问题描述
所以，我们的目标是为我写出对应于此装配C code：
So the goal is for me to write out the C code that corresponds to this assembly :
0: 85 f6 test %esi,%esi
2: 78 13 js 17 <part3+0x17>
4: 83 fe 07 cmp $0x7,%esi
7: 77 14 ja 1d <part3+0x1d>
9: 8d 0c f5 00 00 00 00 lea 0x0(,%rsi,8),%ecx
10: 48 d3 ff sar %cl,%rdi
13: 48 89 f8 mov %rdi,%rax
16: c3 retq
17: b8 00 00 00 00 mov $0x0,%eax
1c: c3 retq
1d: b8 00 00 00 00 mov $0x0,%eax
22: c3 retq
我有点困惑，因为第一个循环测试％ESI寄存器结束第二循环结束了。
I am a little confused because the first loop testing the %esi register ends before the second loop ends.
时的if语句比较％ESI至7的第一个循环内的第二次？或者这是一个如果，否则，如果情??
Is the second if statement comparing %esi to 7 inside the first loop? or is this a if , else if situation??
推荐答案
让我总结一下，什么已经说过
Let me sum up, what's already been said
0: 85 f6 test %esi,%esi
2: 78 13 js 17 <part3+0x17>
这是如果（ESI℃下）转到17;
this is " if (esi < 0) goto 17; "
4: 83 fe 07 cmp $0x7,%esi
7: 77 14 ja 1d <part3+0x1d>
这是如果（ESI> 7）转到1D;
this is " if (esi >7) goto 1d; "
9: 8d 0c f5 00 00 00 00 lea 0x0(,%rsi,8),%ecx
CX = 8 * RSI//不是很明显它的只是一个乘法）
"cx = 8*rsi" // not that obvious it's "just" a multiplication)
10: 48 d3 ff sar %cl,%rdi
RDI >> CL; //没有CX，但CX是安全的，是＆LT; = 7 * 8，所以这是相同的
13：48 89 F8 MOV％RDI，RAX％
16：C3 retq
rdi >> cl; // not cx, but cx is safe to be <= 7*8, so that's the same 13: 48 89 f8 mov %rdi,%rax 16: c3 retq
返回RDI;
17: b8 00 00 00 00 mov $0x0,%eax
1c: c3 retq
17返回0
1d: b8 00 00 00 00 mov $0x0,%eax
22: c3 retq
1D：另一个返回0
1d: another "return 0"
所以C- code是：
so the C-Code is:
{
if (esi < 0) return 0;
if (esi > 7) return 0;
return rdi >> ( 8 * rsi );
}
PS：2返回0（第17和1D）表示他们不会在一个组合的如果
PSS：有C code的显然不是编译优化：P
PS: the 2 "return 0" (17 and 1d) indicate they're NOT combined in one "if"PSS: there C Code was obviously not compiled with optimization :P
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