Old school ascii shellcode

Question

This part of ascii shellcode set eax at zero with the and instruction (%...). In the debugger and in pratice this PIC code work all the time but why? the AND instruction algorithme is:

Operand target ← Operand target ∩ Operande Source
flag CF ← 0
flag OF ← 0 

It's possible that the eax register is not set to zero if the previous value of eax is not good?

#include <stdlib.h>

int main()
{
    asm("\
        and eax, 0x454e4f4a;\
        and eax, 0x3a313035;\
        ");

    return 0;
}

compilation line:

gcc -m32  -W -Wall -std=gnu99 -masm=intel -g eaxZero.c -o eaxZero

gdb instruction:

    (gdb)   b 5
..
    (gdb)   r
..
    (gdb)   x/i $eip
..
    (gdb)   x/x $eax
..
    (gdb)   nexti
..
    (gdb)   x/x $eax
..
    (gdb)   nexti
..
    (gdb)   x/x $eax
..