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I was doing a project in ASM about pascal triangle using NASM
so in the project you need to calculate pascal triangle from line 0 to line 63
my first problem is where to store the results of calculation -> memory
second problem what type of storage I use in memory, to understand what I mean I have 3 way first declare a full matrices so will be like this way
memoryLabl: resd 63*64 ; 63 rows of 64 columns each
but the problem in this way that half of matrices is not used that make my program not efficient so let's go the second method is available
which is declare for every line a label for memory
for example :
line0: dd 1
line1: dd 1,1
line2: dd 1,2,1 ; with pre-filled data for example purposes
...
line63: resd 64 ; reserve space for 64 dword entries
this way of doing it is like do it by hand,
some other from the class try to use macro as you can see here
but i don't get it
so far so good
let's go to the last one that i have used
which is like the first one but i use a triangle matrices , how is that,
by declaring only the amount of memory that i need
so to store line 0 to line 63 line of pascal triangle, it's give me a triangle matrices because every new line I add a cell
I have allocate 2080 dword for the triangle matrices how is that ??
explain by 2080 dword:
okey we have line0 have 1 dword /* 1 number in first line */
line1 have 2 dword /* 2 numbers in second line */
line2 have 3 dword /* 3 numbers in third line */
...
line63 have 64 dword /* 64 numbers in final line*/
so in the end we have 2080 as the sum of them
I have give every number 1 dword
okey now we have create the memory to store results let's start calculation
first# in pascal triangle you have all the cells in row 0 have value 1
I will do it in pseudo code so you understand how I put one in all cells of row 0:
s=0
for(i=0;i<64;i++):
s = s+i
mov dword[x+s*4],1 /* x is addresses of triangle matrices */
second part in pascal triangle is to have the last row of each line equal to 1
I will use pseudo code to make it simple
s=0
for(i=2;i<64;i++):
s = s+i
mov dword[x+s*4],1
I start from i equal to 2 because i = 0 (i=1) is line0 (line1) and line0 (line1)is full because is hold only one (tow) value as I say in above explanation
so the tow pseudo code will make my rectangle look like in memory :
1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
1 1
...
1 1
now come the hard part is the calculation using this value in triangle to fill all the triangle cells
let's start with the idea here
let's take cell[line][row]
we have cell[2][1] = cell[1][0]+cell[1][1]
and cell[3][1]= cell[2][0]+cell[2][1]
cell[3][2]= cell[2][1]+cell[2][2]
in **general** we have
cell[line][row]= cell[line-1][row-1]+cell[line-1][row]
my problem I could not break this relation using ASM instruction because i have a
triangle matrices which weird to work with can any one help me to break it using a relation or very basic pseudo code or asm code ?
TL:DR: you just need to traverse the array sequentially, so you don't have to work out the indexing. See the 2nd section.
To random access index into a (lower) triangular matrix, row r starts after a triangle of size r-1. A triangle of size n has n*(n+1)/2 total elements, using Gauss's formula for the sum of numbers from 1 to n-1. So a triangle of size r-1 has (r-1)*r/2 elements. Indexing a column within a row is of course trivial, once we know the address of the start of a row.
Each DWORD element is 4 bytes wide, and we can take care of that scaling as part of the multiply, because lea lets us shift and add as well as put the result in a different register. We simplify n*(n-1)/2 elements * 4 bytes / elem to n*(n-1) * 2 bytes.
The above reasoning works for 1-based indexing, where row 1 has 1 element. We have to adjust for that if we want zero-based indexing by adding 1 to row indices before the calculation, so we want the size of a triangle
with r+1 - 1 rows, thus r*(r+1)/2 * 4 bytes. It helps to put the linear array index into a triangle to quickly double-check the formula
0
4 8
12 16 20
24 28 32 36
40 44 48 52 56
60 64 68 72 76 80
84 88 92 96 100 104 108
The 4th row, which we're calling "row 3", starts 24 bytes from the start of the whole array. That's (3+1)*(3+1-1) * 2 = (3+1)*3 * 2; yes the r*(r+1)/2 formula works.
