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Razi Expert Cheater Reputation: 1
Joined: 17 Jan 2018 Posts: 202
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Posted: Sat Jun 02, 2018 2:44 am Post subject: How to get the result 1,2,4,5,7,8,10, three times? |
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How to get the result 1,2,4,5,7,8,10, three times in the next code? From 1 to 10, but without 3,6,9. How do I need to change formula?
Code: | for x = 1, 21 do
local offset = 0x3*((x-1) // 2) + ((x-1) % 2) + 1
print (offset)
end |
In result must be: 1,2,4,5,7,8,10, 1,2,4,5,7,8,10, 1,2,4,5,7,8,10.
or How to get the result 0,1,3,4,6,7,9, three times? From 0 to 9, but without 2,5,8.
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Dark Byte Site Admin Reputation: 457
Joined: 09 May 2003 Posts: 25262 Location: The netherlands
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Posted: Sat Jun 02, 2018 3:08 am Post subject: |
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I don't have the context in where you need it, or if you actually want to use that formula or not, or if 'if' is allowed, but this does return the result you describe
Code: |
for x=1,32 do
local offset=x%11
if (offset % 3)>0 then
print(offset)
end
end
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as for 0,1,3,4,6,7,9 , have you tried:
Code: |
print(0)
print(1)
print(3)
print(4)
print(6)
print(7)
print(9)
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lol jk
but perhaps
Code: |
for x=0,29 do
local offset=x%10
if (offset+1)%3~=0 then print(offset) end
end
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Razi Expert Cheater Reputation: 1
Joined: 17 Jan 2018 Posts: 202
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Posted: Sat Jun 02, 2018 3:33 am Post subject: |
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I'm using the following code: Code: | for x = 1, 21 do
local offset1 = 0x84*((x-1) // 7)
a = bAnd(0xFFFF, readInteger(0x009E8646+offset1))
local bitss = ((x-1) % 7) + 1
local bit = ((a >> (bitss-1)) & 0x01)
if bit == 1 then
UDF1["CECheckbox"..x].Checked = true
elseif bit == 0 then
UDF1["CECheckbox"..x].Checked = false
end
end |
I need for CECheckbox1 to have x=1 (I mean: bitss = 1)
for CECheckbox2 to have x=2 (bitss = 2)
for CECheckbox3 to have x=4 (bitss = 4)
for CECheckbox4 to have x=5
for CECheckbox5 to have x=7
for CECheckbox6 to have x=8
for CECheckbox7 to have x=10
then
for CECheckbox8 to have x=1
for CECheckbox9 to have x=2
for CECheckbox10 to have x=4 and so on.
or
I need for CECheckbox1 to have x=0 (I mean: bitss = 0)
for CECheckbox2 to have x=1 (bitss = 1)
for CECheckbox3 to have x=3 (bitss = 3)
for CECheckbox4 to have x=4
for CECheckbox5 to have x=6
for CECheckbox6 to have x=7
for CECheckbox7 to have x=9
then
for CECheckbox8 to have x=0
for CECheckbox9 to have x=1
for CECheckbox10 to have x=3 and so on.
I need exactly 21 in the code: for x = 1, 21 do
and the result must be 1,2,4,5,7,8,10, three times. I do not know if it possible.
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