# Assembly logic and/or...

Would you please tell me are there instructions in Assembly I can use to logically and/or/xor two number?

I only know and or ...but they are bitwise.

Good Luck
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Commented:
There are no logic binary operations in assembler.. you have to evaluate every single expression of your code... in most cases you can combine several expressions into one using binary logic.

AND-instructions must be evaluated for each expression while OR and XOR can be combined.. if you take a look at the examples below you'll get the point why..

here is one example where AND binary doesn't work:

int a = 16;
int b = 32;

if (a&&b)
{
// will be called because a && b is true... note: a&b would evaluate to zero!
blah();
}

that would translate into: (x86 code assumed) (remember: this is wrong code!)

mov eax, 16
mov ebx, 32
and eax, ebx
jnz .label
call blah
..label:

however, if you rewrite it this way it works:

if (a) if (b)
{
blah();
}

translated:

mov eax, 16
mov ebx, 32
test eax, eax
jz   .label1
test ebx, ebx
jz   .label1
call blah
..label:

for the logical or operation you can simply use the binary or since or preserves bits:

int a = 16;
int b = 32;

if (a||b)
{
blah();
}

translates nicely into:

mov eax, 16
mov ebx, 32
or  eax, ebx
jz  .label1
call blah
..label1:

Logical or operations can always be evaluated using the binary or instruction (unless you do more arithmetics with the result.. see below)...

xor will always be evaluated binary.. there is no logical version of xor.

there are still some things to keep in mind: the compiler use a internal representation of the "true" state..

for example:

int a = 2;
int b = 2;
printf ("%d\n", (a&&b)|1);

this will print a 1, not a 3 because (a&&b) evaluates to true which is 0x01.. (I'm not 100% sure if it depends on the compiler).

if you have code that does such bad things you have to use the 1 for logical arithmetics in your assembly code.

mov ecx, 0
mov eax, 2
mov ebx, 2
test eax, eax
jz   .label1
mov  ecx, 1
..label1:
test ebx, ebx
jz   .label2
mov  ecx, ecx
..label2:

; now the or-part:
or   ecx, 1
push ecx
call print_integer

I hope this helped,
Nils
0

Commented:
ehm. the last example was incorrect..

try this one..

; ecx will hold the result:
mov ecx, 0
mov eax, 2
mov ebx, 2
test eax, eax
jz   .label1
mov  ecx, 1
..label1:
test ebx, ebx
jz   .label2
mov  ecx, 1
..label2:

; now the or-part:
or   ecx, 1
push ecx
call print_integer

0

Commented:
Assembly is CPU-specific.  What CPU?

>> logically and/or/xor two number?
*   *   *
>> I only know and or ...but they are
>> bitwise.

logical and/or/xor doesn't apply to numbers, only to boolean (true/false) values.   Bitwise and/or/xor applies to numbers.  If you represent your boolean values with true as 1 and false as 0, then you use the bitwise operators to perform logical (boolean) evaluations.
0
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