產生變數四則運算的 x86 組合語言

這個難很多。

add_1.c

 1 int main()
 2 {
 3   int i;
 4   int j;
 5 
 6   //i=23;
 7   //j=5678;
 8   //printf("i = %d\n", i+j);
 9   //printf("j = %d\n", j);
10   i=2;
11   j=1;
12   //printf("i+j = %d\n", i+j);
13 
14   printf("i = %d\n", i);
15   printf("j = %d\n", j);
16   //printf("i+5 = %d\n", i+5);
17   printf("5+i = %d\n", 5+i);
18 
19   //(1+2)+((3+4)+(5+6));
20   //printf("1+2 = %d\n", 1+2);
21   //printf("(1+2)+(3+4) = %d\n", (1+2)+(3+4));
22   //printf("1+2*3 = %d\n", 1+2*3);
23   //printf("xx 5*(6+7) = %d\n", 5*(6+7));
24   //printf("yy %d\n", 1+2);
25   //(2*3*4) * (5*(7*(8*9)));
26   // (2*3)*(5*6);
27   
28   //3*9;
29   //4*(2*3);
30   //2-1-6;
31 
32   // toplevel if (child[0]->is_leaf() && child[1]->is_leaf() != true)
33   // toplevel if (child[0]->is_leaf() != true && child[1]->is_leaf())
34   //1+2+5;
35   //5+(1+2);
36 
37   //printf("%d\n", (1+2+3)+ (5+6+7)+(9+10));
38 
39   // if (child[0]->is_leaf() != true && child[1]->is_leaf())
40   //(9+10) + ((1+2+3)+ (5+6+7));
41   //(1+2)+9+(3+5)+(7+8);
42 
43   // (child[0]->is_leaf() && child[1]->is_leaf() != true)
44   //(3+(5+6)+9) + (1+2);
45   //(1+2) + (9 + (3 +(5+6)));
46 
47   //(1+2)+(3+4+5);
48   //(1+2)+(3+4)+(5+6) + (7+8);
49 
50   //1+2+3+4+5;
51   //1+2;
52   //1+(2+3);
53   //1+2+3;
54   //(1+2)+(3+4);
55 }

r.S

 1 .section .rodata
 2 .LC1: 
 3     .string "i = %d\n"
 4 .section .rodata
 5 .LC2: 
 6     .string "j = %d\n"
 7 .section .rodata
 8 .LC3: 
 9     .string "5+i = %d\n"
10 .text
11 .globl main
12 .type main, @function
13 main:
14 pushl %ebp
15 movl %esp, %ebp
16 subl $4, %esp # reserve temp object size
17 subl $8, %esp # reserve local variable
18 mov $2 ,%eax
19 movl %eax, (%ebp)
20 mov $1 ,%eax
21 movl %eax, -4(%ebp)
22 pushl (%ebp)
23 pushl $.LC1
24 call printf
25 addl $8, %esp
26 pushl -4(%ebp)
27 pushl $.LC2
28 call printf
29 addl $8, %esp
30 movl $5, %eax
31 ADD (%ebp), %eax
32 pushl %eax
33 popl %eax
34 pushl %eax
35 pushl $.LC3
36 call printf
37 addl $8, %esp
38 leave
39 ret