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I need 2 bits with offset 6 from bit array.

mov eax, [bitarray]; // get address of bit array
shr eax, 6; // clear first 6 bits
shl eax, 30 // clear last 30 bits
shr eax, 30; // move that 2 bits back to start

now in eax is these 2 bits i need, right?

When i have memory started from 0 (one unit is one bit), then bit on position 0 will be after load into register eax in most right place or most left place?

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2 Answers 2

Reset to default
3

Instead of the two shifts at the end you could have use a bitwise AND:

AND EAX,3  ; Keep the original value of the two least significant bits; all
           ; other bits in EAX are cleared.

The left-most bit is the most significant one, and the right-most bit the least significant one.

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4 Comments

yes i know left-most is the most significant, but if there is little endian, then in memory that left-most bit in register is where? On position 0 or position 7?
Endianness doesn't affect the order of bits within a byte. It affects the order of bytes with a word/dword/etc. The most significant bit of a byte is bit 7.
oh i see it now, if you are doing AND with 3
michael: yes, so first bit in memory isn't first bit in eax?
0

mov eax, [bitarray]

bitmask?

mov eax,[bitarray]
mov ebx,C0         ;11000000 binary
and ax,bx
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1 Comment

i think this is bad, it should be ANDed with value 3 (00000011)

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