Exceeding unsigned boundary

Question

#include <stdio.h>
int main ()
{
unsigned char a=250,b=20, c;
c=a+b;
printf ("%d\n",c);
return 0;
}

What can I do for it to show 270? Besides declaring c as int.

Why must `a`, `b`, and `c` be defined as `unsigned char`? What other restrictions haven't you told us about? `puts("270")` solves the problem as stated; if that doesn't answer your question, you'll need to state the problem more clearly. — Keith Thompson, Aug 15 '13 at 23:22
@chris: No, it doesn't; the operands of `+` are promoted to `int`. — Keith Thompson, Aug 15 '13 at 23:24
Another silly solution: Compile and run the program on a system with `CHAR_BIT > 8`. (Good luck finding one!) — Keith Thompson, Aug 15 '13 at 23:25
@KeithThompson, Oh, shoot, I keep forgetting it does that for types smaller than `int`. If you did this with `unsigned int + unsigned int`, it would stay that. — chris, Aug 15 '13 at 23:27
@AssuraTheorycraft: Now help us by clarifying just what you were asking for. — Keith Thompson, Aug 15 '13 at 23:41
At a job interview I received this problem, they wanted to see if I could bypass the boundary restrictions of char, I wasn't allowed to modify the types, or at least not within the declaration part. — Assura Theorycraft, Aug 15 '13 at 23:48
@DeadMG: Actually `#define char int`, though it's a horribly bad idea, doesn't cause undefined behavior unless it occurs prior to the inclusion of standard header or to the expansion of any macro defined in a standard header. C11 7.1.2p4. Did I mention that it's still a horribly bad idea? — Keith Thompson, Aug 16 '13 at 17:43

score 3 · Accepted Answer · edited Aug 15 '13 at 23:21

3

Unsigned char's (assuming char is only eight bits) can only represent 2⁸ numbers, from 0 to 255. You'll need to use another type such as int to represent this.

edited Aug 15 '13 at 23:21

Keith Thompson

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answered Aug 15 '13 at 23:19

charmlessCoin

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i was hoping for something a little more clever, as I stated, I don't wish to define c as int – Assura Theorycraft Aug 15 '13 at 23:22
3

More clever? `puts("270")` is the best answer for a _clever_ response. `printf("%d\n", a + b);` is the next best, as it should convert the result to an int for you anyway. – charmlessCoin Aug 15 '13 at 23:27

score 1 · Answer 2 · answered Aug 15 '13 at 23:22

1

I'm sorry but you just can't , an unsigned char can hold a maximum of 255

answered Aug 15 '13 at 23:22

Farouq Jouti

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It's not clear that the value `270` ever has to be stored in an object of type `unsigned char`. One comment suggests `puts("270");`; another suggests `printf("%d\n", a+b);`. Either will work. – Keith Thompson Aug 15 '13 at 23:23
is type casting useful for example? after declaring can I convert the unsigned char to an int? or some other method? – Assura Theorycraft Aug 15 '13 at 23:24
@KeithThompson I know but I'm talking about the possibility of storing that number in an unsigned char (and you didn't have to vote down to say that) – Farouq Jouti Aug 15 '13 at 23:28
@FaroukJouti: Could have been someone else. (It’s best to leave votes out of things for the most part) – Ry- Aug 15 '13 at 23:29
@AssuraTheorycraft no that won't work either because in this case their will be a loss of information – Farouq Jouti Aug 15 '13 at 23:30
@FaroukJouti: In that case, no cast is necessary, since the result of `a + b` is already of type `int`. (And I didn't downvote your answer; please don't make assumptions.) – Keith Thompson Aug 15 '13 at 23:43
@FaroukJouti: It's not related to gcc; it's required by the C standard. (Actually `a + b` is of type `unsigned int` if `UCHAR_MAX > INT_MAX`, but that's impossible for a system with 8-bit bytes.) – Keith Thompson Aug 15 '13 at 23:55
@KeithThompson Yeah I just realized how wrong (stupid ?) I am , it took this much to realize , you're absolutely right the sum of both variables is an int :) – Farouq Jouti Aug 15 '13 at 23:59

Exceeding unsigned boundary

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