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習題二
wukunru edited this page Jun 10, 2022
·
6 revisions
#include <stdio.h>
int main()
{
int n;
int i;
n=0;
i=10;
do {
n = n + 1;
} while (n <= i);
printf("%d\n",n);
return 0;
}
enum {
Num = 128, Fun, Sys, Glo, Loc, Id,
Char, Else, Enum, If, Int, Return, Sizeof, While, Do,
Assign, Cond, Lor, Lan, Or, Xor, And, Eq, Ne, Lt, Gt, Le, Ge, Shl, Shr, Add, Sub, Mul, Div, Mod, Inc, Dec, Brak
};
int compile(int fd) {
int i, *t;
p = "char else enum if int return sizeof while do "
"open read write close printf malloc free memset memcmp exit void main";
i = Char; while (i <= Do) { next(); id[Tk] = i++; }
i = OPEN; while (i <= EXIT) { next(); id[Class] = Sys; id[Type] = INT; id[Val] = i++; }
next(); id[Tk] = Char;
next(); idmain = id;
if (!(lp = p = malloc(poolsz))) { printf("could not malloc(%d) source area\n", poolsz); return -1; }
if ((i = read(fd, p, poolsz-1)) <= 0) { printf("read() returned %d\n", i); return -1; }
p[i] = 0;
return prog();
}
./c6.exe dowhile.c
11
exit(0) cycle = 195
./c6.exe -s dowhile.c
1: #include <stdio.h>
2:
3: int main()
4: {
5: int n;
6: int i;
7: n=0;
1:660054 ENT 2
3:66005C LEA -1
5:660064 PSH
6:660068 IMM 0
8:660070 SI
8: i=10;
9:660074 LEA -2
11:66007C PSH
12:660080 IMM 10
14:660088 SI
9: do {
10: n = n + 1;
15:66008C LEA -1
17:660094 PSH
18:660098 LEA -1
20:6600A0 LI
21:6600A4 PSH
22:6600A8 IMM 1
24:6600B0 ADD
25:6600B4 SI
11: } while (n <= i);
26:6600B8 LEA -1
28:6600C0 LI
29:6600C4 PSH
30:6600C8 LEA -2
32:6600D0 LI
33:6600D4 LE
34:6600D8 BZ 38:6600E8
36:6600E0 JMP 15:66008C
12: printf("%d\n",n);
38:6600E8 ADDR 0:6A0058
40:6600F0 PSH
41:6600F4 LEA -1
43:6600FC LI
44:660100 PSH
45:660104 PRTF
46:660108 ADJ 2
13: return 0;
48:660110 IMM 0
50:660118 LEV