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Code explanation

Name: Anonymous 2012-01-11 12:49

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;
}

Name: Anonymous 2012-01-12 8:37

>>199
architecture dependent != undefined behavior
This is incorrect, read the standard.
undefined behavior == non-deterministic behavior
Whether this is correct or not is undefined, read the standard.

You must understand, the program in OP features both undefined behavior in the form of architecture (compiler, compiler flag, OS) dependencies and undefined behavior in the form of non-determinism.

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