What do these 2 lines of assembly code do? - memory

Question

I am in the middle of phase 2 for bomb lab and I can't seem to figure out how these two lines of assembly affect the code overall and how they play a role in the loop going on.
Here is the 2 lines of code:
add -0x24(%ebp,%ebx,4),%eax
cmp %eax,-0x20(%ebp,%ebx,4)
and here is the entire code:
Dump of assembler code for function phase_2:
0x08048ba4 <+0>: push %ebp
0x08048ba5 <+1>: mov %esp,%ebp
0x08048ba7 <+3>: push %ebx
0x08048ba8 <+4>: sub $0x34,%esp
0x08048bab <+7>: lea -0x20(%ebp),%eax
0x08048bae <+10>: mov %eax,0x4(%esp)
0x08048bb2 <+14>: mov 0x8(%ebp),%eax
0x08048bb5 <+17>: mov %eax,(%esp)
0x08048bb8 <+20>: call 0x804922f <read_six_numbers>
0x08048bbd <+25>: cmpl $0x0,-0x20(%ebp)
0x08048bc1 <+29>: jns 0x8048be3 <phase_2+63>
0x08048bc3 <+31>: call 0x80491ed <explode_bomb>
0x08048bc8 <+36>: jmp 0x8048be3 <phase_2+63>
0x08048bca <+38>: mov %ebx,%eax
0x08048bcc <+40>: add -0x24(%ebp,%ebx,4),%eax
0x08048bd0 <+44>: cmp %eax,-0x20(%ebp,%ebx,4)
0x08048bd4 <+48>: je 0x8048bdb <phase_2+55>
0x08048bd6 <+50>: call 0x80491ed <explode_bomb>
0x08048bdb <+55>: inc %ebx
0x08048bdc <+56>: cmp $0x6,%ebx
0x08048bdf <+59>: jne 0x8048bca <phase_2+38>
0x08048be1 <+61>: jmp 0x8048bea <phase_2+70>
0x08048be3 <+63>: mov $0x1,%ebx
0x08048be8 <+68>: jmp 0x8048bca <phase_2+38>
0x08048bea <+70>: add $0x34,%esp
0x08048bed <+73>: pop %ebx
0x08048bee <+74>: pop %ebp
0x08048bef <+75>: ret
I noticed the inc command that increments %ebx by 1 and using that as %eax in the loop. But the add and cmp trip me up every time. If I had %eax as 1 going into to the add and cmp what %eax comes out? Thanks! I also know that once %ebx gets to 5 then the loop is over and it ends the entire code.

Answer 1

You got a list of 6 numbers. This means you can compare at most 5 pairs of numbers. So the loop that uses %ebx does 5 iterations.
In each iteration the value at the lower address is added to the current loop count, and then compared with the value at the next higher address. As long as they match the bomb won't explode!
This loops 5 times:
add -0x24(%ebp,%ebx,4),%eax
cmp %eax,-0x20(%ebp,%ebx,4)
These numbers are used:
with %ebx=1 numbers are at -0x20(%ebp) and -0x1C(%ebp)
with %ebx=2 numbers are at -0x1C(%ebp) and -0x18(%ebp)
with %ebx=3 numbers are at -0x18(%ebp) and -0x14(%ebp)
with %ebx=4 numbers are at -0x14(%ebp) and -0x10(%ebp)
with %ebx=5 numbers are at -0x10(%ebp) and -0x0C(%ebp)

Answer 2

Those two instructions are dealing with memory at two locations, indexed by ebp and ebx. In particular, the add instruction is keeping a running total of all the numbers examined so far, and the comparison instruction is checking whether that is equal to the next number. So something like:
int total = 0;
for (i=0; ..., i++) {
total += array[i];
if (total != array[i+])
explode_bomb();
}