Convert a binary number, represented as a string (e.g. '101010'), to its decimal equivalent using first principles.
Implement binary to decimal conversion. Given a binary input string, your program should produce a decimal output. The program should handle invalid inputs.
Decimal is a base-10 system.
A number 23 in base 10 notation can be understood as a linear combination of powers of 10:
So: 23 => 2*10^1 + 3*10^0 => 2*10 + 3*1 = 23 base 10
Binary is similar, but uses powers of 2 rather than powers of 10.
So: 101 => 1*2^2 + 0*2^1 + 1*2^0 => 1*4 + 0*2 + 1*1 => 4 + 1 => 5 base 10
.
All of Computer Science http://www.wolframalpha.com/input/?i=binary&a=*C.binary-_*MathWorld-
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
(ns binary-test
(:require [clojure.test :refer [deftest is]]
binary))
(deftest binary-1-is-decimal-1
(is (= 1 (binary/to-decimal "1"))))
(deftest binary-10-is-decimal-2
(is (= 2 (binary/to-decimal "10"))))
(deftest binary-11-is-decimal-3
(is (= 3 (binary/to-decimal "11"))))
(deftest binary-100-is-decimal-4
(is (= 4 (binary/to-decimal "100"))))
(deftest binary-1001-is-decimal-9
(is (= 9 (binary/to-decimal "1001"))))
(deftest binary-11010-is-decimal-26
(is (= 26 (binary/to-decimal "11010"))))
(deftest binary-10001101000-is-decimal-1128
(is (= 1128 (binary/to-decimal "10001101000"))))
(deftest invalid-binary-is-decimal-0
(is (= 0 (binary/to-decimal "carrot"))))
(ns binary)
;; I wrote this function to split a number in its digits but then I
;; saw that the tests all use strings. So this is not used.
;; (defn digits [n]
;; (loop [n n digits '()]
;; (if (zero? n)
;; digits
;; (recur (quot n 10) (conj digits (mod n 10))))))
(defn pow [x n]
(int (Math/pow x n)))
(defn parse-digits [n]
(try
(doall (map #(Long/parseLong (str %)) n))
(catch NumberFormatException _)))
(defn to-decimal [n]
(let [digits (parse-digits n)]
(reduce + (map-indexed #(* %2 (pow 2 %1)) (reverse digits)))))
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