; y.bar = 8; z.bar = x.bar + y.bar; if (z.bar < x.bar) { //Overflow occurred }
In that case, even what Salem posted wont work directly either.
Um, this is for two's complement right? (don't want to make any other wrong assumptions)
Last edited by iMalc; 09-12-2007 at 01:47 AM.
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Advice: Take only as directed - If symptoms persist, please see your debugger
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How about this:Code:struct foo { int bar : 4; }; foo x, y, z; x.bar = -6; y.bar = -5; z.bar = x.bar + y.bar; if ((int)z.bar != (int)x.bar + (int)y.bar) { //Overflow occurred }
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Advice: Take only as directed - If symptoms persist, please see your debugger
Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"
