Any Assembly Programmers?

[Tweezy]

I'm using the Intel x86 framework for my Assembly Programming at University. To be honest I am having quite a few issues understanding it. For example, the stack commands, moving address into memory or adding register values onto others. For example:

 

mov ecx, 0
mov ecx, 27
sub ecx, 15

Give the final value in ecx in HEXADECIMAL (ignore leading zeroes)

 

Assume the following instructions are executed:
mov ax,22
mov bx,29
mov cx,32
push bx
push cx
push ax
pop ax
pop bx
pop cx
The result in ax (in DECIMAL) would be
The result in bx (in DECIMAL) would be
The result in cx (in DECIMAL) would be 

 

Assume that the following instructions are executed:


mov eax, 0
mov eax, 4eh
add eax, d7h

Give the final value in eax in HEXADECIMAL (ignore leading zeroes).

 

These are just some example questions in a practice exam. I really don't understand them...

[Bob Loblaw]

I'm using the Intel x86 framework for my Assembly Programming at University. To be honest I am having quite a few issues understanding it. For example, the stack commands, moving address into memory or adding register values onto others. For example:

 

mov ecx, 0
mov ecx, 27
sub ecx, 15

Give the final value in ecx in HEXADECIMAL (ignore leading zeroes)

 

Assume the following instructions are executed:
mov ax,22
mov bx,29
mov cx,32
push bx
push cx
push ax
pop ax
pop bx
pop cx
The result in ax (in DECIMAL) would be
The result in bx (in DECIMAL) would be
The result in cx (in DECIMAL) would be 

 

Assume that the following instructions are executed:


mov eax, 0
mov eax, 4eh
add eax, d7h

Give the final value in eax in HEXADECIMAL (ignore leading zeroes).

 

These are just some example questions in a practice exam. I really don't understand them...

 

I don't know the specific language you are using, but the first one should be 12. By the looks of it, you move 0 into register exc, then move 27 into it (overwriting 0), then sub 15 from exc = 12.

 

Second you have 22 in ax, 29 in bx, 32 in cs. You push bx -> cx -> ax onto the stack. And stacks are LIFO. Then you pop into ax -> bx -> cx.

 

So ax is the one on top, then you pop it back into ax, then you pop cx into bx, and bx into cx. So the other *should* be:

22 (ax), 32 (bx), 29 (cx).

 

Last one is 293.

We are used to base 10, but basically the number 25 (to us) means 2 * 10^1 + 5 * 10^0 = (20) + (5).

 

Hex is the same, but base 16.

 

So 4eh (h just means the number is already in hex I believe), so 4 * 16^1 + e * 16^0. a:10, b:11, c:12, d:13, e:14, f:15, and since it's base 16, that's as high as it goes (because then it overflows into next column, same reason why 9 is the highest number in base 10).

 

So 4e = 14 * 16^0 = 14.

4 = 4 * 16^1 = 64

64 + 14 = 78.

 

d7h (d7) = 7 * 16^0 + d * 16^1

(7) + 13 * 16^1 = (7) + (208)

= 215.

 

These 2 are added together via add eax so 215 + 78 = 293.

 

 

edit: Answer to be left in hex so it's even easier.

 

4e

+d7

-----

e+7 = 5 carry the 1.

4 + d + 1 = 2 carry 1 again.

 

So you get 125h (not 125 in decimal, in hex -ie. 293 in decimal).

[fanatiik]

glad to see I'm not the only one that doesn't understand x86!

[MPQC]

0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90 0x90.

[Tweezy]

I don't know the specific language you are using, but the first one should be 12. By the looks of it, you move 0 into register exc, then move 27 into it (overwriting 0), then sub 15 from exc = 12.

 

Second you have 22 in ax, 29 in bx, 32 in cs. You push bx -> cx -> ax onto the stack. And stacks are LIFO. Then you pop into ax -> bx -> cx.

 

So ax is the one on top, then you pop it back into ax, then you pop cx into bx, and bx into cx. So the other *should* be:

22 (ax), 32 (bx), 29 (cx).

 

Last one is 293.

We are used to base 10, but basically the number 25 (to us) means 2 * 10^1 + 5 * 10^0 = (20) + (5).

 

Hex is the same, but base 16.

 

So 4eh (h just means the number is already in hex I believe), so 4 * 16^1 + e * 16^0. a:10, b:11, c:12, d:13, e:14, f:15, and since it's base 16, that's as high as it goes (because then it overflows into next column, same reason why 9 is the highest number in base 10).

 

So 4e = 14 * 16^0 = 14.

4 = 4 * 16^1 = 64

64 + 14 = 78.

 

d7h (d7) = 7 * 16^0 + d * 16^1

(7) + 13 * 16^1 = (7) + (208)

= 215.

 

These 2 are added together via add eax so 215 + 78 = 293.

 

 

edit: Answer to be left in hex so it's even easier.

 

4e

+d7

-----

e+7 = 5 carry the 1.

4 + d + 1 = 2 carry 1 again.

 

So you get 125h (not 125 in decimal, in hex -ie. 293 in decimal).

 

You're a life saver. I understand!