 |
Cheat Engine
The Official Site of Cheat Engine
|
| View previous topic :: View next topic |
| Author |
Message |
bknight2602
Grandmaster Cheater
![]() Reputation: 0
Joined: 08 Oct 2012
Posts: 586
|
 Posted: Wed Feb 02, 2022 6:45 pm Post subject: Odd way to display numbers |
 |
|
suppose I have a number say 1234567890 The game I have breaks these numbers into
Address 1 would contain values 1-99, in this case 90
Address 2 would contain values 100-9999 in this case 78
Address 3 would contain values 10000-999999 in this case 56
Address 4 would contain numbers 1 *1^6-99999999 in this case 34
Address 5 would contain values 1 *10*-9999999999 in this case 12.
The values in each address all have the range of 1-99.
I've been going over various formulas and running into numbers that are not correct. Any coding ideas?
|
|
| Back to top |
|
 |
LeFiXER
Grandmaster Cheater Supreme
![]() Reputation: 20
Joined: 02 Sep 2011
Posts: 1069
Location: 0x90
|
| Posted: Thu Feb 03, 2022 5:57 am Post subject: |
|
|
| I don't want you to think you are being ignored, but I feel your initial post lacks information; it's just too vague. I would advise providing actual data rather than a hypothetical scenario.
|
|
| Back to top |
|
|
bknight2602
Grandmaster Cheater
![]() Reputation: 0
Joined: 08 Oct 2012
Posts: 586
|
| Posted: Thu Feb 03, 2022 6:31 am Post subject: |
|
|
I never believe I was ignored, but I'm searching for coding scheme that will take a number and chop it into 10^2 10^4 etc. and display those two digits that represent those value.
345 would be 0 0 0 3 45
5684 would be 0 0 0 56 84
672301 would be 0 0 67 23 01
Is that clearer?
|
|
| Back to top |
|
|
++METHOS
I post too much
![]() Reputation: 92
Joined: 29 Oct 2010
Posts: 4197
|
| Posted: Thu Feb 03, 2022 7:16 am Post subject: |
|
|
| I know that you can get something very close in Excel with a formula, I am just not sure how you would account for the extra zeros. Probably easy for Excel experts.
|
|
| Back to top |
|
|
panraven
Grandmaster Cheater
![]() Reputation: 62
Joined: 01 Oct 2008
Posts: 958
|
| Posted: Thu Feb 03, 2022 8:00 am Post subject: |
|
|
You can 'split' a number using divide and modulus operation like,
math.floor(12345 / 100 ) -> 123
12345 % 100 -> 45
Here a sample iterator (use in 'for ... in .. do' ) to generate such numbers,
| Code: |
function repeatDivMod(M, D)
local n, d, r = M, D or 100
return function()
if n>0 and d>0 then
n, r = math.floor(n/d), n % d
return n, r
end
end
end
-- test
local todisplay = {}
for n,r in repeatDivMod(71234567890,100)do
todisplay[1+#todisplay] = r
end
print(table.concat(todisplay,', '))
-- output
90, 78, 56, 34, 12, 7
|
Not work for negative numbers.
_________________
- Retarded. |
|
| Back to top |
|
|
bknight2602
Grandmaster Cheater
![]() Reputation: 0
Joined: 08 Oct 2012
Posts: 586
|
| Posted: Thu Feb 03, 2022 9:07 am Post subject: |
|
|
| ++METHOS wrote: | | I know that you can get something very close in Excel with a formula, I am just not sure how you would account for the extra zeros. Probably easy for Excel experts. |
That's the problem because the number we are dealing with has a large range 0-9999999999
Any formula needs to be able to deal with all those possibilities.
|
|
| Back to top |
|
|
AylinCE
Grandmaster Cheater Supreme
 Reputation: 37
Joined: 16 Feb 2017
Posts: 1526
|
| Posted: Thu Feb 03, 2022 9:13 am Post subject: |
|
|
| bknight2602 wrote: | I never believe I was ignored, but I'm searching for coding scheme that will take a number and chop it into 10^2 10^4 etc. and display those two digits that represent those value.
345 would be 0 0 0 3 45
5684 would be 0 0 0 56 84
672301 would be 0 0 67 23 01
Is that clearer? |
I will not do complex calculations.
