Interpreting an assembly jump table

Question

I am trying to interpret line-by-line what is this assembly code doing but I found myself really confused when presented with this jump table which is in assembly.This is taken from the textbook exercise question 3.63 but there is no explanation on it - hence why I am asking it here. The goal is to reverse engineer provided assembly listing and write C code which could generate it (feel switch statement body). Please help:(

The textbook is: Randal E. Bryant, David R. O'Hallaron - Computer Systems. A Programmer's Perspective [3rd ed.] (2016, Pearson)

qn 3.63

long switch_prob(long x, long n) { 
    long result = x;  
    switch(n) {  
            /* Fill in code here */  
    }  
    return result;  
}  

I am not sure how to 'decode' it or how to know where it is pointing to.

0000000000400590 <switch_prob>:  
    400590: 48 83 ee 3c             sub $0x3c,%rsi  
    400594: 48 83 fe 05             cmp $0x5,%rsi  
    400598: 77 29                     ja 4005c3 <switch_prob+0x33>  
    40059a: ff 24 f5 f8 06 40 00     jmpq *0x4006f8(,%rsi,8)  
    4005a1: 48 8d 04 fd 00 00 00     lea 0x0(,%rdi,8),%rax  
    4005a8: 00  
    4005a9: c3                         retq  
    4005aa: 48 89 f8                 mov %rdi,%rax  
    4005ad: 48 c1 f8 03             sar $0x3,%rax  
    4005b1: c3                         retq  
    4005b2: 48 89 f8                 mov %rdi,%rax  
    4005b5: 48 c1 e0 04             shl $0x4,%rax  
    4005b9: 48 29 f8                 sub %rdi,%rax  
    4005bc: 48 89 c7                 mov %rax,%rdi  
    4005bf: 48 0f af ff             imul %rdi,%rdi  
    4005c3: 48 8d 47 4b             lea 0x4b(%rdi),%rax  
    4005c7: c3                         retq  

The jump table resides in a different area of memory. We can see from the indirect jump on line 5 that the jump table begins at address 0x4006f8. Using the GDB debugger, we can examine the six 8-byte words of memory comprising the jump table with the command x/6gx 0x4006f8. GDB prints the following:

(gdb) x/6gx 0x4006f8
0x4006f8: 0x00000000004005a1 0x00000000004005c3
0x400708: 0x00000000004005a1 0x00000000004005aa
0x400718: 0x00000000004005b2 0x00000000004005bf

I understand that this line 40059a: ff 24 f5 f8 06 40 00 jmpq *0x4006f8(,%rsi,8)
is jumping to the table but I am unsure about how to
1)interpret the jump table [what does each address correspond to, what does each of the 6 values
mean/hold]
2) reverse engineer it to get the different cases of the switch statement.

Any help is appreciated, thank you:)

Answer 1

There are apparently (5 or) 6 case s of consecutive values, and the omnipresent default .

The jump table contains one address per case, and you will find these addresses in your listing.

For example, 0x00000000004005a1 is the address of this part:

    4005a1: 48 8d 04 fd 00 00 00     lea 0x0(,%rdi,8),%rax  
    4005a8: 00  
    4005a9: c3                         retq  

Because the second entry in the table points to the same address as the default (detected by cmp $0x5,%rsi and ja 4005c3 <switch_prob+0x33> ), we can assume that this case is not explicitly listed. That's why it might be just 5 case s.

The subtracted value 0x3c might be the character '<' in ASCII. As well you might like to interpret it in decimal.

The interpretation of each branch of the switch is left as an exercise for you, as this seems to be homework.