Assembly - moving data 組合語言，移動資料

從記憶體中讀資料
load date from memory into register
把暫存器中存資料存進記憶體
store register data into memory

               Figure 1. integer registes, 
               source:https://www.cs.cmu.edu/~fp/courses/15213-s07/misc/asm64-handout.pdf

mov operand combinations(mov的運算整理)
mov imm to register, mov $0x11,%eax -> var_a=0x11;
mov imm to mem, mov $0x39,(%eax) -> *p=0x39;
mov reg to reg, mov %edx,%eax -> var_a=var_b;
mov reg to mem, mov %eax,(%edx) -> *p=var_a;
mov mem to reg, mov (%eax),%eax -> var_a=*p;

*can not do memory to memory transfer in single instruction.

Here is an simple swap code:

void swap(int *xp, int *yp)
{
   int a=*xp;
   int b=*yp;

   *xp=b;
   *yp=a;
}

Let's see how above code done by assembly.
Variables store in which memory place and registers is assigned by complier.
讓我們來看上面的程式碼,如何使用組合語言完成,
變數會放在哪個register對應到哪個memory位址，是由complier決定的。


gcc -g -c mov.c -m32 -o mov32

00000000 :
   0:   55                      push   %ebp
   1:   89 e5                   mov    %esp,%ebp
   3:   83 ec 10                sub    $0x10,%esp
//start-- int a=*xp; 
//move Memory[ebp+0x8](xp) to register exa.
//move memory content(*xp) to register eax
//move value(*xp) in eax to Memory[ebp-0x4](variable a).
   6:   8b 45 08                mov    0x8(%ebp),%eax
   9:   8b 00                   mov    (%eax),%eax
   b:   89 45 fc                mov    %eax,-0x4(%ebp)
//end int a=*xp;


//  int b=*yp;
// *yp會存在記憶體中ebp+0xC的位址,先把這位址存到eax中
// 再把exa記憶體中對應的值取出來存到,eax
//接著把exa的值,存到b變數所在的記憶體ebp-0x8中
   e:   8b 45 0c                mov    0xc(%ebp),%eax
  11:   8b 00                   mov    (%eax),%eax
  13:   89 45 f8                mov    %eax,-0x8(%ebp)
//end   int b=*yp;
  16:   8b 45 08                mov    0x8(%ebp),%eax
  19:   8b 55 f8                mov    -0x8(%ebp),%edx
  1c:   89 10                   mov    %edx,(%eax)
  1e:   8b 45 0c                mov    0xc(%ebp),%eax
  21:   8b 55 fc                mov    -0x4(%ebp),%edx
  24:   89 10                   mov    %edx,(%eax)
  26:   c9                      leave
  27:   c3                      ret

X86-64
 gcc -g -c mov.c -o mov64

0000000000000000 :
   0:   55                      push   %rbp
   1:   48 89 e5                mov    %rsp,%rbp
   4:   48 89 7d e8             mov    %rdi,-0x18(%rbp)
   8:   48 89 75 e0             mov    %rsi,-0x20(%rbp)
   c:   48 8b 45 e8             mov    -0x18(%rbp),%rax
  10:   8b 00                   mov    (%rax),%eax
  12:   89 45 fc                mov    %eax,-0x4(%rbp)
  15:   48 8b 45 e0             mov    -0x20(%rbp),%rax
  19:   8b 00                   mov    (%rax),%eax
  1b:   89 45 f8                mov    %eax,-0x8(%rbp)
  1e:   48 8b 45 e8             mov    -0x18(%rbp),%rax
  22:   8b 55 f8                mov    -0x8(%rbp),%edx
  25:   89 10                   mov    %edx,(%rax)
  27:   48 8b 45 e0             mov    -0x20(%rbp),%rax
  2b:   8b 55 fc                mov    -0x4(%rbp),%edx
  2e:   89 10                   mov    %edx,(%rax)
  30:   5d                      pop    %rbp
  31:   c3                      retq

reference:
https://www.cs.cmu.edu/~fp/courses/15213-s07/misc/asm64-handout.pdf