I will only take it superficially and give an example:
| Code: | function convert1(num)
b = (tostring(num)):reverse()
lnn="0 "
a = b:gsub("..", "%1 "):sub(1,-1)
--print("a:"..a)
a=(a):reverse():sub(1,-1)
if (a):sub(1,1)==" " then a=(a):sub(2, -1) end
--print("a1:"..a)
a1=string.len(a) / 3
a1=math.floor(5 - tonumber(a1))
--print(a1)
ln=string.rep(lnn, a1) .. a
--print(ln)
return ln
end
--Format by splitting:
function convert2(num,fmt)
num1=math.floor(tonumber(num) / tonumber(fmt))
print("num1:" .. num1)
b = (tostring(num1)):reverse()
lnn="0 "
a = b:gsub("..", "%1 "):sub(1,-1)
--print("a:"..a)
a=(a):reverse():sub(1,-1)
if (a):sub(1,1)==" " then a=(a):sub(2, -1) end
--print("a1:"..a)
a1=string.len(a) / 3
a1=math.floor(5 - tonumber(a1))
--print(a1)
ln=string.rep(lnn, a1) .. a
--print(ln)
return ln
end
print("convert1.3: " .. convert1(345))
print("convert1.7: " .. convert1(6723012)) --opps! max=9
print("convert2/2: " .. convert2(123456789,2))
print("convert2/4: " .. convert2(123456789,4)) |
_________________
|
|
| Back to top |
|
|
bknight2602
Grandmaster Cheater
![]() Reputation: 0
Joined: 08 Oct 2012
Posts: 586
|
| Posted: Thu Feb 03, 2022 9:30 am Post subject: |
|
|
| panraven wrote: | You can 'split' a number using divide and modulus operation like,
math.floor(12345 / 100 ) -> 123
12345 % 100 -> 45
Here a sample iterator (use in 'for ... in .. do' ) to generate such numbers,
| Code: |
function repeatDivMod(M, D)
local n, d, r = M, D or 100
return function()
if n>0 and d>0 then
n, r = math.floor(n/d), n % d
return n, r
end
end
end
-- test
local todisplay = {}
for n,r in repeatDivMod(71234567890,100)do
todisplay[1+#todisplay] = r
end
print(table.concat(todisplay,', '))
-- output
90, 78, 56, 34, 12, 7
|
Not work for negative numbers. |
That works great but I don't understand the scheme, so I'll ask for some expansion of the idea (which is great)
I need to be able to take these outputs and set them into specific addresses in my table. Now for my questions.
So n would be described as n1, n2 , n3 etc.?
It looks like the last n will always be tens digits so n-1 would be thousands etc., correct?
All of the larger address values would be zero in a number like 5799. There would be only two significant entries the rest would be 0, i.e. 99 57 0 0 0. How to set those specific addresses to 0?
|
|
| Back to top |
|
|
AylinCE
Grandmaster Cheater Supreme
Reputation: 37
Joined: 16 Feb 2017
Posts: 1526
|
| Posted: Thu Feb 03, 2022 10:58 am Post subject: |
|
|
| bknight2602 wrote: | I can't double post yet but this is my question in a quote.
Another great example that I don't understand the scheme nor how to arrive at each element used in setting values in specific addresses.
Your convert2/2 and 2/4 but we don't know the number in advance which scheme would work for an unknown variable 0-9999999999 and then how would I obtain each output to assign values in specific address? |
I still don't understand the logic of splitting a plain value into Aobs-style bytes.
I think that's why I posted a plain replacement code.
As far as I understand from the question, you can turn to any of the following stages (ideas):
| Code: | function searchAddrValue(search)
rst=""
local aobs1=AOBScan(search)
if aobs1~=nil then
aobs2=readInteger(aobs1)
if (string.len(aobs2))==3 then
rst=rst .. convert2(aobs2,1)
end
if (string.len(aobs2))==4 then
rst=rst .. convert2(aobs2,2)
end
if (string.len(aobs2))==5 then
rst=rst .. convert2(aobs2,3)
end
if (string.len(aobs2))==6 then
rst=rst .. convert2(aobs2,4)
end
if (string.len(aobs2))==7 then
rst=rst .. convert2(aobs2,4)
end
if (string.len(aobs2))==8 then
rst=rst .. convert2(aobs2,5)
end
if (string.len(aobs2))==9 then
rst=rst .. convert2(aobs2,5)
end
end
return rst
end
src=searchAddrValue(tostring("00 00 80 3F 00 ?? ?? ?? FF"))
print(src) |
Now you can expand the question a little more to understand better. 
