Subj : Re: possible overflow?
To   : comp.programming
From : Roger Willcocks
Date : Sun Aug 21 2005 07:09 pm

<mr_semantics@hotmail.com> wrote in message 
news:1124626129.497709.106800@g49g2000cwa.googlegroups.com...
>I am still fiddling with bitwise operations - this time I have been
> working
> on a program that accepts input (an unsigned integral) from the user,
> and
> then after some 'conversion' outputs that number in binary format.  I
> have
> one problem that is vexing me though:
>
> #include <stdio.h>
>
> int main(void)
> {
>        unsigned int input = 8;
>        unsigned long int i;
>        unsigned long int j = 0;
>        size_t number_of_bits = sizeof(j) * 8;
>
>
>        j = (1 << (number_of_bits - 1));
>        for (i = 1; i != 0; i <<= 1) {
>                if (input & j) {
>                       (void) putchar('1');
>                } else {
>                       (void) putchar('0');
>                }
>                j >>= 1;
>        }
>        puts("");
>        return 0;
> }
>
> That is  a working program that gives an example of my problem - its
> output
> is:
>
> 0000000000000000000000000000111111111111111111111111111111111000
>
> instead of:
>
> 0000000000000000000000000000000000000000000000000000000000001000
>
> If I declare i and j as unsigned char, short or int, I get results as I
> would expect, output of 8, 16, or 32 bits respectively.  But long is
> not
> working.  I have sent it through the debugger, to no avail.  I am
> stumped.

Since you're looking at underlying bit patterns, it doesn't really matter 
whether the variables are signed or unsigned:

#include <stdio.h>
int main(int argc, char* argv[])
{
    int i;
    long long foo = 0x123456;
    for (i = 0; i < 8 * sizeof(foo); i++, foo += foo)
        putchar('0' + (foo < 0));
    return 0;
}

--

:-) Roger

.