Create an ASCII-to-hex table for Mark Watney

Question

Inspired by this SciFi.SE question.

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Background (with minor spoiler):

In the movie The Martian, protagonist Mark Watney uses an ASCII table to look up hexadecimal values of ASCII characters so he can attempt to communicate back to Earth.*

Challenge

With no input, output the following ASCII table exactly like this:

Dec Hex    Dec Hex    Dec Hex  Dec Hex  Dec Hex  Dec Hex   Dec Hex   Dec Hex  
  0 00 NUL  16 10 DLE  32 20    48 30 0  64 40 @  80 50 P   96 60 `  112 70 p
  1 01 SOH  17 11 DC1  33 21 !  49 31 1  65 41 A  81 51 Q   97 61 a  113 71 q
  2 02 STX  18 12 DC2  34 22 "  50 32 2  66 42 B  82 52 R   98 62 b  114 72 r
  3 03 ETX  19 13 DC3  35 23 #  51 33 3  67 43 C  83 53 S   99 63 c  115 73 s
  4 04 EOT  20 14 DC4  36 24 $  52 34 4  68 44 D  84 54 T  100 64 d  116 74 t
  5 05 ENQ  21 15 NAK  37 25 %  53 35 5  69 45 E  85 55 U  101 65 e  117 75 u
  6 06 ACK  22 16 SYN  38 26 &  54 36 6  70 46 F  86 56 V  102 66 f  118 76 v
  7 07 BEL  23 17 ETB  39 27 '  55 37 7  71 47 G  87 57 W  103 67 g  119 77 w
  8 08 BS   24 18 CAN  40 28 (  56 38 8  72 48 H  88 58 X  104 68 h  120 78 x
  9 09 HT   25 19 EM   41 29 )  57 39 9  73 49 I  89 59 Y  105 69 i  121 79 y
 10 0A LF   26 1A SUB  42 2A *  58 3A :  74 4A J  90 5A Z  106 6A j  122 7A z
 11 0B VT   27 1B ESC  43 2B +  59 3B ;  75 4B K  91 5B [  107 6B k  123 7B {
 12 0C FF   28 1C FS   44 2C ,  60 3C <  76 4C L  92 5C \  108 6C l  124 7C |
 13 0D CR   29 1D GS   45 2D -  61 3D =  77 4D M  93 5D ]  109 6D m  125 7D }
 14 0E SO   30 1E RS   46 2E .  62 3E >  78 4E N  94 5E ^  110 6E n  126 7E ~
 15 0F SI   31 1F US   47 2F /  63 3F ?  79 4F O  95 5F _  111 6F o  127 7F DEL

The final newline is optional. With the newline, the md5 of the output is 58824a1dd7264c0410eb4d727aec54e1. Without, it is 41b6ecde6a3a1324be4836871d8354fe.

In case it helps, this is the output from the ascii Linux command with the usage info at the top chopped off. You can recreate this on Ubuntu as follows:

sudo apt-get install ascii
ascii | tail -n+7

You may not use the ascii utility (or similar) in your answers.

Because ASCII characters are small Enough with this silly meme already!

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I am aware this is similar to Print the ASCII table, but I believe the formatting of the output in this question is significantly more complex and warrants a different question.

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*Note, I have not yet seen this movie.

Answer 1

JavaScript (ES6), 323 332 353

Edit I managed to shorten this a bit
Step 1, believe it or not, using an array of 16 chars is shorter than toString + toUpperCase
Step 2, fiddling around to use .map instead of for, so that I can transform it all to a single expression function, avoding console.log and/or return.

Edit 2 Moved DEL at place 0, idea borrowed from Cole Cameron

H=x=>(x=[...'0123456789ABCDEF']).map((d,r)=>x[S='slice'](0,8).map(c=>(o+=`   ${w=c*16+r} ${c+d} `[S](-z)+("DELNULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US ".substr((w+1&127)*3,3)||String.fromCharCode(w)),z=c<5?8:9,'Dec Hex   '[S](0,c<2?z+2:z)),o+=`
`,z=7).join` `,o='')[0]+o

