Map inputted ASCII characters

Question

Be sure to see the other challenge, Reverse ASCII character map!

The ASCII charset (American Standard Code for Information Interchange) is the most widely-used character encoding standard. ASCII codes represent text in computers, telecommunications equipment, and other devices.

Challenge

Your challenge is to print a mapping of the ASCII character set as the user inputs them. GIF:

After the user enters every ASCII character, the output should look like this:

Mapping

Each character has an assigned position on a 16x6 logical grid, starting with the space character in the top left position, and wrapping such that the digit 0 appears below it.

When printable ASCII input is received, print that ASCII character at its assigned screen location without deleting any of the characters currently onscreen.

Rules

• Your program only needs to map out the printable ASCII characters, 0x20 to 0x7E.
• Your program must not terminate and continue to map characters to the screen until all printable ASCII characters have been inputted. From here, your program can either terminate or run off into Neverland.
• Your program can map characters any way you like, e.g. to a spreadsheet, table, console window, or a graphical window.
• No matter how you display the mapping, it 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.

Help

Here is the pseudocode algorithm I used to generate the GIF:

loop forever
    c = input
    y_coord = c / 16
    x_coord = c - y * 16
    if c is printable
        print c at (x_coord * 2 + 1, y_coord + 1)
    end if
end loop

There may be another way to achieve the required output. You can choose to use my algorithm or your own, but the output must be the same regardless.

Here's a useful ASCII table reference.

Scoring

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

Answer 1

JavaScript (ES6) + HTML, 114 + 16 = 130 bytes

Saved 16 bytes thanks to @Shaggy

a=Array(96).fill` `;onkeypress=k=>(a[k.key.charCodeAt()-32]=k.key,O.innerText=a.join` `.match(/.{1,32}/g).join`
`)

<pre id=O></pre>

It's so unbelievably satisfying to just mash the keyboard...

Answer 2

QBasic 4.5, 81 85 bytes

Added 4 bytes to comply with the spacing-rule.

DO
LOCATE 7,1
LINE INPUT A$:i=ASC(A$)
LOCATE i\16-1,(i MOD 16+1)*2
?CHR$(i)
LOOP

And the output will look like this (NOTE: Old screenshot, now every character is separated by a space):

QBasic has the LOCATE command, which comes in handy here. A breakdown of this code:

DO                          Starts an infinite loop
LOCATE 7,1                  Moves the cursor out of the way
LINE INPUT A$:i=ASC(A$)     LINE INPUT gets user input; we need LINE INPUT instead of regular input
                            for support of <space> and <comma>. The ASC() function then takes the
                            ASCII value of the first character in the input, so it can deal with
                            inputs like 'Hello' - it will take ASC('H') and assign that to 'i'
LOCATE i\16-1               Here's the cool bit: LOCATE takes a row and a column to put the cursor on.
    ,(i MOD 16+1)*2         Row is determined by dividing the ASC value by 16, minus 1 (SPACE, ASC 32 
                            is placed on row 1), and for columns we take the modulo plus 1 (Again, SPACE 
                            mod 16 = 0, plus 1 = column 1). Multiplied by 2 gives us the spacing. 
                            We move the cursor to 1,2
?CHR$(i)                    PRINT a cast of the ASCII value to CHR at the specified location.
LOOP                        Ad infinitum

Answer 3

Java 8, 143 bytes

o->{for(;;){char c=System.console().readPassword()[0];if(c>31&c<127)System.out.println(String.format("\u001B[%d;%df%c",c/16+1,(c%16+1)*2,c));}}

Uses the ANSI control code CSI n ; m f to set the cursor position and Console.readPassword() to read the user input silently. Output of some characters:

Answer 4

BrainFuck, 355 Bytes

>>++++[->++++[->+>++++++<<]<]>>->[-<[-<<<<++++[->++++++++<]]>>>[<<<<<++++++++[->>++<<<+>]>>-<<<++>>]<<[>>>>>>[>>>]>+<<<<<[<<<]<<-]>>>>>>[>>>]++++[-<++++++++>]>[-<+>]<<[<<<]>>]<[-]<<,[[->+>+<<],[-]++++[->>--------<<]>>[>>>>[>>>]+[<<<]<-]>>>>[>>>]<<[-]<[<<<]<<[>>>>>[>>>]<<+<[<<<]<<-]>>>>>[>>>]<<<[[-]<<<]>[.>.>>]++++++++++[->+>++<<]>>[-<.>]<[-]<<<<[<<<]<,]

BrainFuck's options are pretty limited, so output is in the terminal and the screen is "cleared" with 20 newlines. Input should be the ASCII characters, separated by newlines.

