Assembly Code and Switch Statement Cases

Question

I'm having a bit of trouble understanding how the cases 50, 52, etc were determined through the assembly language.

From what I understand, the jump table corresponds to the actions to do in each case and that the check that edx > 5 means that the case ranges from 0 to 5? I believe 1 is left out because it is the default case, but why is 5 left out?

I feel like there should be a case 55: where result *= result, no?

If anyone could help explain, that would be great. Thank you!

int switch_prob(int x) 
{
    int result = x;
    switch (x)
    {
        case 50:
        case 52:
            result <<= 2;
            break;
        case 53:
            result >>= 2;
            break;
        case 54:
            result *= 3;
            break;
        default:
            result += 10;   
    }
    return result;
}
Figure 3.38 shows the disassembled object code for the procedure. We are only interested in the part of the code shown on lines 4 through 16. We can see on line 4 that parameter x (at offset 8 relative to %ebp) is loaded into register %eax, corresponding to program variable result. The “lea 0x0(%esi), %esi” instruction on line 11 is a nop instruction inserted to make the instruction on line 12 start on an address that is a multiple of 16

The jump table resides in a different area of memory. Using the debugger GDB, we can examine the six 4-byte words of memory starting at address 0x8048468 with the command x/6w 0x8048468. GDB prints the following:
(gdb) x/6w
0x8048468: 0x080483d5 0x080483eb 0x080483d5 0x0x80483e0
0x8048478: 0x080483e5 0x080483e8
(gdb)
Assembly Code:
1: 080483c0 <switch_prob>:
2: 80483c0: push %ebp
3: 80483c1: mov %esp,%ebp
4: 80483c3: mov 0x8(%ebp),%eax // X is copied into eax ; eax = x
5: 80483c6: lea 0xffffffce(%eax),%edx // placeholder 
6: 80483c9: cmp $0x5, %edx // Compare edx (3rd argument) with 5; Edx - 5 // clearly, edx is x
7: 80483cc: ja 80483eb <switch_prob+0x2b> // if edx - 5 > 0, Jump into line 16 (default)
8: 80483ce: jmp *0x8048468(,%edx,4) // Go into the jump table
9: 80483d5: shl $0x2, %eax // eax << 2
10: 80483d8: jmp 80483ee <switch_prob+0x2e> // Jump to line 17
11: 80483da: lea 0x0(%esi),%esi // esi = esi  NOP. Filling in N bytes
12: 80483e0: sar $0x2, %eax // eax >> 2
13: 80483e3: jmp 80483ee <switch_prob+0x2e> // Jump to line 17
14: 80483e5: lea (%eax, %eax, 2), %eax // eax = eax + 2(eax)
15: 80483e8: imul %eax, %eax // eax *= eax
16: 80483eb: add $0xa, %eax // eax += 10
17: 80483ee: mov %ebp, %esp // esp = ebp
18: 80483f0: pop %ebp
19: 80483f1: ret

Shouldn't I have 6 since Case 5 corresponds to 0x80483e8? @JoachimPileborg — Crowning
– Crowning, Commented Oct 19, 2015 at 5:52
Zero-based counting, from 0 to 5 (inclusive) there are six numbers. :) — Some programmer dude
– Some programmer dude, Commented Oct 19, 2015 at 6:01
@JoachimPileborg. Yeah, but from my example the only ones are 50, 52, 53, 54, and 51 for default. It's missing 55? Or is that absorbed into 54 — Crowning
– Crowning, Commented Oct 19, 2015 at 6:10
As you are probably aware lea 0xffffffce(%eax),%edx effectively adds -50(decimal) to eax (variable x) and stores it in edx (edx=-50+x). — Michael Petch
– Michael Petch, Commented Oct 19, 2015 at 7:18

Brendan · Accepted Answer · 2015-10-19 07:07:39Z

4

The assembly does not match the source code. It matches something more like this:

int switch_prob(int x) 
{
    int result = x;
    switch (x)
    {
        case 50:
        case 52:
            result <<= 2;
            break;
        case 53:
            result >>= 2;
            break;
        case 54:
            result *= 3;
            // WARNING: Falls through
        case 55:
            result *= result;
            // WARNING: Falls through
        default:
            result += 10;   
            break;
    }
    return result;
}

This is likely caused by human error (e.g. updating the source code in the question so it's not identical to the question last year's students got, but forgetting to update the assembly to match).

Never assume teachers/professors are non-human...

answered Oct 19, 2015 at 7:07

Brendan

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2 Comments

Michael Petch Over a year ago

I'd agree human error, sounds like the answer that was provided was incorrect (the answer was probably based on a similar exercise from a previous semester and not properly updated)

Crowning Over a year ago

Thank you! This makes much more sense. The problem was from a Harvard assignment too Lol

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