MD5: 41B6ECDE6A3A1324BE4836871D8354FE

Pixel perfect, I'd say

LESS GOLFED

H=x=>(
  x=[...'0123456789ABCDEF'],
  a="DELNULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US ",
  o='', // the string o will contain the body (16 rows)
  x.map((d,r)=> ( // main loop, r is thr row number, d is the second hex digit to print
    o+=`\n`, z=7,
    x.slice(0,8).map(c=> // loop on 8 columns, c is both column number and digit to print
      (
        // append the column to o
        o += `   ${w=c*16+r} ${c+d} `.slice(-z) + (a.substr((w+1&127)*3,3)||String.fromCharCode(w)), 
        z=c<5?8:9, // adjust the column size
        'Dec Hex   '.slice(0,c<2?z+2:z) // column head, right size
      ) // the .map result is the 8 columns heading
    ).join` ` // join the heading in a single string
  ))[0] // any element of the result map is the heading
  + o // concatenate the body
)

Test

H=x=>(x=[...'0123456789ABCDEF']).map((d,r)=>x[S='slice'](0,8).map(c=>(o+=`   ${w=c*16+r} ${c+d} `[S](-z)+("DELNULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US ".substr((w+1&127)*3,3)||String.fromCharCode(w)),z=c<5?8:9,'Dec Hex   '[S](0,c<2?z+2:z)),o+=`
`,z=7).join` `,o='')[0]+o

/* TEST */ 
console.log=x=>O.textContent=x

console.log(H())

<pre id=O></pre>

Answer 2

C, 307 310 308 307 305 bytes

Finally working 100%.

i,j,z=127;main(){for(;j++<8;)printf("Dec Hex%*s",j<3?4:j<6||j>7?2:3,"");for(;i<143&&putchar(!i|i>z?10:32);i+=16)i=i>z?i%z:i,printf("%*d %02X ",i>95?4:3,i,i),i%z>31?putchar(i):printf("%.3s","DELNULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US "+(i+1)%128*3);}

Output:

$ ./a.out
Dec Hex    Dec Hex    Dec Hex  Dec Hex  Dec Hex  Dec Hex   Dec Hex   Dec Hex
  0 00 NUL  16 10 DLE  32 20    48 30 0  64 40 @  80 50 P   96 60 `  112 70 p
  1 01 SOH  17 11 DC1  33 21 !  49 31 1  65 41 A  81 51 Q   97 61 a  113 71 q
  2 02 STX  18 12 DC2  34 22 "  50 32 2  66 42 B  82 52 R   98 62 b  114 72 r
  3 03 ETX  19 13 DC3  35 23 #  51 33 3  67 43 C  83 53 S   99 63 c  115 73 s
  4 04 EOT  20 14 DC4  36 24 $  52 34 4  68 44 D  84 54 T  100 64 d  116 74 t
  5 05 ENQ  21 15 NAK  37 25 %  53 35 5  69 45 E  85 55 U  101 65 e  117 75 u
  6 06 ACK  22 16 SYN  38 26 &  54 36 6  70 46 F  86 56 V  102 66 f  118 76 v
  7 07 BEL  23 17 ETB  39 27 '  55 37 7  71 47 G  87 57 W  103 67 g  119 77 w
  8 08 BS   24 18 CAN  40 28 (  56 38 8  72 48 H  88 58 X  104 68 h  120 78 x
  9 09 HT   25 19 EM   41 29 )  57 39 9  73 49 I  89 59 Y  105 69 i  121 79 y
 10 0A LF   26 1A SUB  42 2A *  58 3A :  74 4A J  90 5A Z  106 6A j  122 7A z
 11 0B VT   27 1B ESC  43 2B +  59 3B ;  75 4B K  91 5B [  107 6B k  123 7B {
 12 0C FF   28 1C FS   44 2C ,  60 3C <  76 4C L  92 5C \  108 6C l  124 7C |
 13 0D CR   29 1D GS   45 2D -  61 3D =  77 4D M  93 5D ]  109 6D m  125 7D }
 14 0E SO   30 1E RS   46 2E .  62 3E >  78 4E N  94 5E ^  110 6E n  126 7E ~
 15 0F SI   31 1F US   47 2F /  63 3F ?  79 4F O  95 5F _  111 6F o  127 7F DEL$ ./a.out > file.txt
$ md5sum file.txt
41b6ecde6a3a1324be4836871d8354fe  file.txt

Ungolfed:

/* some variables for the trip */
i,j,z=127;

main()
{
    /* print header row */
    for(;j++<8;)
            printf("Dec Hex%*s", j<3?4:j<6||j>7?2:3, "");

    /* Iterate through ASCII values, print a space after every column, newline after every 8th value */
    for(;i<143 && putchar(!i|i>z ? 10 : 32); i+=16)
    {
            /* print dec/hex value */
            printf("%*d %02X ", i>95?4:3, i, i=i>z?i%z:i);

            /* print character or friendly name for non-printable characters */
            i%z>31
                    ? putchar(i)
                    : printf("%.3s", "DELNULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US "+(i+1)%128*3);
    }
}

Try it on Ideone.

Edit: 2 more bytes. Many thanks to Dan Allen and Digital Trauma.

Answer 3

Bubblegum, 535 bytes

0000000: e0 05 2f 02 0f 5d 00 22 19 48 62 01 d3 1f 78 e2  ../..].".Hb...x.
0000010: 9a a0 8e 4e 5d d1 b4 c1 77 aa 32 58 ca 97 55 7e  ...N]...w.2X..U~
0000020: a8 01 87 7d db e4 00 55 8f c0 49 67 b6 56 02 5e  ...}...U..Ig.V.^
0000030: ae b2 4d e2 a9 f4 7f 99 a8 56 9e b7 4c 60 a4 79  ..M......V..L`.y
0000040: 6a 76 54 11 90 72 d6 b7 19 df 2f 57 39 2d 21 c0  jvT..r..../W9-!.
0000050: d1 4c 5e d6 21 29 c8 ed 7c a9 7b 8c 85 dc 62 a1  .L^.!)..|.{...b.
0000060: 65 98 e1 0b a7 36 83 c8 ca 88 0c 57 22 f6 56 1e  e....6.....W".V.
0000070: 45 03 b6 74 21 a8 39 52 e9 71 b4 98 ed 71 38 9f  E..t!.9R.q...q8.
0000080: 2d dc 21 d7 bf 60 41 cc bb bd a7 cb 0b 17 8d 65  -.!..`A........e
0000090: 05 13 04 0f 6c bb 67 62 aa c7 ad 6b be 9e 46 77  ....l.gb...k..Fw
00000a0: 35 b9 91 85 f5 47 31 2f c7 ec da c0 00 0e a6 48  5....G1/.......H
00000b0: 01 ba 8b cd b0 34 81 c4 74 9f 4e 3b c3 d0 f7 10  .....4..t.N;....
00000c0: 46 a0 55 8d 49 5d b7 b0 c9 79 ac e5 5f ef 49 f2  F.U.I]...y.._.I.
00000d0: b0 1b 71 3a e1 30 7a fc ce a7 a8 d5 c3 9a 35 1a  ..q:.0z.......5.
00000e0: 4e 27 92 40 4b b5 9b c4 0d 5c e8 cd 71 00 bd c1  N'.@K....\..q...
00000f0: ca aa d2 05 dc e1 0f d9 19 1d 6f 14 87 b3 e4 e8  ..........o.....
0000100: 9e 82 64 d8 e4 76 e7 24 0a 0e 88 72 a1 12 44 95  ..d..v.$...r..D.
0000110: d4 78 82 bd da 71 f3 fb 03 00 d1 4b c8 80 cb 49  .x...q.....K...I
0000120: 0b 98 be 26 ba 3e e8 82 e2 14 9b ba 1a cf bf bc  ...&.>..........
0000130: 30 4e c4 e8 7e b4 d5 46 e6 bc 73 97 c5 ed a6 e2  0N..~..F..s.....
0000140: 06 02 e7 1b 74 4d da 73 fb 15 68 50 c0 ed 32 9b  ....tM.s..hP..2.
0000150: 0d d7 49 d5 c1 a2 e9 07 2c 77 81 6c d3 8d 59 26  ..I.....,w.l..Y&
0000160: 1c 35 ec 2b 7e cb 3a f1 cc 45 a9 e5 6d 3e 33 ca  .5.+~.:..E..m>3.
0000170: 56 3c 8a 8d f6 13 e9 59 d4 52 07 44 ab 5e bc f4  V<.....Y.R.D.^..
0000180: 1f ed f8 9c 8b 48 e1 c4 6c fd 47 d5 04 cc 6e aa  .....H..l.G...n.
0000190: 3f 54 b8 cc cd 09 01 6d 20 3c 42 c9 44 da b1 c1  ?T.....m <B.D...
00001a0: 69 80 12 26 6b 65 e1 4d 1c c3 48 36 2b 14 00 61  i..&ke.M..H6+..a
00001b0: 04 6b 9a 59 2a 53 e3 64 a7 4f dd cc be 2c 20 5e  .k.Y*S.d.O..., ^
00001c0: f7 c7 64 34 e6 12 a6 44 c1 69 35 76 05 db 13 ab  ..d4...D.i5v....
00001d0: 52 10 b5 8e da 8e c5 3c 4c d0 69 0b 19 18 67 ef  R......<L.i...g.
00001e0: 44 1c 7b 70 63 98 95 40 28 6e 3d e7 44 cb 24 83  D.{pc..@(n=.D.$.
00001f0: 88 62 63 3c 02 1c e7 db db 02 56 ae cd 9c e0 9c  .bc<......V.....
0000200: 1c a1 c1 ae d1 dd 7b b7 e6 bd 5b 38 ee 75 c5 6c  ......{...[8.u.l
0000210: 06 16 6c b2 fb 00 00                             ..l....

