Reverse ASCII character map

Question

In Map inputted ASCII characters, I asked you to .. map inputted ASCII characters. This time, you will do the opposite, by immediately mapping all ASCII characters, and then mapping spaces over them.

This program will have practical uses, such as seeing what characters you have left in a restricted-source challenge submission.

Challenge

Your job is to write a program that un-maps inputted ASCII characters. As soon as the program is run, it will display this exact map (except color, font, size, etc.):

Each time the user inputs a printable ASCII character, you will print a space over it.

Rules

• The initial output must be exactly as shown in the first image above: a 6x16 grid with ASCII characters in their respective positions, with the space (32) at the top left and the tilde (126) at one character left from the bottom right.
• Your program only needs to display the printable ASCII characters, 0x20 to 0x7E.
• Your program must not terminate and continue to print spaces over characters until all printable ASCII characters have been overwritten. From here, your program can either terminate or run off into Neverland.
• Your reverse map must be updated in realtime (as soon as it receives user input).
• If your program does not read input silently, it must put the cursor out of the way, so the text won't get in the way of the map.

Here's a useful ASCII table for reference.

Scoring

The answer with the least bytes in each language wins. Have fun!

Answer 1

x86-16 Machine Code, 62 bytes

Hex dump:

BC0100B101B07EBAFE0581EAE00080EA02B402CD1085E4750830E439C5750A30C0B40ACD1084C074094884D275E085D275D831C031E4CD1630E489C5EBC7

Assembly:

mov sp, 0x0001           ; Stack pointer is used as a flag; 0 - Print all characters, 1 - Delete specific character
mov cl, 0x01             ; Number of characters to print per interrupt call

printString:
  mov al, 0x7E           ; Last character to print
  mov dx, 0x05FE         ; Row: 0x05, Collumn: 0xFE
  printRow:
    sub dx, 0x00E0       ; Decrement row number + 2 extra characters
    printChar:
      sub dl, 0x02       ; Decrement collumn index + 1 space
      mov ah, 0x02       ; Prepare for interrupt call, 0x02 - Set cursor position
      int 0x10           ; BIOS interrupt

      test sp, sp        ; Are we printing all characters or removing specific character
      jnz print          ; In first case just print it and go on
      xor ah, ah         ; Otherwise reset the upper byte of ax (shorter than "and ax, 0x00FF")
      cmp bp, ax         ; Is the current character same as the input character
      jne after          ; If no, continue searching
      xor al, al         ; If yes, remove it

      print:
        mov ah, 0x0A     ; Prepare for print
        int 0x10         ; Print

      test al, al        ; If we found target character
      jz loop            ; then stop searching

      after:
        dec ax           ; Shorter than "dec, al"
        test dl, dl      ; Is it the last character in the row
        jnz printChar    ; If no, continue searching
    test dx, dx          ; Is it last char
    jnz printRow         ; If no, go to next row

loop:
  xor ax, ax             ; Remove "ah" cache
  xor sp, sp             ; Reset sp (it will never become 1 again)
  int 0x16               ; BIOS interrupt for reading keyboard input
  xor ah, ah             ; Again reset "ah", because BIOS may change it
  mov bp, ax             ; Save it in stack base pointer
  jmp printString        ; Remove the character from the list

Answer 2

SOGL V0.12, 23 bytes

] ~Δ8«n"5αx2⁰³⁄¹‘→č@ŗ░T

Try it Here!

Takes input in the input box. I hope that it isn't too big of an issue that characters can be deleted :p

Explanation:

]                   do.. while (popping (which makes the stack not blow up luckily :D))
  ~Δ                push the ascii characters (range("~"))
    8«n             split into lines of length 16
       "...‘        push "inputs.value" (could be 2 bytes less if my dictionary contained the word "inputs". I even added "input", but only now realise that the input box is id=inputs :/)
            →       evaluate as JavaScript, then push the result
             č      chop into characters
              @ŗ    replace each of the characters in the array with space
                ░   clear the output
                 T  output without popping (so do..while continues looping)

Answer 3

C++ (Visual C++), 253 (@Step Hen) 261 bytes

#include<cstdlib>
#include<iostream>
#include<conio.h>
int main(){char a[0x5E];for(int i=0;i<0x5E;i++)a[i]=(char)(i+0x20);while(true){system("cls");for(int i=0;i<0x5E;i++)if(i&&!(i%16))printf("\n%c ",a[i]);else printf("%c ",a[i]);a[_getch()-0x20]=' ';}}

Answer 4

Python 3, 116 bytes

o='\n'.join(''.join(map(chr,range(i,16+i)))for i in range(32,124,16))[:-1]
while 1:print(o);o=o.replace(input()," ")

Try it online!

Answer 5

Python 2, 132 bytes

^{Saved 4 bytes thanks to @alleks!}

s=''
for i in range(32,128,16):s+=' '.join(map(chr,range(i,i+16)))+'\n'
while 1:print s[:-2];i=2*ord(input())-64;s=s[:i]+' '+s[i+1:]

Try it online!