_________________
|
|
| Back to top |
|
|
bknight2602
Grandmaster Cheater
![]() Reputation: 0
Joined: 08 Oct 2012
Posts: 586
|
| Posted: Thu Feb 03, 2022 11:20 am Post subject: |
|
|
| AylinCE wrote: | | bknight2602 wrote: | I never believe I was ignored, but I'm searching for coding scheme that will take a number and chop it into 10^2 10^4 etc. and display those two digits that represent those value.
345 would be 0 0 0 3 45
5684 would be 0 0 0 56 84
672301 would be 0 0 67 23 01
Is that clearer? |
I will not do complex calculations.
I will only take it superficially and give an example:
| Code: | function convert1(num)
b = (tostring(num)):reverse()
lnn="0 "
a = b:gsub("..", "%1 "):sub(1,-1)
--print("a:"..a)
a=(a):reverse():sub(1,-1)
if (a):sub(1,1)==" " then a=(a):sub(2, -1) end
--print("a1:"..a)
a1=string.len(a) / 3
a1=math.floor(5 - tonumber(a1))
--print(a1)
ln=string.rep(lnn, a1) .. a
--print(ln)
return ln
end
--Format by splitting:
function convert2(num,fmt)
num1=math.floor(tonumber(num) / tonumber(fmt))
print("num1:" .. num1)
b = (tostring(num1)):reverse()
lnn="0 "
a = b:gsub("..", "%1 "):sub(1,-1)
--print("a:"..a)
a=(a):reverse():sub(1,-1)
if (a):sub(1,1)==" " then a=(a):sub(2, -1) end
--print("a1:"..a)
a1=string.len(a) / 3
a1=math.floor(5 - tonumber(a1))
--print(a1)
ln=string.rep(lnn, a1) .. a
--print(ln)
return ln
end
print("convert1.3: " .. convert1(345))
print("convert1.7: " .. convert1(6723012)) --opps! max=9
print("convert2/2: " .. convert2(123456789,2))
print("convert2/4: " .. convert2(123456789,4)) |
|
By inspection one knows that 345 has length of three and this example should output 0 0 0 3 45
again, your second example 6723012, has a length of 7 and it output should be 0 6 72 30 12
Your third example 123456789 has a length of 9 and its output should be 1 23 45 67 89.
The problem is we don't know the length of the number, so we need a routine that will handle any/all of these variables.
This attempt seems to me to work better than your second. Now you asked why to break any number in this range? I need them to put those parts into specific addresses in my table.
ETA: It seems like one could use a len command to use in your example to solve one question.
|
|
| Back to top |
|
|
AylinCE
Grandmaster Cheater Supreme
Reputation: 37
Joined: 16 Feb 2017
Posts: 1526
|
| Posted: Thu Feb 03, 2022 11:55 am Post subject: |
|
|
Here is an example that you can edit and expand:
| Code: | function convert2(num,fmt)
num1=math.floor(tonumber(num) / tonumber(fmt))
print("num1:" .. num1)
b = (tostring(num1)):reverse()
lnn="0 "
a = b:gsub("..", "%1 "):sub(1,-1)
--print("a:"..a)
a=(a):reverse():sub(1,-1)
if (a):sub(1,1)==" " then a=(a):sub(2, -1) end
--print("a1:"..a)
a1=string.len(a) / 3
a1=math.floor(5 - tonumber(a1))
--print(a1)
ln=string.rep(lnn, a1) .. a
--print(ln)
return ln
end
function searchAddrValue(search)
rst=""
--local aobs1=AOBScan(search)
local aobs2=search
if aobs2~=nil then
--aobs2=readInteger(aobs1)
if (string.len(aobs2))==3 then
rst=rst .. convert2(aobs2,1)
end
if (string.len(aobs2))==4 then
rst=rst .. convert2(aobs2,2)
end
if (string.len(aobs2))==5 then
rst=rst .. convert2(aobs2,3)
end
if (string.len(aobs2))==6 then
rst=rst .. convert2(aobs2,4)
end
if (string.len(aobs2))==7 then
rst=rst .. convert2(aobs2,4)
end
if (string.len(aobs2))==8 then
rst=rst .. convert2(aobs2,5)
end
if (string.len(aobs2))==9 then
rst=rst .. convert2(aobs2,5)
end
if (string.len(aobs2))==10 then
rst=rst .. convert2(aobs2,5)
end
end
print(rst)
return rst
end
--src=searchAddrValue(tostring(123))
--src=searchAddrValue(tostring(1234))
--src=searchAddrValue(tostring(12345))
src=searchAddrValue(tostring(9999999999)) --len-10 |
If you don't want to split, type 1 in the code equivalent:
| Code: | if (string.len(aobs2))==9 then
rst=rst .. convert2(aobs2,1)
end |
Otherwise, you can specify how much you want to shrink in the template:
| Code: | if (string.len(aobs2))==10 then
rst=rst .. convert2(aobs2,2)
end |
If you don't want to split, type 1 in the code equivalent:
Otherwise, you can specify how much you want to shrink in the template:
You can export these results to certain address values (in your Table):
No; I don't know where you got these address values from.