;; given a row number in EDI, and column in ESI (zero-extended into RSI)
;; load triangle[row][col] into eax
lea ecx, [2*rdi + 2]
imul ecx, edi ; ecx = r*(r+1) * 2 bytes
mov eax, [triangle + rcx + rsi*4]
This assuming 32-bit absolute addressing is ok (32-bit absolute addresses no longer allowed in x86-64 Linux?). If not, use a RIP-relative LEA to get the triangle base address in a register, and add that to rsi*4. x86 addressing modes can only have 3 components when one of them is a constant. But that is the case here for your static triangle, so we can take full advantage by using a scaled index for the column, and base as our calculated row offset, and the actual array address as the displacement.
Calculating the triangle
The trick here is that you only need to loop over it sequentially; you don't need random access to a given row/column.
You read one row while writing the one below. When you get to the end of a row, the next element is the start of the next row. The source and destination pointers will get farther and farther from each other as you go down the rows, because the destination is always 1 whole row ahead. And you know the length of a row = row number, so you can actually use the row counter as the offset.
global _start
_start:
mov esi, triangle ; src = address of triangle[0,0]
lea rdi, [rsi+4] ; dst = address of triangle[1,0]
mov dword [rsi], 1 ; triangle[0,0] = 1 special case: there is no source
.pascal_row: ; do {
mov rcx, rdi ; RCX = one-past-end of src row = start of dst row
xor eax, eax ; EAX = triangle[row-1][col-1] = 0 for first iteration
;; RSI points to start of src row: triangle[row-1][0]
;; RDI points to start of dst row: triangle[row ][0]
.column:
mov edx, [rsi] ; tri[r-1, c] ; will load 1 on the first iteration
add eax, edx ; eax = tri[r-1, c-1] + tri[r-1, c]
mov [rdi], eax ; store to triangle[row, col]
add rdi, 4 ; ++dst
add rsi, 4 ; ++src
mov eax, edx ; becomes col-1 src value for next iteration
cmp rsi, rcx
jb .column ; }while(src < end_src)
;; RSI points to one-past-end of src row, i.e. start of next row = src for next iteration
;; RDI points to last element of dst row (because dst row is 1 element longer than src row)
mov dword [rdi], 1 ; [r,r] = 1 end of a row
add rdi, 4 ; this is where dst-src distance grows each iteration
cmp rdi, end_triangle
jb .pascal_row
;;; triangle is constructed. Set a breakpoint here to look at it with a debugger
xor edi,edi
mov eax, 231
syscall ; Linux sys_exit_group(0), 64-bit ABI
section .bss
; you could just as well use resd 64*65/2
; but put a label on each row for debugging convenience.
ALIGN 16
triangle:
%assign i 0
%rep 64
row %+ i: resd i + 1
%assign i i+1
%endrep
end_triangle:
I tested this and it works: correct values in memory, and it stops at the right place. But note that integer overflow happens before you get down to the last row. This would be avoided if you used 64-bit integers (simple change to register names and offsets, and don't forget resd to resq). 64 choose 32 is 1832624140942590534 = 2^60.66.
The %rep block to reserve space and label each row as row0, row1, etc. is from my answer to the question you linked about macros, much more sane than the other answer IMO.
You tagged this NASM, so that's what I used because I'm familiar with it. The syntax you used in your question was MASM (until the last edit). The main logic is the same in MASM, but remember that you need OFFSET triangle to get the address as an immediate, instead of loading from it.
I used x86-64 because 32-bit is obsolete, but I avoided too many registers, so you can easily port this to 32-bit if needed. Don't forget to save/restore call-preserved registers if you put this in a function instead of a stand-alone program.
Unrolling the inner loop could save some instructions copying registers around, as well as the loop overhead. This is a somewhat optimized implementation, but I mostly limited it to optimizations that make the code simpler as well as smaller / faster. (Except maybe for using pointer increments instead of indexing.) It took a while to make it this clean and simple. :P
Different ways of doing the array indexing would be faster on different CPUs. e.g. perhaps use an indexed addressing mode (relative to dst) for the loads in the inner loop, so only one pointer increment is needed. But if you want it to run fast, SSE2 or AVX2 vpaddd could be good. Shuffling with palignr might be useful, but probably also unaligned loads instead of some of the shuffling, especially with AVX2 or AVX512.
But anyway, this is my version; I'm not trying to write it the way you would, you need to write your own for your assignment. I'm writing for future readers who might learn something about what's efficient on x86. (See also the performance section in the x86 tag wiki.)
How I wrote that:
I started writing the code from the top, but quickly realized that off-by-one errors were going to be tricky, and I didn't want to just write it the stupid way with branches inside the loops for special cases.