This above program uses LZMA compression. Try it online!

Answer 4

JavaScript ES6 432 405 398 385

o="Dec Hex  "
o=`${o}  ${o}  ${o+o+o+o} ${o} ${o}\n`
for(i=0;i<16;i++){for(j=0;j<8;j++){k=i+j*16
o+=k>9&&k<16?' ':k<96||k>99?'  ':'   '
o+=k+" "+(k<16?0:'')+k.toString(16).toUpperCase()+" "
o+=k>31?String.fromCharCode(k):"NULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US ".substr(k*3,3)
}o+="\n"}
console.log(o.substr(0,o.length-2)+'DEL')

Answer 5

C, 385 384 358 bytes

i,n,j;char a[100]="NULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US DEL";z(o){printf("%3d %02X %.*s ",j,j,3,a+o);}main(){for(;i<8;i++){printf("%3s %3s   ","Dec","HEX ");}printf("\n");for(;n<16;n++){for(j=n;j<=n+112;j+=16){if(j==127)z(96);else j<32?z(j*3):printf("%3d %02X %c   ",j,j,j);}printf("\n");}return 0;}

The guy above beat me to the punch but I still wanted to submit because I enjoyed this one.

De-golfed:

#include<stdio.h>
i,n,j;
char a[100] = "NULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US DEL";
z(o){printf("%3d %02X %.*s ",j,j,3,a+o);}
main(){
   for(;i<8;i++){printf("%3s %3s   ","Dec","HEX ");}printf("\n");
   for(;n<16;n++){
      for(j=n;j<=n+112;j+=16){
         if(j==127)z(96);
         else j<32?z(j*3):printf("%3d %02X %c   ",j,j,j);
      }
      printf("\n");
   }
   return 0;
}

UPDATE : replaced a var with j. Saved a byte ;)
UPDATE2: Trimmed a few extra thingys and functionized a print thingy to save some bytes.

Answer 6

Python 2.7, 389 Bytes

Probably not going to try and trim this down anymore, but it was fun to get it this far.

r=range
c='NULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US '
h=''
p=['']*16
o=[11,11,9,9,9,10,10,9]
e=['%3s '%str(i)+('%02X ')%(i)+('DEL',c[i*3:i*3+3].strip() if i<32 else chr(i))[i<127] for i in r(128)]
for i in r(8):
    l=e[i*16:(i+1)*16]
    h+='Dec Hex'.ljust(o[i])
    p=[p[j]+l[j].ljust((0,o[i])[i<7]) for j in r(16)]
print h+'\n'+'\n'.join(p)

Answer 7

Golfscript, 225 bytes

"\x04\x04\x02\x02\x02\x03\x03\x02"{"Dec Hex"" ":s@*}%n[128,{3s*\+-4>s+.~.96<@>\256+16base{10,"ABCDEF"1/+=}%1>{+}/}%"\x1f\xbb\x89\xbc\xaf\x82=F\xd7U%\x80\x8a\xf6\xc7U\xa6al)A\xf6\x19\xfe\x03\x83\xef-\x9f\xe7m\xeb\x82h\xf3\xbfEm6V\x1fD\x8c\xd7~\xcb\x95&(\x1e/:\x8e\xc5\xb0\x0b\xd6\xd4\xd09\xdd""\xff\x1e"{base}/{90,65>1342s++1/=}%3/32/"\x7f"{,32>}%1/*]zip{s*}%16/zip{s*n}%