A specific address scan (in Lua Script and a code that finds multiple values)
Found list? (Within too many addresses and values)
Address list (Too many addresses and values)
Just find a specific address, convert it to value with "readInteger("address")" and format the value with "src=searchAddrValue(tostring(9999999999))" and leave it as "oldvalue=src" where you want to write the value in your table.
_________________
|
|
| Back to top |
|
|
bknight2602
Grandmaster Cheater
![]() Reputation: 0
Joined: 08 Oct 2012
Posts: 586
|
| Posted: Thu Feb 03, 2022 12:49 pm Post subject: |
|
|
| AylinCE wrote: | Here is an example that you can edit and expand:
| Code: | function convert2(num,fmt)
num1=math.floor(tonumber(num) / tonumber(fmt))
print("num1:" .. num1)
b = (tostring(num1)):reverse()
lnn="0 "
a = b:gsub("..", "%1 "):sub(1,-1)
--print("a:"..a)
a=(a):reverse():sub(1,-1)
if (a):sub(1,1)==" " then a=(a):sub(2, -1) end
--print("a1:"..a)
a1=string.len(a) / 3
a1=math.floor(5 - tonumber(a1))
--print(a1)
ln=string.rep(lnn, a1) .. a
--print(ln)
return ln
end
function searchAddrValue(search)
rst=""
--local aobs1=AOBScan(search)
local aobs2=search
if aobs2~=nil then
--aobs2=readInteger(aobs1)
if (string.len(aobs2))==3 then
rst=rst .. convert2(aobs2,1)
end
if (string.len(aobs2))==4 then
rst=rst .. convert2(aobs2,2)
end
if (string.len(aobs2))==5 then
rst=rst .. convert2(aobs2,3)
end
if (string.len(aobs2))==6 then
rst=rst .. convert2(aobs2,4)
end
if (string.len(aobs2))==7 then
rst=rst .. convert2(aobs2,4)
end
if (string.len(aobs2))==8 then
rst=rst .. convert2(aobs2,5)
end
if (string.len(aobs2))==9 then
rst=rst .. convert2(aobs2,5)
end
if (string.len(aobs2))==10 then
rst=rst .. convert2(aobs2,5)
end
end
print(rst)
return rst
end
--src=searchAddrValue(tostring(123))
--src=searchAddrValue(tostring(1234))
--src=searchAddrValue(tostring(12345))
src=searchAddrValue(tostring(9999999999)) --len-10 |
If you don't want to split, type 1 in the code equivalent:
| Code: | if (string.len(aobs2))==9 then
rst=rst .. convert2(aobs2,1)
end |
Otherwise, you can specify how much you want to shrink in the template:
| Code: | if (string.len(aobs2))==10 then
rst=rst .. convert2(aobs2,2)
end |
If you don't want to split, type 1 in the code equivalent:
Otherwise, you can specify how much you want to shrink in the template:
You can export these results to certain address values (in your Table):
No; I don't know where you got these address values from.
A specific address scan (in Lua Script and a code that finds multiple values)
Found list? (Within too many addresses and values)
Address list (Too many addresses and values)
Just find a specific address, convert it to value with "readInteger("address")" and format the value with "src=searchAddrValue(tostring(9999999999))" and leave it as "oldvalue=src" where you want to write the value in your table. |
The values are located in a listview, and I select one of them. This represents the variable. Now the outputs will represent the values of fixed address 1, fixed address 2, fixed address 3, fixed address 4 and fixed address 5. The code should be able to insert those values into the five addresses.