What ended up helping was writing the comments for the pre and post conditions on the pointers for the inner loop. That made it clear I needed to enter the loop with eax=0, instead of with eax=1 and storing eax as the first operation inside the loop, or something like that.
Obviously each source value only needs to be read once, so I didn't want to write an inner loop that reads [rsi] and [rsi+4] or something. Besides, that would have made it harder to get the boundary condition right (where a non-existant value has to read as 0).
It took some time to decide whether I was going to have an actual counter in a register for row length or row number, before I ended up just using an end-pointer for the whole triangle. It wasn't obvious before I finished that using pure pointer increments / compares was going to save so many instructions (and registers when the upper bound is a build-time constant like end_triangle), but it worked out nicely.
I am reading and studying The Elements of Computing Systems but I am stuck at one point. Sample chapter skip the next 5 instruction s can be found here.
Anyway, I am trying to implement a Virtual Machine (or a byte code to assembly translator) but I am stuck at skip the next 5 instruction one point.
You can find the assembly notation here.
The goal is to implement a translator that will translate a specific byte code to this assembly code.
An example I have done successfully is for the byte code
push constant 5
which is translated to:
#5
D=A
#256
M=D
As I said, the assembly language for Hack is found in the link I provided but basically:
#5 // Load constant 5 to Register A
D=A // Assign the value in Reg A to Reg D
#256// Load constant 256 to Register A
M=D // Store the value found in Register D to Memory Location[A]
Well this was pretty straight forward. By definition memory location 256 is the top of the stack. So
push constant 5
push constant 98
will be translated to:
#5
D=A
#256
M=D
#98
D=A
#257
M=D
which is all fine..
I also want to give one more example:
push constant 5
push constant 98
add
is translated to:
#5
D=A
#256
M=D
#98
D=A
#257
M=D
#257 // Here starts the translation for 'add' // Load top of stack to A
D=M // D = M[A]
#256 // Load top of stack to A
A=M // A = M[A]
D=D+A
#256
M=D
I think it is pretty clear.
However I have no idea how I can translate the byte code
eq
to Assembly. Definition for eq is as follows:
Three of the commands (eq, gt, lt) return Boolean values. The VM
represents true and false as -1 (minus one, 0xFFFF) and 0 (zero,
0x0000), respectively.
So I need to pop two values to registers A and D respectively, which is quite easy. But how am I supposed to create an Assembly code that will check against the values and push 1 if the result is true or 0 if the result is false?
The assembly code supported for Hack Computer is as follows:
I can do something like:
push constant 5
push constant 6
sub
which will hold the value 0 if 2 values pushed to the stack are equal or !0 if not but how does that help? I tried using D&A or D&M but that did not help much either..
I can also introduce a conditional jump but how am I supposed to know what instruction to jump to? Hack Assembly code does not have something like "skip the next 5 instructions" or etc..
[edit by Spektre] target platform summary as I see it
16bit Von Neumann architecture (address is 15 bits with 16 bit Word access)
Data memory 32KW (Read/Write)
Instruction (Program) memory 32KW (Read only)
native 16 bit registers A,D
general purpose 16 bit registers R0-R15 mapped to Data memory at 0x0000 - 0x000F
these are most likely used also for: SP(R0),LCL(R1),ARG(R2),This(R3),That(R4)
Screen is mapped to Data memory at 0x4000-0x5FFF (512x256 B/W pixels 8KW)
Keyboard is mapped to Data memory at 0x6000 (ASCII code if last hit key?)
It appears there is another chapter which more definitively defines the Hack CPU. It says:
The Hack CPU consists of the ALU specified in chapter 2 and three
registers called data register (D), address register (A), and program
counter (PC). D and A are general-purpose 16-bit registers that can be
manipulated by arithmetic and logical instructions like A=D-1 , D=D|A
, and so on, following the Hack machine language specified in chapter
4. While the D-register is used solely to store data values, the contents of the A-register can be interpreted in three different ways,
depending on the instruction’s context: as a data value, as a RAM
address, or as a ROM address
So apparently "M" accesses are to RAM locations controlled by A. There's the indirect addressing I was missing. Now everything clicks.
With that confusion cleared up, now we can handle OP's question (a lot more easily).
Let's start with implementing subroutine calls with the stack.