Try it online!

Formatted and Documented

These are the debug notes I used to write the program. I used my interpreter which can optionally print the state of the tape at every '~' character for debugging.

[
    run.bf
    codegolf.stackexchange.com/questions/124306/map-inputted-ascii-characters
]


[
    Calculate 16 * 6
    Resulting tape state:
    [0 0 0 0 0 0 16 96 0 0 0 0 ...]
               ^
    Note that, to obtain a 16-by-6 grid, the 16
    immediately to the right is decreased to 15
    (since we will decrease it by 1 each loop
    until we reach 0 and immediately reset)
]
>>>>++++[->++++[->+>++++++<<]<]>~

[
    Our next goal is to make 96 sets of 3 cells each in the pattern [C D 0]
    The first cell will represent an entered character--when the corresponding
    input on the keyboard is pressed, it will change to the entered key.
    The first cell is initialized to 32 (' ').

    The second cell will represent the delimiter after that character.
    Every 16 cells, this should be 10 for '\n'. Otherwise, it should be 32 for ' '.

    The third cell is a buffer cell, used for traversal of the grid. In general,
    it should be only temporarily modified and then reset to 0.
]

>->[-<
    [
       -<<<<++++[->++++++++<]
       [
           The second cell of our 3-set should be 32, so the above line
           writes 32 to the 3rd cell from the beginning of the tape (0-indexed)
       ]
    ]
    >>>
    [
       <<<[ The second cell of our 3-set should be 10, and we must reset the line counter ] 
       <<++++++++[->>++<<<+>]>>-<<<++>>
    ]

    [ At this point, the delimiting cell we need is two cells to the left. ]
    <<[>>>>>>[>>>]>+<<<<<[<<<]<<-]
    
    >>>>>>[>>>]++++[-<++++++++>]
    [ Debug Mode: In the previous loop, add a + in the string of 8 +'s to get visible spaces in the grid ($-signs) ]
    >[-<+>]<<[<<<]>>
]

[ Go back to the beginning of the tape and clear up the residual '15' ]
<[-]~

<<,

[
    [->+>+<<],[-]++++[->>--------<<]

    [
        Take input such that the state of the tape now looks like this:
        [0 0 0 0 0 c c-32 0 32 32 0 32 32 0 32 32 0 ...]
                 ^
        Where 'c' was the entered character.
        We now set up 1's in the buffer zones of the first c-32
        3-sets and clear the character that is currently there.
        All that is left, then, is to copy c to that location.
    ]

    [ Set up the row of 1's. ]
    >>[>>>>[>>>]+[<<<]<-]

    [ Clear the current character. ]
    >>>>[>>>]<<[-]~<[<<<]

    [ Copy the new character. ]
    <<[>>>>>[>>>]<<+<[<<<]<<-]

    [ Clean up the 1's. ]
    >>>>>[>>>]~<<<[[-]<<<]

    [ Print the grid. ]
    >[.>.>>]~

    [ Print a bunch of newlines ]
    ++++++++++[->+>++<<]>>[-<.>]<[-]

    [ Take a new input. ]
    <<<<[<<<]<,
]

Answer 5

Mathematica, 108 bytes

a=" "~Table~16~Table~6;Dynamic@Grid@a
((a[[⌊#/16⌋-1,#~Mod~16+1]]=i)&@ToCharacterCode[i=Input[]];#0[])&[]

Try it online at https://sandbox.open.wolframcloud.com/

When you paste code and press Shift+Enter, a dialog box will pop up, you enter "a" for example for character a. The program runs forever.

Note: On Wolfram sandbox, the font is formatted differently from that in Mathematica in my computer. So the line/column spacing may looks weird.

Answer 6

Python 2, 115 bytes

s='\n'.join([' '*31]*6)
while 1:
 c=input();i=ord(c)
 if 31<i<128:i-=32;i=i%16*2+i//16*32;s=s[:i]+c+s[i+1:];print s

Try it online!

Requires quotation marks (single or double) around the inputted characters (the TIO version does not).