Answer 8

Python 3.4, 216 Bytes

Also valid Python 2.7. Used FryAmTheEggman's idea/suggestion about curses.ascii.controlnames, which saves almost 100 bytes.

import curses.ascii as a
o=''
w=4,4,2,2,2,3,3,1
for x in w:o+='Dec Hex'+' '*x
o+=' '
for n in range(16):
 o+='\n'
 for x in w:o+='%3d %02X %-*s'%(n,n,x,n>31and[chr(n),'DEL'][n>126]or a.controlnames[n]);n+=16
print(o)

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$ python ascii.py | md5
58824a1dd7264c0410eb4d727aec54e1

Answer 9

Ruby (2.2.2p95), 277 295 306 331 364

a='Dec Hex  ';puts"#{a}  "*2+a*4+" #{a}"*2,16.times.map{|i|8.times.map{|j|(k=i+j*16;k==127?'  127 7F DEL':"#{k} #{'%.2X'%k} #{'NULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US  '[k*3..k*3+2]||k.chr}").rjust(j<2?10+j :9+j/6)}.join}

ungolfed

s = "Dec Hex    " * 2 + "Dec Hex  " * 4 + " Dec Hex  " * 2
a = 127.times.map { |i|
  "#{i} #{'%.2X'%i} #{'NULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US  '[i*3..i*3+2]||i.chr}"
}
a << '  127 7F DEL'
x = 16.times.map { |i|
  8.times.map { |j|
    a[i + j * 16].rjust(j < 2 ? 10 + j : 9 + j / 6)
  }.join
}.join "\n"
puts s, x

Answer 10

///, 998 bytes

//\/\////  /Dec Hex/ 0/ 1/ 2/ 3/ 4/ 5/ 6	/ 7/ 8/ 9/0 /1 /2 /3 /4 /5 /6 /7 /8 /9 /A /B /C /D /E /F /
 /  NUL 6DL280 4@ 0P6` 12	pSOH 7DC3! 91 51Q7a 13	qSTX 8DC4" 02 62R8b 14	rETX 9DC5# 13 73S9c 15	sEOT 0DC6$ 24 84T 00d 16	tENQ 1NAK 7% 35 95U 01e 17	uACK 2SYN 8& 46 	06V 02f 18	v	BEL 3ET9' 57	 	1G 7W 03g 19	wBS4CAN 0( 68 	2H 8X 04h 20	xHT5EM1) 79 	3I 9Y 05i 21	y
0L 6SU2* 8: 	4J 0Z 06j 22	z
1VT7ES3+ 9; 	5K 1[ 07k 23	{
2F 8FS4, 0< 	6L 2\\ 08l 24	|
3CR9GS5- 1= 	7M 3] 09m 25	}
4SO0RS6. 2> 	8N 4^ 10n 26	~
5SI1US7\/ 3? 	9O 5_ 11o 27	DEL

Try it online!

Answer 11

PHP, 330 321 bytes

$a=[NUL,SOH,STX,ETX,EOT,ENQ,ACK,BEL,BS,HT,LF,VT,FF,CR,SO,SI,DLE,DC1,DC2,DC3,DC4,NAK,SYN,ETB,CAN,EM,SUB,ESC,FS,GS,RS,US,127=>DEL];foreach($t=[3,3,1,1,1,2,2,0]as$i)echo'Dec Hex ',str_repeat(' ',$i);echo" 
";for($x=0;$x<16;)for($y=$x++;$y<128;$y+=16)printf("% 3d %1$02X % -".$t[$y/16].s.($y<112?' ':"
"),$y,$a[$y]?:chr($y));

Try it online!