; subroutine calling sequence
#returnaddress ; sets the A register
D=A
#subroutine
0 ; jmp
returnaddress:
...
subroutine: ; D contains return address
; all parameters must be passed in memory locations, e.g, R1-R15
; ***** subroutine entry code *****
#STK
AM=M+1 ; bump stack pointer; also set A to new SP value
M=D ; write the return address into the stack
; **** subroutine entry code end ***
<do subroutine work using any or all registers>
; **** subroutine exit code ****
#STK
AM=M-1 ; move stack pointer back
A=M ; fetch entry from stack
0; jmp ; jmp to return address
; **** subroutine exit code end ****
The "push constant" instruction can easily be translated to store into a dynamic location in the stack:
#<constant> ; sets A register
D=A ; save the constant someplace safe
#STK
AM=M+1 ; bump stack pointer; also set A to new SP value
M=D ; write the constant into the stack
If we wanted to make a subroutine to push constants:
pushR2: ; value to push in R2
#R15 ; save return address in R15
M=D ; we can't really use the stack,...
#R2 ; because we are pushing on it
D=M
#STK
AM=M+1 ; bump stack pointer; also set A to new SP value
M=D ; write the return address into the stack
#R15
A=M
0 ; jmp
And to call the "push constant" routine:
#<constant>
D=A
#R2
M=D
#returnaddress ; sets the A register
D=A
#pushR2
0 ; jmp
returnaddress:
To push a variable value X:
#X
D=M
#R2
M=D
#returnaddress ; sets the A register
D=A
#pushR2
0 ; jmp
returnaddress:
A subroutine to pop a value from the stack into the D register:
popD:
#R15 ; save return address in R15
M=D ; we can't really use the stack,...
#STK
AM=M-1 ; decrement stack pointer; also set A to new SP value
D=M ; fetch the popped value
#R15
A=M
0 ; jmp
Now, to do the "EQ" computation that was OP's original request:
EQ: ; compare values on top of stack, return boolean in D
#R15 ; save return address
M=D
#EQReturn1
D=A
#PopD
0; jmp
#EQReturn1:
#R2
M=D ; save first popped value
#EQReturn2
D=A
#PopD
0; jmp
#EQReturn2:
; here D has 2nd popped value, R2 has first
#R2
D=D-M
#EQDone
equal; jmp
#AddressOfXFFFF
D=M
EQDone: ; D contains 0 or FFFF here
#R15
A=M ; fetch return address
0; jmp
Putting it all together:
#5 ; push constant 5
D=A
#R2
M=D
#returnaddress1
D=A
#pushR2
0 ; jmp
returnaddress1:
#X ; now push X
D=M
#R2
M=D
#returnaddress2
D=A
#pushR2
0 ; jmp
returnaddress2:
#returnaddress3 ; pop and compare the values
D=A
#EQ
0 ; jmp
returnaddress3:
At this point, OP can generate code to push D onto the stack:
#R2 ; push D onto stack
M=D
#returnaddress4
D=A
#pushR2
0 ; jmp
returnaddress4:
or he can generate code to branch on the value of D:
#jmptarget
EQ ; jmp
As I wrote in last comment there is a branch less way so you need to compute the return value from operands directly
Lets take the easy operation like eq for now
if I get it right eq a,d is something like a=(a==d)
true is 0xFFFF and false is 0x0000
So this if a==d then a-d==0 this can be used directly
compute a=a-d
compute OR cascade of all bits of a
if the result is 0 return 0
if the result is 1 return 0xFFFF
this can be achieved by table or by 0-OR_Cascade(a)
the OR cascade
I do not see any bit shift operations in your description
so you need to use a+a instead of a<<1
and if shift right is needed then you need to implement divide by 2
So when I summarize this eq a,d could look like this:
a=a-d;
a=(a|(a>>1)|(a>>2)|...|(a>>15))&1
a=0-a;
you just need to encode this into your assembly
as you do not have division or shift directly supported may be this may be better
a=a-d;
a=(a|(a<<1)|(a<<2)|...|(a<<15))&0x8000
a=0-(a>>15);
the lower and greater comparison are much more complicated
you need to compute the carry flag of the substraction
or use sign of the result (MSB of result)
if you limit the operands to 15 bit then it is just the 15th bit
for full 16 bit operands you need to compute the 16th bit of result
for that you need to know quite a bit of logic circuits and ALU summation principles
or divide the values to 8 bit pairs and do 2x8 bit substraction cascade
so a=a-d will became:
sub al,dl
sbc ah,dh
and the carry/sign is in the 8th bit of result which is accessible
If there is any number in the range [0 .. 264] which can not be generated by any XOR composition of one or more numbers from a given set, is there a efficient method which prints at least one of the unreachable numbers, or terminates with the information, that there are no unreachable numbers?
