what will be the addressing mode in assembly code generated by the compiler here ?

[Garrett2011]

Suppose we've got two integer and character variables:

int adad=12345;
char character;

Assuming we're discussing a platform in which, length of an integer variable is longer than or equal to three bytes, I want to access third byte of this integer and put it in the character variable, with that said I'd write it like this:

character=*((char *)(&adad)+2);

Considering that line of code and the fact that I'm not a compiler or assembly expert, I know a little about addressing modes in assembly and I'm wondering theaddress of the third byte(or I guess it's better to say offset of the third byte) here would be within the instructions generated by that line of code themselves, or it'd be in a separate variable whose address (or offset) is within those instructions ?

[tesuji]

hi,

decimal 12345 is hex 0x00003039. So to get a more practical example I changed:

character=*((char *)(&adad)+2);  // 0x00 is stored

to:

character=*((char *)(&adad)+1);  // 0x30 is stored

Assuming Intel processor, the assembly should look like:

lea ebx, adad         ; address points to 0x39 (little Endian)
add ebx, 1            ; address of 2nd byte, points to 0x30
mov al, [ebx]         ; loads 0x30 into reg al
mov character, al     ; stores 0x30 into character

After execution above code:

printf ("character: %x\n", character);

should output character: 30

I hope this is a little help.

Addition: Instead of add ebx,1 and mov al, [ebx] also displacement addressing: mov al, [ebx+1] is possible.

-- tesu

[r.evrard]

I would think that most modern compilers generate flat 32-bit code, don't they?. An int occupies 32 bits of memory. To see that, declare an int array, for example int a[3] and watch &a, &a[1] and &a[2] during debugging.

Here is the disassembly of the code generated by the visual studio compiler for your snippets above:
int main()
{
01171350 push ebp
01171351 mov ebp,esp
01171353 sub esp,0D8h
01171359 push ebx
0117135A push esi
0117135B push edi
0117135C lea edi,[ebp-0D8h]
01171362 mov ecx,36h
01171367 mov eax,0CCCCCCCCh
0117136C rep stos dword ptr es:[edi]
int adad=12345;
0117136E mov dword ptr [adad],3039h
char character;
character=*((char *)(&adad)+2);
01171375 mov al,byte ptr [ebp-6]
01171378 mov byte ptr [character],al

return 0;
This is probably not of much help.
As 12345 = 3039h, its first (most significant) two bytes are zero, so that character = 0. With adad = 1234567 = 12d687h, you get character = 18 = 12h, as expected.