Decimal to Binary function I made

[iPoisonxL]

Hi! I created a cool function that converts decimal integers to binary integers. Here it is.

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function toBinary(int)
	if(type(int)=="number")then
		local current = int
		local bin = {}
		while(current~=0)do
			table.insert(bin, current%2)
			current = math.floor(current / 2)
		end
		local result = ""
		for i,v in pairs(bin) do
			result = result .. v
		end
		return string.reverse(result)
	end
	return "error: string input"
end

Use it however you want, I don't care.

Here's an example

binary.lua

(428 Bytes) Downloaded 165 times

[Roland_Yonaba]

Small addendum, if you want to pad it to 8 characters, you can change the return statement to this.

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return ('%08s'):format(string.reverse(result))

That said, nice.

[iPoisonxL]

Small addendum, if you want to pad it to 8 characters, you can change the return statement to this.

Code: Select all

return ('%08s'):format(string.reverse(result))

That said, nice.

Sweet thanks!

[Ref]

Could shorten a bit.

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function toBinary( int )
	assert( type( int ) == 'number', 'Error: Non-number input to toBinary!' )
	local current	= math.floor( int )
	local result	= ''
	while( current	~= 0 ) do
		result		= result..current % 2
		current		= math.floor( current / 2)
		end
	return ( '%08s' ):format( string.reverse( result ) )
	end

[iPoisonxL]

Could shorten a bit.

Code: Select all

function toBinary( int )
	assert( type( int ) == 'number', 'Error: Non-number input to toBinary!' )
	local current	= math.floor( int )
	local result	= ''
	while( current	~= 0 ) do
		result		= result..current % 2
		current		= math.floor( current / 2)
		end
	return ( '%08s' ):format( string.reverse( result ) )
	end

That's true, thank you. Not much of a code golfer myself

[airstruck]

Could shorten a bit.

Could just as easily shorten it with table.concat, probably better than doing all that string concatenation.

Not sure only converting to binary is very useful, might as well add the extra line or two to convert to arbitrary base.

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local function toBase (base, number)
    local result = {}
    
    while number > 0 do
        local digit = number % base
        result[#result + 1] = digit < 10 and digit or string.char(55 + digit)
        number = math.floor(number / base)
    end
    
    return string.reverse(table.concat(result))
end

[Positive07]

Well I have weird ways of doing things so here is my attempt at this!

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local hex = {
    ["0"] = "0000", ["1"] = "0001", ["2"] = "0010", ["3"] = "0011",
    ["4"] = "0100", ["5"] = "0101", ["6"] = "0110", ["7"] = "0111",
    ["8"] = "1000", ["9"] = "1001", ["A"] = "1010", ["B"] = "1011",
    ["C"] = "1100", ["D"] = "1101", ["E"] = "1110", ["F"] = "1111",
}

local func = function (n) return hex[n] end

tobinary = function (n)
    local result = string.format("%X", n):gsub("(.)", func):gsub("^(0*)(.)","%2")
    return result
end

Explanation:

• First we format the number into an hexadecimal string
• Second we take each character of the string, that is, each number (or letter) and pass them through a function using gsub
• The function looks up the hexadecimal character in the lookup table and returns the binary value associated to that character
• Lastly we trim all the leading zeros because they are not really needed

NOTE: I assign the value to the local variable result so that all the other return values provided by gsub are discarded

[airstruck]

First we format the number into an hexadecimal string

Could use octal and cut that table in half

I assign the value to the local variable result so that all the other return values provided by gsub are discarded

You can wrap the expression in parens to get the same effect:

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return (string.format("%X", n):gsub("(.)", func):gsub("^(0*)(.)","%2"))

[ivan]

Funny enough binary to a number (the reverse of what you are doing) is actually much easier:

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decimal = tonumber(binarysz,2)

where binarysz is a string of 0s and 1s

[bartbes]

Well I have weird ways of doing things so here is my attempt at this!

My attempt at cleaning this up (and using octal, as time thief mentioned):

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do
	local conv = {}

	for i = 0, 7 do
		conv[tostring(i)] = ("%d%d%d"):format(math.floor(i%8/4), math.floor(i%4/2), math.floor(i%2/1))
	end

	function tobinary(n)
		return (("%o"):format(n):gsub(".", conv):match("1.+$"))
	end
end

print(tobinary(4713))

It still doesn't strike me as efficient, though.

EDIT: I just realized the match doesn't quite work for 0, oops. (Just add 'or 0'.)