Does this problem have a name? Is it similar to another problem or do you have any idea, how to solve it?
Each number can be treated as a vector in the vector space (Z/2)^64 over Z/2. You basically want to know if the vectors given span the whole space, and if not, to produce one not spanned (except that the span always includes the zero vector – you'll have to special case this if you really want one or more). This can be accomplished via Gaussian elimination.
Over this particular vector space, Gaussian elimination is pretty simple. Start with an empty set for the basis. Do the following until there are no more numbers. (1) Throw away all of the numbers that are zero. (2) Scan the lowest bits set of the remaining numbers (lowest bit for x is x & ~(x - 1)) and choose one with the lowest order bit set. (3) Put it in the basis. (4) Update all of the other numbers with that same bit set by XORing it with the new basis element. No remaining number has this bit or any lower order bit set, so we terminate after 64 iterations.
At the end, if there are 64 elements, then the subspace is everything. Otherwise, we went fewer than 64 iterations and skipped a bit: the number with only this bit on is not spanned.
To special-case zero: zero is an option if and only if we never throw away a number (i.e., the input vectors are independent).
Example over 4-bit numbers
Start with 0110, 0011, 1001, 1010. Choose 0011 because it has the ones bit set. Basis is now {0011}. Other vectors are {0110, 1010, 1010}; note that the first 1010 = 1001 XOR 0011.
Choose 0110 because it has the twos bit set. Basis is now {0011, 0110}. Other vectors are {1100, 1100}.
Choose 1100. Basis is now {0011, 0110, 1100}. Other vectors are {0000}.
Throw away 0000. We're done. We skipped the high order bit, so 1000 is not in the span.
As rap music points out you can think of the problem as finding a base in a vector space. However, it is not necessary to actually solve it completely, just to find if it is possible to do or not, and if not: give an example value (that is a binary vector) that can not be described in terms of the supplied set.
This can be done in O(n^2) in terms of the size of the input set. This should be compared to Gauss elimination which is O(n^3), http://en.wikipedia.org/wiki/Gaussian_elimination.
64 bits are no problem at all. With the example python code below 1000 bits with a set with 1000 random values from 0 to 2^1000-1 takes about a second.
Instead of performing Gauss elimination it's enough to find out if we can rewrite the matrix of all bits on triangular form, such as: (for the 4 bit version:)
original triangular
1110 14 1110 14
1011 11 111 7
111 7 11 3
11 3 1 1
1 1 0 0
The solution works like this: First all original values with the same most significant bit are places together in a list of lists. For our example:
[[14,11],[7],[3],[1],[]]
The last empty entry represents that there were no zeros in the original list. Now, take a value from the first entry and replace that entry with a list containing only that number:
[[14],[7],[3],[1],[]]
and then store the xor of the kept number with all the removed entries at the right place in the vector. For our case we have 14^11 = 5 so:
[[14],[7,5],[3],[1],[]]
The trick is that we do not need to scan and update all other values, just the values with the same most significant bit.
Now process the item 7,5 in the same way. Keep 7, add 7^5 = 2 to the list:
[[14],[7],[3,2],[1],[]]
Now 3,2 leaves [3] and adds 1 :
[[14],[7],[3],[1,1],[]]
And 1,1 leaves [1] and adds 0 to the last entry allowing values with no set bit:
[[14],[7],[3],[1],[0]]
If in the end the vector contains at least one number at each vector entry (as in our example) the base is complete and any number fits.
Here's the complete code:
# return leading bit index ir -1 for 0.
# example 1 -> 0
# example 9 -> 3
def leadbit(v):
# there are other ways, yes...
return len(bin(v))-3 if v else -1
def examinebits(baselist,nbitbuckets):
