35
\$\begingroup\$

Introduction

Some days ago I needed a metronome for something. I had none available so I downloaded an app from the App Store. The app had a size of 71 MB!!!
71 MB for making tic-toc...?!
So code-golf came into my mind and I was wondering if some of you guys could improve this.

Challenge

Golf some code that outputs some sound. It's pretty irrelevant what kind of sound. If required create some sound file... but a System beep will do the job as well. (Here is some sound I created... nothing special.)

Input: The beats per minute the metronome outputs.

Example

This is a non-golfed Java-version! It's just to show you the task.

public class Metronome {
  public static void main(String[] args) throws InterruptedException {
    int bpm = Integer.valueOf(args[0]);
    int interval = 60000 / bpm;

    while(true) {
        java.awt.Toolkit.getDefaultToolkit().beep();
        // or start playing the sound
        Thread.sleep(interval);
        System.out.println("Beep!");

    }
  }
}

Rules

You may not use external libaries, only tools of the language itself are allowed.
Only the bytes of the source code count... not the sound file.

This is , so the submission with the least amount of bytes wins!

EDIT:

Example output: So something like this would be the output for 120 bps: link

\$\endgroup\$
16
  • 1
    \$\begingroup\$ Can you add a few examples for I/O (record some sound and upload it, post the links here)? \$\endgroup\$ Jan 23, 2016 at 14:25
  • 2
    \$\begingroup\$ Question: when you say "external libraries", does that include the libraries that are suggested with the language? (I won't use this, but an example is in Vitsy wherein I can access shell or JS (but JS is builtin)) \$\endgroup\$ Jan 23, 2016 at 17:48
  • 3
    \$\begingroup\$ Can you add a leaderboard snippet in? \$\endgroup\$ Jan 23, 2016 at 21:53
  • 1
    \$\begingroup\$ I suspect the majority of that app you downloaded is pretty graphics and sound effects. It's like those flashlight apps that do nothing but turn the screen all white but still manage to somehow use up tens of MB... \$\endgroup\$ Jan 24, 2016 at 15:13
  • 1
    \$\begingroup\$ What's the requirement on accuracy? In your sample, both beep() and console output aren't exactly instant IIRC. Neither sleep() is known for beeing accurate. \$\endgroup\$
    – Num Lock
    Jan 25, 2016 at 9:04

27 Answers 27

19
\$\begingroup\$

Mathematica, 26 bytes

Pause[Beep[];60/#]~Do~∞&

Do is normally used as a "for" loop in the narrowest sense: repeat this piece of code for each i from x to y... or even just repeat this piece of code n times. Instead of a number n we can give it infinity though to create an infinite loop. The loop body is Pause[Beep[];60/#] which is just a golfy way of writing Beep[];Pause[60/#] where # is the function argument.

If it's admissible for the solution to blow up the call stack eventually, we can save one byte with a recursive solution:

#0[Beep[];Pause[60/#];#]&
\$\endgroup\$
4
  • \$\begingroup\$ I didn't know that ~Do~∞ was possible. A For loop only got me to 29 bytes. (Also, I personally believe that the 26-byte version is the only valid one.) \$\endgroup\$ Jan 23, 2016 at 18:51
  • \$\begingroup\$ @LegionMammal978 Unfortunately, ~Do~∞ doesn't seem to work when the comes from a variable. (I tried using that when golfing your truth machine.) \$\endgroup\$ Jan 23, 2016 at 18:52
  • 1
    \$\begingroup\$ Attributes[Do] includes HoldAll, so my guess is that _~Do~∞ has a special evaluation pattern. \$\endgroup\$ Jan 23, 2016 at 19:00
  • \$\begingroup\$ @LegionMammal978 It seems more like variables do, because the error message for Do[...,a] where a holds infinity actually shows the call as Do[...,{a}]. \$\endgroup\$ Jan 23, 2016 at 19:01
14
\$\begingroup\$

Pyth, 11 10 9 bytes

Thanks to Adnan for reminding me about #.