|
|
| Back to top |
|
|
panraven
Grandmaster Cheater
![]() Reputation: 62
Joined: 01 Oct 2008
Posts: 958
|
| Posted: Thu Feb 03, 2022 2:52 pm Post subject: |
|
|
It can be match values with your addresses (or memory record identify by description etc) like this:
| Code: |
local addies = {'1ab000','1ab004','1ab008','1ab00c','1ab010'}
local nextmod = repeatDivMod(71234567890,100)
for i=1,#addies do
local _, r = nextmod() -- nextmod return 2 values, we only interested in second one
print(addies[i],r or 0) -- 'or 0' to set default value
end
|
_________________
- Retarded. |
|
| Back to top |
|
|
AylinCE
Grandmaster Cheater Supreme
Reputation: 37
Joined: 16 Feb 2017
Posts: 1526
|
| Posted: Thu Feb 03, 2022 3:57 pm Post subject: |
|
|
We are still far from a scenario, we proceed with assumptions.
| Code: | --We are still stuck on this code.
--Let's rearrange it and set it for 5 outputs:
function convert2(num,fmt)
num1=math.floor(tonumber(num) / tonumber(fmt))
--print("num1:" .. num1)
b = (tostring(num1)):reverse()
lnn="0 "
a = b:gsub("..", "%1 "):sub(1,-1)
--print("a:"..a)
a=(a):reverse():sub(1,-1)
if (a):sub(1,1)==" " then a=(a):sub(2, -1) end
--print("a1:"..a)
a1=string.len(a) / 3
a1=math.floor(5 - tonumber(a1))
--print(a1)
ln=string.rep(lnn, a1) .. a
--print(ln)
return ln
end
--Let's take an address value from the list:
adrr="06D9EC68" --The address is an example!
addrValue=readInteger(adrr) --Get the current value of your address.
--print(addrValue) -- 121863880 (example!)
--Now let's divide them into 5 different formats
--or assign them to 5 addresses:
adrrValrst1=convert2(addrValue,1)
adrrValrst2=convert2(addrValue,2)
adrrValrst3=convert2(addrValue,4)
adrrValrst4=convert2(addrValue,5)
adrrValrst5=convert2(addrValue,7)
--And you choose which 5 address values to write them to;
-- address1.value=adrrValrst1
-- etc ..
-- etc .. |
If you can achieve the results you want with the existing code, you can distribute those results to other addresses.
But I didn't understand how to write this result (output) "1 21 86 38 80" to an address.
I think you'll make it.
EDIT:
| bknight2602 wrote: | The first two lines of your most recent post
num=5799--from the list view
function convert2(num,fmt)
num1=math.floor(tonumber(num) / tonumber(fmt))
What is fmt?
The code gives Error:[string "num=5799
..."]:3: attempt to perform arithmetic on a nil value
Script Error I put 100 into where fmt was and obtained similar errors.
|
addrValue=Address.value
fmt=1 ("fmt" or the division operation that reduces the format value. If you don't want to divide the value (2,3,4, etc.), put 1 there.)
convert2(addrValue,1)
| Code: | hnmel1 = addresslist_getMemoryRecordByDescription(addresslist, "exaple1")
result1=convert2(hnmel1.value,1) -- value / 1 = convert(value) .. value / 2 = convert(new value)..
Where are you going to write the new formatted value?
hnmel1.value=result1 --?
hnmel2 = addresslist_getMemoryRecordByDescription(addresslist, "exaple2")
hnmel2.value=result1 |
_________________
|
|
| Back to top |
|
|
bknight2602
Grandmaster Cheater
![]() Reputation: 0
Joined: 08 Oct 2012
Posts: 586
|
| Posted: Thu Feb 03, 2022 8:23 pm Post subject: |
|
|
| AylinCE wrote: |
after inserting a fixed num = 5799 in the function
and printing
print(adrrValrst1)
print(adrrValrst2)
print(adrrValrst3)
after the
adrrValrst1=convert2(addrValue,1)
adrrValrst2=convert2(addrValue,2)
adrrValrst3=convert2(addrValue,4)
The results of those 3 outputs
are:
0 0 0 57 99
0 0 0 57 99
0 0 0 57 99
This is probably what you were indicating in a previous post.
How the output to the function, which by the way is correct into five discrete
addresses the 99 goes into one, the 57 goes into another and 0 go into 3 more addresses.
we need to break apart the aa bb cc dd ee into five answers.
|
|
|
| Back to top |
|
|
|
|
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You can download files in this forum
|
|
FAQ
Search
Memberlist
Usergroups
Register
Profile
Log in to check your private messages
Log in