Answer 7

str, noncompeting, 18 bytes

Presenting my new semi-esoteric language.

#C;dby16#/~2-~u#pq

#C;dby16#/~2-~u#pq
..;                   preamble
#C                    clear screen
   ...............    main program; each character is pushed to the stack before
   d                  duplicate
    b                 buffer the character
     y                convert to character code
      16#/            divmod by 16 (a / b, a % 6)
          ~2-~        subtract a / b by 2
              u       unbuffer the character
               #p     place that character in the given position
                 q    "quiet"; disable auto-printing

Answer 8

QBIC, 53 57 bytes

Added 4 bytes for the spacing.

{locate 7,1┘_?┘i=asc(A)┘locate i/16-1,(i%16+1)*2┘?chr$(i)

QBIC began development as a shorthand for QBasic, so I thought translating my QBasic answer would demonstrate this nicely. We've saved some 40% in the byte-count for a functionally identical program - and that's even when LOCATE, ASC, and CHR have no QBIC-functions yet. Fortunately, QBIC can pass code directly to QBasic to compensate for this. A side-by-side:

QBIC              QBASIC
------------      ------------
{                 DO
locate 7,1        LOCATE 7,1
                  note that the lower-case alphabet is left unaltered in QBIC.
_?                LINE INPUT A$  (LINE INPUT used instead of INPUT to handle comma's)
i=asc(A)          i=ASC(A$)
locate i/16-1     LOCATE i/16-1
   ,(i%16+1)*2       ,(i MOD 16+1)*2
?chr$(i)          ?CHR$(i)
                  LOOP   (implicitly added to QBIC at EOF)

Answer 9

Haskell, 133 bytes

p=putStr.("\27["++)
g l=do c<-getChar;p"2J";mapM h(c:l);g(c:l)
h c|(y,x)<-divMod(fromEnum c)16=p$show y++';':show(2*x+1)++'H':[c]
g[]

Requires a terminal that understands ANSI escape sequences.

It's shorter to keep a list of all keys pressed so far and clearing the screen before printing all of them each round than turning off the echo in the terminal session. The latter needs import System.IO and hSetEcho stdin(2<1) which costs too many bytes.

Answer 10

C, 101 bytes

c,y,x;f(){while(~(c=getchar()))printf("\e[1;20H"),y=c/16,x=c-y*16,printf("\e[%d;%dH%c",y+1,x*2+1,c);}

This was the program I used to make the graphics. Output is as shown in the GIF. ;)

Answer 11

QBasic, 62 58 bytes

a=ASC(INPUT$(1))
LOCATE a\16-1,1+2*(a MOD 16)
?CHR$(a)
RUN

Tested with QB64. Should work fine on regular QBasic, too, although you may want to modify it to do a CLS on the first run.

Similar to steenbergh's answer, but uses INPUT$(1) to read characters one at a time. This approach is shorter and also displays no prompt. It also uses RUN for the infinite loop, since we don't have to store any state between iterations except the state of the screen.

Answer 12

Ruby, 79 75 71 + 13 = 84 bytes

+13 bytes for -rio/console flag.

loop{$/+=STDIN.getch
97.times{|n|print$/[(n+31).chr]||" ",["
"][n%16]}}

Ungolfed

loop {
  $/ += STDIN.getch
  97.times {|n|
    print $/[(n+31).chr] || " ", ["
"][n%16]
  }
}

Answer 13

Pascal, 112 chars

Uses crt;var c:char;Begin ClrScr;repeat c:=ReadKey;GotoXY(ord(c)and$F*2+1,ord(c)shr 4-1);write(c);until 1<0;End.

As my Mathematica solution takes many bytes in div, mod and ToCharacterCode[Input[]], I try making another answer with Pascal. But without ClrScr my compiler (FPC) left some compile information on the screen. ClrScr; takes 7 bytes.

The *2 used for proper spacing takes another 2 bytes.

Answer 14

LOGO, 90 bytes

cs
rt 90
keyboardon[invoke[setxy 30*modulo ? 16 -30*int ?/16 label char ?]keyboardvalue]pu

Try it on FMSLogo.

After all, my Logo solution is the shortest, compared with my Mathematica and Pascal answer.

Add 3 bytes if the turtle is required to be hidden.