#C7.dc60Q

Forever (#), print Char code 7. Then sleep (.d) 60 seconds divided by (c) input (Q).

\$\endgroup\$
10
  • \$\begingroup\$ @Adnan Forgot about that one. Thanks. \$\endgroup\$ Jan 23, 2016 at 18:26
  • \$\begingroup\$ Do you need the space? \$\endgroup\$
    – lirtosiast
    Jan 23, 2016 at 18:30
  • \$\begingroup\$ @ThomasKwa Yes. IIRC 7. would be parsed as a number. \$\endgroup\$ Jan 23, 2016 at 18:30
  • 5
    \$\begingroup\$ Oh. #pythnoob \$\endgroup\$ Jan 23, 2016 at 18:35
  • 2
    \$\begingroup\$ I couldn't get .d to sleep when i tried. It kept printing unix time \$\endgroup\$
    – busukxuan
    Jan 24, 2016 at 17:50
8
\$\begingroup\$

JavaScript, 36 45 42 41 34 bytes

Saved 1 byte thanks to @RikerW

Saved 1 byte thanks to @ETHproductions

n=>{for(;;sleep(60/n))print("\7")}

This is a function.

If I use `\7`, SpiderMonkey complains octal literals are deprecated.

Alternative, 31 bytes

n=>{for(;;sleep(60/n))print``}

The problem is the unprintables are stripped but this should work.

\$\endgroup\$
9
  • \$\begingroup\$ Dammit, I was just about to post something like this. I'm still going to post it (because it uses node and all) because I use a different approach. \$\endgroup\$ Jan 23, 2016 at 18:18
  • \$\begingroup\$ If you look it from the way I asked the question, the recursive solution would not be possible. Metronomes are made for working and working... not for crashing after some time. \$\endgroup\$
    – PEAR
    Jan 23, 2016 at 19:12
  • \$\begingroup\$ @PEAR this shouldn't crash because no variable is being increments. The only thing that might cause it to crash is the terminal buffer except on modern computers that could take > 50-100 years I think \$\endgroup\$
    – Downgoat
    Jan 23, 2016 at 19:16
  • \$\begingroup\$ What environment does this run under? I've tried chrome and Node.js, but I can't get it to work. \$\endgroup\$ Apr 13, 2016 at 17:07
  • \$\begingroup\$ @starbeamrainbowlabs this uses the JavaScript shell (SpiderMonkey) \$\endgroup\$
    – Downgoat
    Apr 13, 2016 at 22:02
8
\$\begingroup\$

Bash, 53 55 41 bytes

Thanks to @Dennis for shaving off 14 bytes1

Okay, truth time: I'm terrible at golfing bash. Any help would be so very appreciated.

echo " ";sleep `bc -l<<<60/$1`;exec $0 $1
      ^ That's ASCII char 7

1 Holy crap. No wonder nobody can outgolf Dennis.

\$\endgroup\$
6
  • \$\begingroup\$ Is while 1 possible? \$\endgroup\$
    – PEAR
    Jan 23, 2016 at 19:02
  • \$\begingroup\$ @PEAR Nupe - already tried that. \$\endgroup\$ Jan 23, 2016 at 19:03
  • \$\begingroup\$ while printf \\a perhaps? \$\endgroup\$
    – Neil
    Jan 23, 2016 at 20:19
  • \$\begingroup\$ This doesn't work since bash uses integer division. You'll need to use bc. \$\endgroup\$ Jan 23, 2016 at 20:37
  • \$\begingroup\$ 1. The BEL character isn't special to Bash, so you don't need the quotes. 2. If you read the input as a CLA, you don't need read. 3. echo exists with code 0, so you can use that statement instead of true. \$\endgroup\$
    – Dennis
    Jan 24, 2016 at 2:11
7
\$\begingroup\$

JavaScript ES6 (browser), 43 bytes

This may be stretching the rules:

x=>setInterval('new Audio(1).play()',6e4/x)

Give this function a name (e.g. F=x=>...) and enter it in the browser console on this page. Then call the function with your bps, e.g. F(60), and wait for the magic to happen. :-)

Why does this work? Well, b.html is in the same folder as a file named 1, which is the sample sound file from the OP. I'm not sure if this is within the rules (I guess it's like the shell version; it needs to be run in a specific environment), but it was worth a shot.

