汇编一个C switch语句-它是如何工作的? [英] Assembly for a C switch statement - how does it work?

问题描述

我正在读一本关于汇编开关语句的书，当输入n为大小写时，代码具有大小写/分支:100、102、103、104、106.它通过从n减去100来简化跳转表，如果结果大于6，将转到L2中的默认情况，否则将转到与％eax中的值匹配的相应分支.

I am reading a book about assembly switch statement, the code has cases/branches when the input n is case: 100, 102, 103, 104, 106. It simplified the jump table by subtracting 100 from n and then if the result is above 6, it goes to the default case in L2, Otherwise it will go to the corresponding branches that match the value in %eax.

我的问题是:

1. 如果是这样，如果跳转表的索引保存在%eax中，第7行是否不是jmp *.L7(,%eax)?

1. if so, isn't line 7 suppose to be jmp *.L7(,%eax) if the index of jump table is held in %eax?

为什么他们通过执行ja .L2将数字更改为第5行中的无符号?

And why did they change the number into unsigned in line 5 by doing ja .L2?

int x位于%ebp+8，int n位于%ebp+12

 movl 8(%ebp), %edx 
 movl 12(%ebp), %eax 
 subl $100, %eax 
 cmpl $6, %eax 
 ja .L2 
 jmp *.L7(,%eax,4) 
.L2: 
 movl $0, %eax 
 jmp .L8 
.L5: 
 movl %edx, %eax 
jmp .L9 
 .L3: 
 leal (%edx,%edx,2), %eax 
 leal (%edx,%eax,4), %eax 
 jmp .L8 
.L4: 
 leal 10(%edx), %eax 
.L9: 
 addl $11, %eax 
 jmp .L8 
.L6: 
 movl %edx, %eax 
 imull %edx, %eax 
.L8: 

跳转表

.section .rodata
.align 4 Align       
.L7:
.long .L3         //Case 100: loc_A
.long .L2         //Case 101: loc_def
.long .L4         //Case 102: loc_B
.long .L5        //Case 103: loc_C
.long .L6        //Case 104: loc_D
.long .L2       //Case 105: loc_def
.long .L6        //Case 106: loc_D

推荐答案

如果跳转表的索引保存在%eax中，

不是第7行应该是jmp *.L7(,%eax)吗?

isn't line 7 suppose to be jmp *.L7(,%eax) if the index of jump table is held in %eax?

跳转表中的每个条目都是一个long，它是4个字节.因此，eax缩放为4.

Each entry in the jump table is a long, which is 4 bytes. Hence eax is scaled by 4.

为什么他们通过执行ja .L2将数字更改为第5行中的无符号?

And why did they changed the number into unsigned in line 5 by doing ja .L2?

关键是要排除小于100且大于106的任何数字.我想很明显，它如何排除大于106的值.

The point is to exclude any number that's less than 100 and greater than 106. I assume it's obvious how it excludes values greater than 106.

因此，假设n小于100，例如99.如果再减去100，我们将得到-1，当它被视为无符号的32位值时是4294967295，它显然是高于" 6，并且像应该这样跳转到.L2.

So let's say n was less than 100, e.g. 99. If we then subtract 100 from that we get -1, which when viewed as an unsigned 32-bit value is 4294967295, which is obviously "above" 6, and the jump to .L2 is taken like it should.

subl $100, %eax   ; eax = 99-100 == -1
cmpl $6, %eax     ; set flags based on -1 - 6 == -7 => ZF=0 and CF=0
ja .L2            ; jump if ZF=0 and CF=0

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