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;
}

>>235
First of all, g is the callee, but assuming you meant caller: No it doesn't, it simply adds a return from one of the variables in g.
You are probably right about the callee/caller. What is still right however is that leave alters the stack such that the program is undefined.

Any implementation of longjmp that alters the stack results in undefined behaviour, applying your logic.
I never claimed that, that's a lie and it's also incorrect. I stated that the use of __asm to alter the stack causes undefined behavior, I also stated that usage of longjmp in a manner defined by the C standard does not yield undefined behavior.

>>241
Under the assumptions that longjmp on your machine is implemented using __asm__ in GCC I will agree that logic dictates that the __asm__ that is used to implement longjmp is well defined on your machine as long as it is only used exactly like longjmp is defined to work in the standard, any other usage is still undefined.