Safer version, 57 bytes

If the above code isn't allowed for some reason, try this instead:

x=>setInterval('new Audio("//ow.ly/Xrnl1").play()',6e4/x)

Works on any page!

\$\endgroup\$
6
  • \$\begingroup\$ This is an intersting solution. It's even shorter when you download and rename the file, isn't it? \$\endgroup\$
    – PEAR
    Jan 23, 2016 at 18:57
  • \$\begingroup\$ @PEAR That would be shorter, but then it would need its own webpage with the sound file in the same folder to run. \$\endgroup\$ Jan 23, 2016 at 18:57
  • \$\begingroup\$ Oh, it's JavaScript xD... you're right \$\endgroup\$
    – PEAR
    Jan 23, 2016 at 19:10
  • \$\begingroup\$ @PEAR There, I did it. Is this new solution within the rules? \$\endgroup\$ Jan 23, 2016 at 19:10
  • \$\begingroup\$ Huh. You could specify that it is JS with the certain webpage. It's a preexisting interpreter, so it's a valid language. \$\endgroup\$ Jan 23, 2016 at 19:11
6
\$\begingroup\$

05AB1E, 31 bytes

Code:

I60s/[7ç?D.etime.sleep(#.pop())

If I had a built-in for waiting N seconds, this could have been 11 bytes. Unfortunately, this is not the case. Here is the explanation:

I                               # Push input
 60                             # Push 60
   s                            # Swap the top 2 items
    /                           # Divide the top 2 items
     [                          # Infinite loop
      7ç                        # Push the character \x07
        ?                       # Output it, which give a sound
         .e                     # Evaluate the following as Python code
           time.sleep(       )  # Wait for N seconds
                      #         # Short for stack
                       .pop()   # Pop the last item

Uses the ISO 8859-1 encoding.

\$\endgroup\$
1
  • \$\begingroup\$ This must be one of the first 05AB1E answers o.Ô It looks very weird to see the time.sleep and .pop() in the middle of the code like that. ;) \$\endgroup\$ Aug 20, 2019 at 16:54
6
\$\begingroup\$

osascript, 39 bytes

on run a
repeat
beep
delay 60/a
end
end

There is literally a command called beep? Sweeeet!

Runnable only on Mac OS X due to restricted license, but to run, do:

osascript -e "on run a
repeat
beep
delay 60/a
end
end" bpm
\$\endgroup\$
6
\$\begingroup\$

Python, 68 67 57 bytes

Saved 1 byte thanks to @FlagAsSpam

Saved 9 bytes thanks to @Adnan

import time
a=input()
while 1:print"\7";time.sleep(60./a)

Also it took 2 bytes less after converting line endings to UNIX format.

Older version, that actually takes bpm as command line argument (66 bytes):

import sys,time
while 1:print"\7";time.sleep(60./int(sys.argv[1]))
\$\endgroup\$
8
  • 4
    \$\begingroup\$ Can't you do print"\7";? I'm not sure, but I'm pretty sure that works. \$\endgroup\$ Jan 23, 2016 at 18:16
  • \$\begingroup\$ @Andan No, input() requests input from user. I don't know if that's considered a valid input. Also conversion to number is needed anyway. \$\endgroup\$
    – webwarrior
    Jan 23, 2016 at 21:19
  • 1
    \$\begingroup\$ How about a=input() and a replacing int(sys.argv[1])? I've always thought that Python 2 automatically evaluates input and therefore doesn't need the int conversion, but I may be wrong. \$\endgroup\$
    – Adnan
    Jan 24, 2016 at 11:45
  • \$\begingroup\$ @Andan input() actually does auto evaluate. I forgot about that feature. It's rather unpythonic though - probably a legacy from old times. \$\endgroup\$
    – webwarrior
    Jan 24, 2016 at 15:15
  • \$\begingroup\$ Can time.sleep(60./a) be replaced with time.sleep(60./input()), while completely removing a=input()? \$\endgroup\$
    – clapp
    Jan 25, 2016 at 4:47
4
\$\begingroup\$