Answer 15

6502 machine code + Apple //e ROM, 31 bytes

Hex dump:

8000- 20 58 FC 20 0C FD 48 38
8008- E9 A0 48 29 0F 0A 85 24
8010- 68 4A 4A 4A 4A 20 5B FB
8018- 68 20 ED FD 4C 03 80

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 RDKEY    =     $FD0C      ; GETS CHARACTER FROM KEYBOARD, STORES IN ACC
 6          ORG   $8000
 7          JSR   HOME
 8 GETINP   JSR   RDKEY
 9 * POSITION CURSOR
10          PHA              ; PRESERVE ACC
11          SEC              ; MAKE SURE CARRY IS SET TO SUBTRACT
12          SBC   #" "       ; SUBTRACT CHAR CODE OF SPACE
13          PHA              ; SAVE ACC
14          AND   #$0F       ; GET LOWER 4 BITS TO GET CURSOR X POSITION
15          ASL              ; SHIFT LEFT TO MAKE SPACES BETWEEN CHARS
16          STA   HTAB
17          PLA              ; GET OLD ACC
18          LSR              ; SHIFT HIGH NIBBLE
19          LSR              ; INTO LOW NIBBLE
20          LSR              ; TO GET CURSOR Y POSITION
21          LSR
22          JSR   SETVTAB
23          PLA              ; RESTORE ACC
24 *
25          JSR   COUT
26          JMP   GETINP

If the cursor invalidates it, here's a 36-byte version without a cursor:

8000- 20 58 FC AD 00 C0 10 FB
8008- 8D 10 C0 48 38 E9 A0 48
8010- 29 0F 0A 85 24 68 4A 4A
8018- 4A 4A 20 5B FB 68 20 ED
8020- FD 4C 03 80

Answer 16

SmileBASIC 3, 82 bytes

CLS
@L
C$=INKEY$()IF""!=C$THEN V=ASC(C$)-32LOCATE V MOD 16*2,V DIV 16*2?C$;
GOTO@L

In the SmileBASIC character set, ¥ is located where \ normally would be; hopefully this doesn't invalidate this answer completely.

Answer 17

Japt, 63 43 bytes

Seeing as I apparently have a reputation for making Japt do things it shouldn't be able to do (continuous input, in this case), golfing from my my phone, and golfing with a feed of beer inside me, it's high time I combined all 3 of those things!

;$oninput=$@OqE¬ËoOx`Í*s.value` 1Ãú ò16 m¸·

Test it (just start typing in the "Notes" field)

Answer 18

Applesoft BASIC, 134 bytes

0TEXT:HOME:PR#0
1C=PEEK(49152):POKE49168,0:HTAB1:VTAB20:NORMAL:IFC>=128THENC=C-128:INVERSE
4Y=INT(C/16):X=C-Y*16:HTABX*2+1:VTABY+1:IFC>=32THEN PRINTCHR$(C):IFC<32THEN PRINTCHR$(127)
9GOTO1

This is a golfed version of the Apple ][ keyboard test, the program that inspired the challenge.

Answer 19

8086 machine code, MS-DOS .COM, 27 bytes

Runnable in DOSBox. The only assumption about register starting values is AH = 0, which is true for most DOS isotopes. This, and the fact that only BIOS interrupts are used, means it can be run from a boot sector by changing the first instruction to mov ax, 3 (and the org of course) at the terrible, terrible cost of one byte.

Expects only printable ASCII characters as input. Does not terminate.

Binary

00000000 : B0 03 CD 10 31 C0 CD 16 50 2C 20 D4 10 D0 E0 92 : ....1...P, .....
00000010 : B4 02 CD 10 58 B4 0E CD 10 EB E9                : ....X......

Assembly

org 0x100
cpu 8086
    
    mov al, 3       ; Set video mode to 80x25 text
    int 0x10        ; Has the desired side-effect of clearing the screen
    
main_loop:
    xor ax, ax      ; Read keystroke into AX
    int 0x16        ; AH = scan code, AL = ASCII
    
    push ax         ; Save ASCII code for later
    sub al, 32      ; Make ASCII code into range 0..94
    aam 0x10        ; AH = AL / 16 (row), AL = AL % 16 (col)
    shl al, 1       ; Double column for spacing
    
    xchg ax, dx     ; Move cursor to proper position
    mov ah, 2
    int 0x10
    
    pop ax          ; Retrieve ASCII code
    mov ah, 0x0e    ; Output at cursor position
    int 0x10
    
    jmp main_loop