C pointers and bit setting question

Question

As background to this question this was a test program I wrote to investigate some strange behavior we were seeing in a messaging program I work on. We have various structures that use bit flags to hold states about the messages, some of the flags use char, some use shorts. A utility we have to dump the messages to flat files was using a char* on these flag values to extract the settings, hence my deliberate use of a different pointer on flag2.

When this program is run on AIX UNIX 6.1 why is the output:

2 set
4 set
8 set
16 set
64 set

What happened to values 1 and 32 ?

#include <stdio.h>

#define SET(x,y) y |= (x)
#define UNSET(x,y) y &= (~x)

int main ( int argc, char *argv[] )
{
    unsigned char *g;
    unsigned short *h;
    unsigned long *i;

    char flag1; 
    short flag2; 
    long flag3;

    g = (char*) &flag2;

    SET(1, flag2);
    if ( 1 & *g )
        printf("1 set\n");
    UNSET(1, flag2);

    SET(2, flag2);
    if ( 2 & *g )
        printf("2 set\n");
    UNSET(2, flag2);

    SET(4, flag2);
    if ( 4 & *g )
        printf("4 set\n");
    UNSET(4, flag2);

    SET(8, flag2);
    if ( 8 & *g )
        printf("8 set\n");
    UNSET(8, flag2);

    SET(16, flag2);
    if ( 16 & *g )
        printf("16 set\n");
    UNSET(16, flag2);

    SET(32, flag2);
    if ( 32 & *g )
        printf("32 set\n");
    UNSET(32, flag2);

    SET(64, flag2);
    if ( 64 & *g )
        printf("64 set\n");
    UNSET(64, flag2);

    return 0;
}

Answer 1

AIX is a big endian architecture, that means that g points to the most significant bytes of flag. This means any changes to the least significant bytes are invisible to g.

flag2 is also never initialized so it will contain some arbirary values. So it turns out your think some bit setting worked when it was just the arbirary values that was there in the beginning.

If you run on a little endian machine like x86, things should behave as expected.