AutoIt, 56 bytes

Func _($0)
While Sleep(6e4/$0)
Beep(900,99)
WEnd
EndFunc
\$\endgroup\$
4
\$\begingroup\$

Vitsy, 14 bytes

a6*r/V1m
<wVO7

Verbose mode (interpreter coming soon):

0:                              // a6*r/V1m
push a; // 10
push 6;
multiply top two; // 60
reverse stack; // bpm on top
divide top two; // bpm/60
save/push permanent variable; 
push 1;
goto top method; // goes to 1
1:                              // <wVO7
go backward; // infinite loop, from the bottom of 1
wait top seconds;
save/push permanent variable; // pushes the bpm in terms of seconds of delay
output top as character;
push 7;

Basically, I use the w operator to wait a certain number of seconds as specified by bpm/60, wrapped in an infinite loop. Then, I make noise with the terminal output of ASCII character 7 (BEL).

\$\endgroup\$
2
  • \$\begingroup\$ Looks nice, but how can I test this? :) \$\endgroup\$
    – PEAR
    Jan 23, 2016 at 18:58
  • \$\begingroup\$ @PEAR You'll have to download the interpreter (forgot to link it in the title). Save it in a file and run it with java -jar Vitsy.jar <filename>. \$\endgroup\$ Jan 23, 2016 at 19:03
4
\$\begingroup\$

C#, 118 bytes

class A{static int Main(string[]a){for(;;System.Threading.Thread.Sleep(60000/int.Parse(a[0])))System.Console.Beep();}}

Basic solution.

\$\endgroup\$
3
  • \$\begingroup\$ Why not print ASCII char 7? \$\endgroup\$ Jan 23, 2016 at 19:05
  • \$\begingroup\$ @FlagAsSpam It's longer: The system beep uses System.Console.Beep();, and printing the character uses System.Console.Write('<\a character>');. \$\endgroup\$ Jan 23, 2016 at 19:09
  • \$\begingroup\$ Woah. That's a lot to write a character. \$\endgroup\$ Jan 23, 2016 at 19:10
4
\$\begingroup\$

Java, 103 82 bytes

Thanks to @Justin for shaving off 21 bytes!

Oh, geez.

void x(int b)throws Exception{for(;;Thread.sleep(60000/b))System.out.print('\7');}

Method and golfed version of the sample program.

\$\endgroup\$
4
  • \$\begingroup\$ Why not System.out.print('\7'); instead of the java.awt.Toolkit.getDefaultToolkit().beep();? \$\endgroup\$
    – Justin
    Jan 24, 2016 at 1:52
  • \$\begingroup\$ @Justin \ is solely for escaping regex characters. \$\endgroup\$ Jan 24, 2016 at 1:55
  • 1
    \$\begingroup\$ no the backslash is an escape sequence. '\7' is the bell character, which makes a sound when it is printed out \$\endgroup\$
    – Justin
    Jan 24, 2016 at 1:55
  • \$\begingroup\$ @Justin Huh. I've always thrown errors on that (when using double quotes). My mistake. Thanks! :D \$\endgroup\$ Jan 24, 2016 at 11:42
3
\$\begingroup\$

GMC-4 Machine Code, 21.5 bytes

The GMC-4 is a 4-bit computer by a company called Gakken to teach the principles of assembly language in a simplified instruction set and computer. This routine takes input in data memory addresses 0x5D through 0x5F, in big-endian decimal (that is, one digit per nibble).

