# 1+1 = 10, 1+2 = 3

Write a function or program than can do simple arithmetic (addition, subtraction, multiplication and division) in both base 10 and base 2.

The function will take a mathematical expression as input, and output the correct result in the correct base. The input will be n numbers separated by one or several operators (+ - * /).

If all input values contain only 0 and 1, all values are considered to be binary. If at least one digit is 2-9, all values are considered to be base 10.

Rules:

• You can assume there will only be one operator between numbers (10*-1 will not appear)
• You can assume there will be no parentheses.
• Normal operator precedence (try the expression in the google calculator if you're in doubt).
• You can not assume there will only be integers
• There will be no leading zeros in input or output
• You can assume only valid input will be given
• You can assume all input values are positive (but the minus operator may make negative output possible, 1-2=-1 and 10-100=-10)
• REPL is not accepted
• You may choose to take the input as separate arguments, or as a single argument, but the input has to be in the correct order.
• I.e. you may represent 1-2 with the input arguments 1, -, 2, but not 1, 2, -.
• You must accept the symbols + - * / in the input, not plus, minus etc.
• You must support floating point values (or up to the maximum limit of your language, however supporting only integers is not accepted).
• eval is accepted

Examples:

1+1
10

1010+10-1
1011

102+10-1
111

1+2+3
6

10*10*10
1000

11*11*11
11011

10*11*12+1
1321

10.1*10.1
110.01

20.2*20.2
408.04

10/5
2

110/10
11

Also accepted (optional line or comma-separated input):
10
+
10
-
1
11    <-- This is the output


This is code golf, so the shortest code in bytes will win.

• In the case of 110/10, is 11.0 acceptable? – isaacg Jan 3 '16 at 15:16
• @isaacg yes, that's ok :-) – Stewie Griffin Jan 3 '16 at 15:56
• The downvote... why? – Stewie Griffin Jan 4 '16 at 8:55

## Japt, 7772626062*6059 51 bytes

OvUf"[2-9]" ?U:"({Ur"[\\d.]+""º$&e14+P n2 /2pE¹"})¤  Explanation (more or less the same as for the JS answer): Ov //eval... Uf"[2-9]" //if input contains the digits 2 to 9 U: //then it's base 10, just compute Ur"[\\d.]+" //otherwise replace all the numbers "º$&e14+P n2 /2pE¹"  //with their base 10 equivalents
//I.e., take every number, multiple by 10^14, convert to
//base 10 and divide by 2^14
// º and ¹ are multiple brackets
¤                    //means "s2", i.e. convert the result to binary



Try it online!

* didn't divide properly

• As of 5 days ago, JS's eval is assigned to Ox. I'll see if it can be further shortened. – ETHproductions Jan 3 '16 at 16:27
• @Eth Thanks, this saves 5 bytes. – nicael Jan 3 '16 at 16:43
• 5 more bytes saved: OxUf"[2-9]" ?U:({Ur"\\d+(\\.\\d+)?""(($&e14+P).n(2)/16384)"}).s(2) It's probably possible to make it generate Japt code instead of JS, then use Ov to evaluate that. – ETHproductions Jan 3 '16 at 16:48 • Yup, 62 bytes :) OvUf"[2-9]" ?U:Ur"\\d+(\\.\\d+)?""~~[$&e14+P n2 /16384]" +" s2 The ~~[...] is necessary because mismatched parentheses in a string mess with the transpiler. – ETHproductions Jan 3 '16 at 16:53
• 51: OvUf"[2-9]" ?U:"({Ur"[\\d.]+""º$&e14+P n2 /2pE¹"})¤ – ETHproductions Jan 3 '16 at 19:12 ## JavaScript ES6, 87971100210631021019810049388 86 bytes e=>eval(e.match[2-9]?e:(${e.replace(/[\d.]+/g,"('0b'+$&e14)/16384")}).toString(2))  Demo + explanation: function c(e){ return eval( e.match[2-9]? //check if there are numbers 2 to 9 e: //if there're, just compute the result "("+ e.replace( //otherwise replace... /[\d.]+/g, //any number... "(('0b'+$&e14)/16384)"    //...with itself converted to base 10
)
+").toString(2)"               //compute the result and convert it to binary
)
}

document.write(
c("1.1*1.1")+"<br>"+
c("1010+10-1")+"<br>"+
c("102+10-1")+"<br>"+
c("1+2+3")+"<br>"+
c("10*10*10")+"<br>"+
c("11*11*11")+"<br>"+
c("10*11*12+1")+"<br>"+
c("10.1*10.1")+"<br>"+
c("20.2*20.2")+"<br>"+
c("10/5")+"<br>"+
c(10
+
10
-
1)
)