Answer 6

JavaScript (ES6) + HTML, 139 136 116 + 10 16 = 132 bytes

-3 bytes thanks to @Shaggy.
-14 bytes inspired by @Arnauld.

for(i=32;i<127;)O[h="innerText"]+=String.fromCharCode(i++)+(i%16?" ":`
`);onkeypress=e=>O[h]=O[h].replace(e.key," ")

<pre id=O></pre>

Closing pre tag is required in this case, since we need the innerText value to start totally empty.

Answer 7

QBasic, 107 bytes

An anonymous function that takes as keystrokes, and erases an ASCII table

FOR x=32TO 126
L x
?CHR$(x)
NEXT
DO
L ASC(Input$(1))
?" "'<- `"` included for highlighting only
LOOP
SUB L(x)
LOCATE x\16-1,2*(x MOD 16)+1
END SUB

-8 bytes thanks to @DLosc

Answer 8

C# (.NET Core), (138 + using System;) 151 bytes

()=>{var j="";for(int i=32;i<127;i++){if(i%16<1)j+='\n';j+=(char)i+" ";}while(1>0){Console.Write(j);j=j.Replace(Console.ReadLine()," ");}}

Try it online!

Answer 9

Node.js 233 212 bytes

Saved 21 bytes thanks to @thePirateBay

This works, I'm still trying to find an online option as all the repls I've found hijack stdin.

a=[];l=_=>console.log(a.join``);for(i=32;i<127;i++)(x=String.fromCharCode(i)),a.push(i%16?x:x+'\n');l();b=require('readline');b.createInterface(process.stdin).on('line',c=>(a=a.join``.replace(c,' ').split``),l())

Try it online

Answer 10

65c02 machine code + Apple //e ROM, 52 (47?) bytes

Hex dump:

8000- 20 58 FC A9 A0 20 20 80
8008- 20 ED FD 1A C9 FF D0 F5
8010- AD 00 C0 10 FB 8D 10 C0
8018- 20 20 80 20 57 DB 80 F0
8020- 48 38 E9 A0 48 29 0F 0A
8028- 85 24 68 4A 4A 4A 4A 20
8030- 5B FB 68 60

Commented assembly:

 1 HTAB     =     $24        ; HORIZONTAL POSITION OF CURSOR
 2 SETVTAB  =     $FB5B      ; SETS VERTICAL POSITION OF CURSOR FROM ACC
 3 COUT     =     $FDED      ; OUTPUTS CHARACTER IN ACC
 4 HOME     =     $FC58      ; CLEARS SCREEN
 5 OUTSP    =     $DB57      ; APPLESOFT ROUTINE THAT OUTPUTS A SPACE
 6 KBD      =     $C000      ; KEY LAST PRESSED
 7 KBSTROBE =     $C010      ; ACCESS TO RESET "NEW KEY PRESSED" INDICATOR
 8          ORG   $8000
 9          JSR   HOME
10 * PRINT INITIAL CHARACTER MAP
11          LDA   #" "
12 LOOP     JSR   CHARPOS
13          JSR   COUT
14          INC              ; INCREMENT ACCUMULATOR
15          CMP   #"~"+1
16          BNE   LOOP
17 * WAIT FOR KEYPRESS
18 GETCH    LDA   KBD        ; GET LAST KEY PRESSED
19          BPL   GETCH      ; READ AGAIN IF KEYPRESS IS NOT NEW
20          STA   KBSTROBE   ; RESET "NEW KEYPRESS" INDICATOR
21          JSR   CHARPOS
22          JSR   OUTSP
23          BRA   GETCH
24 * SUBROUTINE TO POSITION CURSOR TO PRINT OVER CHARACTER IN ACCUMULATOR
25 CHARPOS  PHA              ; PRESERVE ACC
26          SEC              ; MAKE SURE CARRY IS SET TO SUBTRACT
27          SBC   #" "       ; SUBTRACT CHAR CODE OF SPACE
28          PHA              ; SAVE ACC
29          AND   #$0F       ; GET LOWER 4 BITS TO GET CURSOR X POSITION
30          ASL              ; SHIFT LEFT TO MAKE SPACES BETWEEN CHARS
31          STA   HTAB
32          PLA              ; GET OLD ACC
33          LSR              ; SHIFT HIGH NIBBLE
34          LSR              ; INTO LOW NIBBLE
35          LSR              ; TO GET CURSOR Y POSITION
36          LSR
37          JSR   SETVTAB
38          PLA              ; RESTORE ACC
39          RTS

This has no cursor whatsoever. I also have a 47 byte version which might be valid, depending on what is meant by "put the cursor out of the way, so the text won't get in the way of the map":

8000- 20 58 FC A9 A0 20 1B 80
8008- 20 ED FD 1A C9 FF D0 F5
8010- 20 0C FD 20 1B 80 20 57
8018- DB 80 F5 48 38 E9 A0 48
8020- 29 0F 0A 85 24 68 4A 4A
8028- 4A 4A 20 5B FB 68 60

This puts a cursor at the character after the character you type (which is the space between characters), so it won't overwrite any of the actual characters in the map. Whether this is valid or not is up to the creator of the challenge.

Answer 11

Python 2, 96 bytes

from textwrap import*
a=fill(str(bytearray(range(32,127))),16)
while 1:a=a.replace(input(a),' ')

Try it online! Output looks iffy on TIO (because input is passed from a file), but it’s fine interactively. I hope 'A'\n, 'B'\n etc is an okay input format.