# index 1 is least significant bit.
# index 0 represent the value 0
bitbuckets=[[] for x in range(nbitbuckets+1)]
for j in baselist:
bitbuckets[leadbit(j)+1].append(j)
for i in reversed(range(len(bitbuckets))):
if bitbuckets[i]:
# leave just the first value of all in bucket i
bitbuckets[i],newb=[bitbuckets[i][0]],bitbuckets[i][1:]
# distribute the subleading values into their buckets
for ni in newb:
q=bitbuckets[i][0]^ni
lb=leadbit(q)+1
if lb:
bitbuckets[lb].append(q)
else:
bitbuckets[0]=[0]
else:
v=2**(i-1) if i else 0
print "bit missing: %d. Impossible value: %s == %d"%(i-1,bin(v),v)
return (bitbuckets,[i])
return (bitbuckets,[])
Example use: (8 bit)
import random
nbits=8
basesize=8
topval=int(2**nbits)
# random set of values to try:
basel=[random.randint(0,topval-1) for dummy in range(basesize)]
bl,ii=examinebits(basel,nbits)
bl is now the triangular list of values, up to the point where it was not possible (in that case). The missing bit (if any) is found in ii[0].
For the following tried set of values: [242, 242, 199, 197, 177, 177, 133, 36] the triangular version is:
base value: 10110001 177
base value: 1110110 118
base value: 100100 36
base value: 10000 16
first missing bit: 3 val: 8
( the below values where not completely processed )
base value: 10 2
base value: 1 1
base value: 0 0
The above list were printed like this:
for i in range(len(bl)):
bb=bl[len(bl)-i-1]
if ii and len(bl)-ii[0] == i:
print "example missing bit:" ,(ii[0]-1), "val:", 2**(ii[0]-1)
print "( the below values where not completely processed )"
if len(bb):
b=bb[0]
print ("base value: %"+str(nbits)+"s") %(bin(b)[2:]), b
I am trying to decipher some assembly code that involves multiple left rotations on an 8-bit binary number.
For reference, the code is:
lab: rol dl,1
rol dl,1
dec ecx
jnz lab
The dec and jnz isn't an issue, but is there to show that the 2 rols are executed several times.
What I am trying to do is figure out a mathematical equivalent of this code, such as a formula. I'm certainly not looking for a complete formula to tell me the whole code, but I would like to know if there is a formula that gives the equivalent (in denary) of a single left rotation.
I've tried figuring this out with a couple of different numbers, but cannot see a link between the two results. For example: if the start number is 115 it comes out as 220, but if the start number is 99 it comes out as 216.
Given your sample results, I assume we are treating the 8-bit quantity as unsigned.
The 7 low-order bits are shifted left, multiplying that part of the number by 2; and the high-order bit is swapped around to the beginning.
Thus, (x % 128) * 2 + (x / 128), using the usual integer div/mod operators.
Shifting a byte containing number X by one bit (position) left is equal to multiplying the number X by 2:
x << 1 <==> x = x * 2
I've been trying to reverse engineer a function of a game but I'm kinda confused. I'm pretty new to reverse engineering (I'm using ollydbg btw) so I don't really know about all the tricks and details yet.
Anyway here's my problem. This function is called when you pick up any Item in the game. It then calculates the value of the item and adds this value to your score. Before the function is called, a value is pushed which I'm quite confident is the ID of the item.
This is the code that confuses me:
SHL ESI,7
MOV CX,WORD PTR DS:[EDX+ESI+42]
ESI = the ID of the item
EDX = constant value FE56A0
I was guessing that EDX (FE56A0) was the start of an array of items, ESI was the index of the item somehow and 42 would be the index of the value the item holds. This would be kinda weird though since your bit shifting ESI to the left by 7. As ESI increases, it's bit shifted value doesn't grow linearly.
So if EDX represent the start of an array and ESI would be an index, the items in the array wouldn't be of equal size.
The meaning of this code is puzzling me.
Anyone got an idea what this code could represent?
The array might hold 128 byte long structures. Shifting by 7 multiplies the ID by 128, giving the offset required to access the structure for that ID. 42 would be the offset into the structure.
This works because multiplication actually increases the multiplied index linearly:
0 << 7 == 0
1 << 7 == 128
2 << 7 == 256
3 << 7 == 384
etc.
This code snippet simply accesses a member of a structure stored in an array.
It could be that EDX points to the start of some structure which the array is part of. The data that comes before the array requires 42 bytes, and each element in the array requires 128 bytes. (1<<7 is 128 - shifting is often used as a quick way to multiply by a power of two.) For example, something like this:
// EDX points here
struct GameItems
{
int numItems;
int stuff;
int moreStuff;
char[30] data;
GameItem[MAX_ITEMS] items; // offset 42 bytes from start
};
struct GameItem
{
// 128-bit structure
}