Code explanation

Can somebody explain why this code outputs what it does?

// tested with Core 2 Duo, Core 2 Quad and Xeon
// tested with gcc4.1.2 gcc4.4.3 and gcc4.6.1
// compile with: gcc -O0 -m32
#include <stdio.h>
#include <stdlib.h>
#include <signal.h>
#include <setjmp.h>

jmp_buf p;
void (*q)();

const char *data =
    "\x8b\x44\x24\x04\x8b\x5c\x24\x08"
    "\x8b\x00\x8b\x1b\x31\xc3\x31\xd8"
    "\x31\xc3\x8b\x4c\x24\x04\x89\x01"
    "\x8b\x4c\x24\x08\x89\x19\xc3\x90"
    "\x55\x89\xe5\x8b\x45\x04\xc9\xc3"
    "\x55\x90\x90\x89\xe5\x90\x90\x90"
    "\x8b\x45\x08\x89\x45\x04\xc9\xc3"
    "\x60\x09\x0e\x13\x14\x01\x0c\x0c"
    "\xc0\x07\x05\x0e\x14\x0f\x0f\x60"
    "\x00\x67\x6f\x74\x6f\x20\x63\x6f"
    "\x6e\x73\x69\x64\x65\x72\x65\x64"
    "\x20\x68\x61\x72\x6d\x66\x75\x6c"
    "\x6c\x00\x90\x90\x1c\x1b\x0a\x20";

int f(int x)
{
    static int b = 0; static int s = 0;
    int a = 0, t;
    if (!s) {
        a = b; b = x;
    } else {
        a = x; t = b;
        do {
            a ^= b;
            b = (a^b) & b;
            b <<= 1;
        } while (b);
        b = t;
    }
    s = (s+1) % 2;
    return a;
}

int g(int i, int *j)
{
    *j = i;
    i = (int) putchar;
    if (*j == (48 << 1)) 
        __asm volatile (
                "movl 8(%ebp),%eax;"
                "leave;"
                "ret"
                );
    return (int) puts;
}

void h(int i)
{
    int b;
    q = (void(*)()) g(i++[data],&b);
    for (f(b);*(data+i)!=b;++i,f(b))
        q(f(i[data])%0xff);
}

void sh(int s)
{
    if (s == 010)
        ((void(*)())g(s,&s))("F");
    longjmp(p,s);
}

int main(void)
{
    int base, addr = 0xffffffff, offs = 16;
    int a = 11, b = 32, i = 25;
    int s = 8, t = 1, u = 4;
    ((void(*)()) data)(&a,&b);
    ((void(*)()) data)(&b,&t);
    ((void(*)()) data)(&t,&s);
    addr ^= a;
       a ^= addr;
    addr ^= a;
    base = ((int(*)())data+addr)();
    if (a == -1)
        goto over;
    puts("A");

    base = (1<<3) | ((f(addr) + f(offs)) & ~0xff);
    h(base+addr+offs);
    exit(0);

over:
    signal(t,sh);signal(s,sh);signal(u,sh);

    if (!(s = setjmp(p))) {
        q = (void(*)()) g(0x30, &a);
        q(data + a + i);
        s = a / (b-1);
        puts("B");
    } else if (s == 0xb) {
        puts("C");
        ((int(*)(int)) data+addr+(offs/2))(base);
    } else {
        puts("D");
        *((int*) base+s) = 0xffffffff;
    }
   
    puts("E");
    return 1;
}

>>207
But OP specified it to "when ran on the proper architecture". So dismissing it entirely as non-deterministic is simply over-simplifying it. As I said, architecture dependent code does not imply non-deterministic behavior.
It doesn't imply anything beyond that of being undefined, which means that you can't guarantee that it's deterministic or non-deterministic, especially since GCC is allowed to do anything it wants you can't guarantee that GCC doesn't use undefined behavior to compile it, so that it will sometimes compile to the same program and other times it won't.

Yes I know, but he also never claimed it to be proper C anyway, in which case he wouldn't have mixed in inline assembly or running machine code.
Compiling with -pedantic won't point out the use of __asm since that is valid gnu89. The things that it does point out however is the use of undefined procedures like conversion between object types and function pointer types and arithmetic involving function pointer types.

But even if this was an extended C which had a stack, had __asm and where you were allowed to do conversion between object types and function pointer types and you could do arithmetic involving function pointer types the use of __asm leaves the program undefined as the GCC page specifically states that significant side-effects (like alteration of the stack as in the example) causes undefined behavior.

No matter how twist and turn this will you able to guarantee that the program will output the same thing no matter how many compiler flags you're using, you can't even guarantee that it will compile to the same program.