The algorithm is basically adding the input to memory and waiting 0.1s, until it's at least 600, and then subtracting 600 and beeping, in an infinite loop. Since the GMC-4 has a bunch of register swap functions but no register copy functions, this is done the hard way.

In hex (second line is position in memory):

A24A14A04 80EC AF5A2EF AE5A1EF AD5A0EF 8A6 F2AF09 86ADEEE9F09
012345678 9ABC DEF0123 4567890 ABCDEF0 123 456789 ABCDEF01234

In assembly:

    tiy 2     ;ld y, 0x2
    AM        ;ld a, [0x50 + y]
    tiy 1
    AM
    tiy 0
    AM
start:
    tia 0     ;ld a, 0x0
    cal timr  ;pause for (a+1)*0.1 seconds
    tiy F
    MA        ;ld [0x50 + y], a
    tiy 2
    cal DEM+  ;add a to [0x50 + y]; convert to decimal and carry.
    tiy E     ;do the same for the second digit
    MA
    tiy 1
    cal DEM+
    tiy D     ;and the third.
    MA
    tiy 0
    cal DEM+
    tia A
    M+
    jump beep
    jump start
beep:
    tia 6
    tiy D
    cal DEM-
    cal SHTS  ;'play short sound'
    jump start

Disclaimer:

I don't actually own a GMC-4. I've meticulously checked this program with documentation from online, but I may have made a mistake. I also don't know the endianness. It looks like the GMC-4 is big-endian, but I'm not sure. If anyone owns a GMC-4 and can verify this/tell me the endianness of the GMC-4, I'd much appreciate it.

\$\endgroup\$
3
\$\begingroup\$

C, 48 bytes

void f(int b){while(printf(""))Sleep(60000/b);}
                            ^ literal 0x07 here

A Windows-only solution (Sleep() function, to be specific).

I also (ab)used the fact that printf() returns the number of characters printed to use it as infinite loop condition.

There IS a character between double-quotes in printf() call, but it is not displayed here for some reason. If in doubt, copy and paste into Sublime Text 2 or Notepad++, the character will be displayed as BEL.

This started as a C++ solution but it kinda fell into the C-subset of C++ (because, you know, Sleep() is a bit shorter than std::this_thread::sleep_for(std::chrono::milliseconds())) and printf() is shorter than std::cout<<).

\$\endgroup\$
3
\$\begingroup\$

AppleScript 94 bytes

I know I'm pretty late, and this is my first post here, but whatever.

display dialog""default answer""
set x to 60000/result's text returned
repeat
beep
delay x
end

Ungolfed:

display dialog "" default answer ""
set x to 60000 / (result's text returned)
repeat
    beep
    delay x
end repeat
\$\endgroup\$
3
  • \$\begingroup\$ Hey, new answers :) Unfortunatly I'm unable to try your post unless I have no Mac ;) - but thanks a lot \$\endgroup\$
    – PEAR
    Apr 13, 2016 at 16:32
  • \$\begingroup\$ @PEAR You're welcome. :) \$\endgroup\$
    – You
    Apr 13, 2016 at 16:35
  • \$\begingroup\$ Welcome to Programming Puzzles and Code Golf. This is good answer, +1. Please keep answering! \$\endgroup\$
    – wizzwizz4
    Apr 13, 2016 at 17:09
2
\$\begingroup\$

VBScript, 113 66 bytes

a=InputBox("")
Do
WScript.Echo(Chr(7))
WScript.Sleep(60000/a)
Loop

This program is simple enough; it takes input, echoes the BEL character, and waits. Thanks to Niel for shaving off almost half the program!

\$\endgroup\$
2
  • \$\begingroup\$ What's wrong with WScript.Echo CHR(7)? Also, did you mean 60000? \$\endgroup\$
    – Neil
    Jan 23, 2016 at 20:18
  • \$\begingroup\$ @Neil Ah, yes. forgot about those.; \$\endgroup\$ Jan 23, 2016 at 21:00
2
\$\begingroup\$

Ruby, 37 33 bytes

m=->b{loop{puts"\7"
sleep 6e1/b}}

Pretty straightforward.