2 - again floats problem: parseInt floors binary so I have to multiply by 1e14 and then divide by 16384
3 - hope that's achieved the given task, now start to golf :D
4 - there was a bug with dividing

• 106^3 bytes? That's more than a million! – ETHproductions Jan 3 '16 at 14:28
• @Eth I knew such a comment was coming ;D – nicael Jan 3 '16 at 14:29
• You can change e.match(/[2-9]/g) to e.match[2-9]. – user81655 Jan 3 '16 at 15:20
• @user Thanks :) – nicael Jan 3 '16 at 15:22
• @nicael ('0b'+$&*1e14)/1638 this should work but I'm not 100% sure – Downgoat Jan 3 '16 at 17:13 # Jolf, 31 bytes, noncompeting I added a decent amount of functions inspired by this challenge, and, as thus, it is considered noncompeting. I'm happy because I finally implemented unary functions (like (H,S,n)=>val in ES6, but are supported in ES5!) ? hi"[2-9]"~eiB~epT mpvid|m'H2H ? hi"[2-9]" if the input contains any of 2..9 ~ei evaluate i (implicitly print) else _mpvid map the input split into number groups m'H2 to H as a binary float | H (or keep H, if that doesn't work) pT join by spaces ~e evaluate B convert to binary (implicitly print)  • Your code says "hi" to its readers! – Cyoce Jan 4 '16 at 7:48 • Wow. I didn't notice that! :D – Conor O'Brien Jan 4 '16 at 12:00 # Bash, 60 bytes [ -z tr -dc 2-9<<<$1 ]&&s='obase=2;ibase=2;';bc -l<<<$s$1


Example Run:

$./bin_dec_add.sh 1+1 10$ ./bin_dec_add.sh 1+2
3

• @Pietu1998 dc would require reverse polish ordering of the operations, which is disallowed by the challenge. – Tyzoid Jan 5 '16 at 15:52

ë(ïđ/[2-9]⎞?ï:(⦃ïē/[\d.]+⌿,↪(Յ+$*ḊⁿḎ)/Ẁ²)})ⓑ  Try it here (Firefox only). # Explanation ë(ïđ/[2-9]⎞?ï:(⦃ïē/[\d.]+⌿,↪(Յ+$*ḊⁿḎ)/Ẁ²)})ⓑ // implicit: ï=input, Ḋ=10, Ḏ=14, Ẁ=128
ë(                                              // eval
ïđ/[2-9]⎞?                                    // does ï have 2-9?
ï                                   // if so, eval input
:                                  // else, eval:
(⦃ïē/[\d.]+⌿,                    // replace the binary numbers
// with their base 10 equivalents:
↪(Յ+                // translates to 0b
$*ḊⁿḎ // matched number * 10^14 )/Ẁ² // divided by 128^2 )})ⓑ // converted to binary // implicit output  # PowerShell, 107 bytes param($e)iex(("{0}String($($e-replace'(\d+)','{0}Int32("$1",2)'),2)"-f'[Convert]::To'),$e)[$e-match'[2-9]']  # Ungolfed param($e) # Accept single argument
Invoke-Expression # Eval
( # Expression, resulting in an array of 2 elements
(
"{0}String( # Binary
$( # Inline subexpression$e -replace'(\d+)', '{0}Int32("$1",2)' # "1010+10-1" becomes "{0}Int32("1010",2)+{0}Int32("10",2)-{0}Int32("1",2)" ) ,2)" -f '[Convert]::To' # "{0}Int32("1010",2)+{0}Int32("10",2)-{0}Int32("1",2)" becomes "[Convert]::ToString([Convert]::ToInt32("1010",2)+[Convert]::ToInt32("10",2)-[Convert]::ToInt32("1",2),2)" ),$e # Plain
)
[\$e-match'[2-9]'] # Return 1st element of array if regex matches, else 0


# Example

PS > .\Calc.ps1 1010+10-1
1011

PS > .\Calc.ps1 20.2*20.2
408,04
`