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;
}

>>212
I'm not >>1-215, this is my first post in this thread.
But so does longjmp, therefore longjmp must also be undefined behavior, right?
setjmp and longjmp mention no alteration of the stack, setjmp saves the environment in a jmp_buf, longjmp restores the environment saved in the jmp_buf.
They may be implemented as certain alterations of the stack, but that doesn't mean it is the only way to do it. Their behaviour is well-defined.
http://pubs.opengroup.org/onlinepubs/7908799/xsh/longjmp.html
http://pubs.opengroup.org/onlinepubs/7908799/xsh/setjmp.html