This is a lambda function. If you wanted 60 bpm, you'd do: m[60].

\$\endgroup\$
3
  • \$\begingroup\$ Theoretically $><<?\a should also work for the beep. And no need to give a name for your proc (all JavaScript solutions with fat arrow function also leave it unassigned), you can call it anonymously too: ->b{loop{$><<?\a;sleep 6e1/b}}[60]. \$\endgroup\$
    – manatwork
    Jan 24, 2016 at 14:32
  • \$\begingroup\$ @manatwork I only have Ruby 2.x, so I couldn't test the ?\a; do you have Ruby 1.x? If so, can you test that this works? \$\endgroup\$
    – Justin
    Jan 24, 2016 at 17:24
  • \$\begingroup\$ Well, I have a Ruby 1.9.3 and the code raises no error with it. But I have another problem with the testing: no beep on my machine. Neither Ruby nor anything else. Set something once, no idea what. \$\endgroup\$
    – manatwork
    Jan 24, 2016 at 17:35
2
\$\begingroup\$

Japt, 30 bytes

6e4/U i`?w Au¹o('../1').play()

The ? should be the literal byte 9A. Test it online! (Sorry about the pop-up delaying the first few beats; this will be removed soon.)

How it works

6e4/U i"new Audio('../1').play()  // Implicit: U = input bps
6e4/U                             // Calculate 60000 / U.
      i                           // Set a timed event every that many milliseconds,
       "new Audio('../1').play()  // running this code every time.
                                  // ../1 is the path to the file used in my JS entry.
\$\endgroup\$
2
\$\begingroup\$

Mumps, 18 bytes

R I F  H 60/I W *7

Read the BPM into variable I, then F {with two spaces after} is an infinate loop. Halt for 60 seconds / BPM, then write $CHR(7) {Ascii: BEL} to standard output, giving the audio output required, then restart at the infinite loop.

\$\endgroup\$
2
\$\begingroup\$

Java, 321 chars

Sounds very good. Works only on systems with MIDI support.

import javax.sound.midi.*;import java.util.*;class A{public static void main(String[] a) throws Exception{int d=new Scanner(System.in).nextInt();Synthesizer b=MidiSystem.getSynthesizer();b.open();MidiChannel c=b.getChannels()[0];c.programChange(116);while(true){c.noteOn(0,100);Thread.sleep((int)(d/.06));c.noteOff(0);}}}

.

\$\endgroup\$
4
  • \$\begingroup\$ Looks nice, but this does not work for me: pastebin.com/0CbGYkU0 \$\endgroup\$
    – PEAR
    Jan 30, 2016 at 21:21
  • \$\begingroup\$ @PEAR fixed. I forgot a cast. \$\endgroup\$ Jan 31, 2016 at 12:24
  • \$\begingroup\$ @PEAR and an import \$\endgroup\$ Jan 31, 2016 at 12:33
  • \$\begingroup\$ @PEAR, i had swapped some ops because of no sound \$\endgroup\$ Jan 31, 2016 at 19:02
2
\$\begingroup\$

ChucK, 90 bytes

White noise that is turned on and off every two ticks.

60./Std.atoi(me.arg(0))*1000=>float s;while(1){Noise b=>dac;s::ms=>now;b=<dac;s::ms=>now;}

Explanation

60./Std.atoi(me.arg(0)) //Convert the input to an int and divide 60 by it
*1000                   //Multiply by 1000 (in order to avoid s::second)
=>float s;              //Store it as a float in variable s
while(1)                //Forever,
{Noise b=>dac;          //Connect a noise generator b to the audio output
s::ms=>now;             //Wait for s milliseconds
b=<dac;                 //Disconnect b from the audio output
s::ms=>now;}            //Wait for s milliseconds

This is made to turn on the sound on a beat, then turn it off on the beat after.

98 93 byte version (fancier)

White noise played for 10 milliseconds per tick.

