Had my software final exams recently, one of the last questions had me thinking for a while after the exam had finished.


IEEE754 numbers are according to the below layout Layout for a 32 bit floating point number

The exponent is stored as n + 127 so for an exponent of 20 the exponent in binary would be: 127 + 20 = 147 then 10010011

The Challange

Write a program that when given text representing a binary floating point number to the IEEE754 standard, returns the exponent given as a decimal value.


Input: 00111101110011001100110011001101 (0.1)

Isolated exponent bits: 01111011

Convert to binary: 0 + 64 + 32 + 16 + 8 + 0 + 2 + 1 = 123

Remove offset: 123 - 127 = -4

Output -4

Input: 10111111110111101011100001010010 (-1.64)

Isolated exponent bits: 01111111

Convert to binary: 0 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127

Remove offset: 127 - 127 = 0

Output 0


  • Input must be text/string/equivalent indexable datatype, eg. strings not integer or numerical
  • Output can be any type, as long as it is decimal
  • Aim for shortest solutions

Adapted from the 2020 Software Design and Development HSC Paper: 30) d) ii

Design an algorithm that takes in a string of 32 characters representing a floating point number, and displays the exponent as its signed decimal value.

Any feedback is appreciated as this is my first question

  • 1
    \$\begingroup\$ Can we take the input as a list of 1s and 0s? \$\endgroup\$ – Adám Nov 5 '20 at 9:50
  • 1
    \$\begingroup\$ All challenges here must have an objective scoring criterion, for which we usually recommend code-golf unless you have something else in mind. \$\endgroup\$ – Bubbler Nov 5 '20 at 9:50
  • 1
    \$\begingroup\$ Also, if you have another challenge idea, please use the sandbox to gather feedback from other users and fix potential problems. It is recommended to leave a challenge there for at least 3 days before posting it to main. \$\endgroup\$ – Bubbler Nov 5 '20 at 10:03
  • 1
    \$\begingroup\$ Btw, welcome to the site! Nice first challenge. In the future, consider using the Sandbox. Had it been there, I might have suggested taking a variable length input representing a binary float with 16/32/64/128/265 bits. \$\endgroup\$ – Adám Nov 5 '20 at 10:05
  • 6
    \$\begingroup\$ @Adám I would of submitted it there but was unable to as i have less than 5 reputation \$\endgroup\$ – Maximilian Rose Nov 5 '20 at 10:25

36 Answers 36


Japt, 8 bytes

Note the unprintable (charcode 127) at the end.

ů8 Í-#

Try it


Whitespace, 121 bytes

[S S S N
_Push_0][S S S T    T   T   N
_Push_7][S N
S _Duplicate_7][T   N
T   S _Read_STDIN_as_character][N
_Create_Label_LOOP][S N
S _Duplicate][S N
S _Duplicate][T N
T   S _Read_STDIN_as_character][T   T   T   _Retrieve_input][S S S T    T   S S S S N
Push_48][T  S S T   _Subtract][S T  S S T   S N
_Copy_0-based_2nd][S S S T  S N
_Push_2][T  S S N
_Multiply][T    S S S _Add][S N
T   _Swap_top_two][S S S T  N
_Push_1][T  S S T   _Subtract][S N
S _Duplicate][N
T   T   S N
_Create_Label_PRINT_RESULT][S N
_Discard_top][S S S T   T   T   T   T   T   T   N
_Push_127][T    S S T   _Subtract][T    N
S T _Print_as_integer]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Integer r = 0
Integer n = 7
Read STDIN as character
Start LOOP:
  Integer c = STDIN as character
  c = c - 48
  r = r * 2
  r = c + r
  n = n - 1
  If(n < 0):
    Jump to label PRINT_RESULT
  Go to next iteration of LOOP

  r = r - 127
  Print r as integer to STDOUT

Husk, 8 bytes


Try it online!

Straightforward implementation following the steps of the question. Accepts list of digits as argument.

       t        # tail: remove first element
     ↑8         # get first 8 remaining elements
    ḋ           # interpret as binary digits
-127            # subtract 127

Julia 0.7, 25 bytes


Try it online!


C# (Visual C# Interactive Compiler), 40 bytes


Try it online!


Zsh, 20 bytes


Try it online!


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