60./Std.atoi(me.arg(0))*1000-9=>float s;while(1){Noise b=>dac;10::ms=>now;b=<dac;s::ms=>now;}

This is made to be a click instead of constant noise being turned on and off.

\$\endgroup\$
2
\$\begingroup\$

Perl 5, 36 bytes

{{$|=print"\a";sleep 60/$_[0];redo}}

A subroutine; use it as

sub{{$|=print"\a";sleep 60/$_[0];redo}}->(21)
\$\endgroup\$
2
  • \$\begingroup\$ sleep is in seconds, so you can't have more than 60 beeps per minute, not sure if that's a requirement. Also, you can probably keep the same byte count but have a full program by doing something like: $|=<>;{print"\a";sleep 60/$|;redo} (can't test it right now). \$\endgroup\$
    – ChatterOne
    Jan 25, 2016 at 20:18
  • \$\begingroup\$ @ChatterOne, according to its documentation, you're right about sleep. But it worked for me. \$\endgroup\$
    – msh210
    Jan 28, 2016 at 15:47
1
\$\begingroup\$

Jolf, 7 bytes, noncompeting

I added sounds after this very fine challenge was made.

TΑa/Αaj
T       set an interval
 Αa      that plays a short beep (Α is Alpha)
   /Αaj  every 60000 / j (the input) seconds. (Αa returns 60000)

If you so desire to clear this sound, take note of the output. Say that number is x. Execute another Jolf command ~CP"x", and the interval will be cleared.

\$\endgroup\$
1
\$\begingroup\$

Zsh, 32 bytes

<<<$'\a'
sleep $[60./$1]
. $0 $1

Based on the leading bash answer, but sources instead of execs. The TIO link sources $0:a because of how the original file is executed, but it will work without it.

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ You're late to the party but this looks like really fine solution! \$\endgroup\$
    – PEAR
    Aug 18, 2019 at 9:28
  • \$\begingroup\$ I know I'm late, but I just felt like golfing today. Decided to check on the music tag for fun, and found this challenge. Good one, btw! \$\endgroup\$ Aug 18, 2019 at 9:45
1
\$\begingroup\$

Stax, 17 bytes

ü7»♥O╚⌂╥☻≈OyM╜Δ∩`

or, unpacked:

4|A48*x/W2|A]pc{| }*

The program outputs bytes that, when fed through the command line tool aplay with default setting, produce a metronome noise. The input is used as bpm

example:

example-stax-interpreter metronome.stax -i "60" | aplay

You should hear a horrible beeping noise at the desired bpm

\$\endgroup\$
0
\$\begingroup\$

Bash + bc + ><>, 44 bytes

Playing on the fact that the ><> interpreter lets you define a tick time :

python fish.py -t $(bc -l<<<"2/$1/60") -c 7o

The ><> code is 7o and should output the BEL character, producing a system beep. It will loop until interrupted.
The -t value is set to (2 / RPM ) / 60 so that the whole code is played RPM * 60 times per second.

\$\endgroup\$
2
  • \$\begingroup\$ Thanks a lot for a new answer after some amount of time after publishing. Doesn't work for me :( Not sure if a problem of my system or something else. I downloaded the fish.py from GitHub and executed your commad (openSUSE). Got this error: (standard_in) 1: syntax error usage: fish.py [-h] (<script file> | -c <code>) [<options>] fish.py: error: argument -t/--tick: expected one argument \$\endgroup\$
    – PEAR
    Apr 13, 2016 at 16:40
  • \$\begingroup\$ Have you got bc installed? It looks like the $(bc -l<<<"2/$1/60") did not produce any output. I'll add it to the list of languages of the answer. I haven't been able to fully test my answer yet, so there might be some kind of error too. \$\endgroup\$
    – Aaron
    Apr 13, 2016 at 16:53
0
\$\begingroup\$

SmileBASIC, 26 bytes

INPUT B$BGMPLAY@8T+B$+"[C]

It can play any general midi instrument, though anything above 9 will use more bytes.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.