Less golfy:

$a=[NUL,SOH,STX,ETX,EOT,ENQ,ACK,BEL,BS,HT,LF,VT,FF,CR,SO,SI,DLE,
DC1,DC2,DC3,DC4,NAK,SYN,ETB,CAN,EM,SUB,ESC,FS,GS,RS,US,127=>DEL];

$t=[3,3,1,1,1,2,2,0];

foreach( $t as $i )
    echo 'Dec Hex ', str_repeat( ' ', $i );

echo" \n";

for( $x=0; $x<16; ) {
    for( $y=$x++; $y<128; $y+=16 ) {
        printf( "% 3d %1$02X % -" . $t[$y/16] . 's' . ( $y<112 ?' ' : "\n" ),
            $y, $a[$y] ?: chr($y) );
    }
}

Answer 12

Microscript II, 1314 bytes

Probably far from optimal.

"Dec Hex    "pp"Dec Hex  "ppps"Dec Hex   "ppoP"  0 00 NUL  16 10 DLE  32 20    48 30 0  64 40 @  80 50 P   96 60 `  112 70 p\n  1 01 SOH  17 11 DC1  33 21 !  49 31 1  65 41 A  81 51 Q   97 61 a  113 71 q\n  2 02 STX  18 12 DC2  34 22 \"  50 32 2  66 42 B  82 52 R   98 62 b  114 72 r\n  3 03 ETX  19 13 DC3  35 23 #  51 33 3  67 43 C  83 53 S   99 63 c  115 73 s\n  4 04 EOT  20 14 DC4  36 24 $  52 34 4  68 44 D  84 54 T  100 64 d  116 74 t\n  5 05 ENQ  21 15 NAK  37 25 %  53 35 5  69 45 E  85 55 U  101 65 e  117 75 u\n  6 06 ACK  22 16 SYN  38 26 &  54 36 6  70 46 F  86 56 V  102 66 f  118 76 v\n  7 07 BEL  23 17 ETB  39 27 '  55 37 7  71 47 G  87 57 W  103 67 g  119 77 w\n  8 08 BS   24 18 CAN  40 28 (  56 38 8  72 48 H  88 58 X  104 68 h  120 78 x\n  9 09 HT   25 19 EM   41 29 )  57 39 9  73 49 I  89 59 Y  105 69 i  121 79 y\n 10 0A LF   26 1A SUB  42 2A *  58 3A :  74 4A J  90 5A Z  106 6A j  122 7A z\n 11 0B VT   27 1B ESC  43 2B +  59 3B ;  75 4B K  91 5B [  107 6B k  123 7B {\n 12 0C FF   28 1C FS   44 2C ,  60 3C <  76 4C L  92 5C \\  108 6C l  124 7C |\n 13 0D CR   29 1D GS   45 2D -  61 3D =  77 4D M  93 5D ]  109 6D m  125 7D }\n 14 0E SO   30 1E RS   46 2E .  62 3E >  78 4E N  94 5E ^  110 6E n  126 7E ~\n 15 0F SI   31 1F US   47 2F /  63 3F ?  79 4F O  95 5F _  111 6F o  127 7F DEL"

Answer 13

JavaScript, 415 413 423¹ 411 406 402 414² 412 bytes

x=>eval('a=`${d="Dec Hex  "}  `[r="repeat"](2)+d[r](3)+`${d} `[r](2)+d+`
`;z="NULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US ".match(/.{3}/g);z[127]="DEL";for(j=i=0;i<128;i++){a+=("  "+((b=i%8*16)+j)).slice(-3)+" "+(0+(c=b+j).toString(16).toUpperCase()).slice(-2)+" "+(c<32||i==127?z[c]:String.fromCharCode(c))+(b==112?`
`:(b < 80 ? " " : "  "));if(b==112)j++}a')

I couldn't figure out how to print the chars prior to char code 32, so just listed them as a string.

The hash I've got seems to match (41b6ecde6a3a1324be4836871d8354fe).

Demo + Ungolfed:

function t() {
  a = `${d="Dec Hex  "}  `.repeat(2) + d.repeat(3) + `${d} `.repeat(2) + d + "\n";
  z = "NULSOHSTXETXEOTENQACKBELBS HT LF VT FF CR SO SI DLEDC1DC2DC3DC4NAKSYNETBCANEM SUBESCFS GS RS US ".match(/.{3}/g);
  z[127] = "DEL";
  for (j = i = 0; i < 128; i++) {
    a += ("  " + ((b = i % 8 * 16) + j)).slice(-3) + " " + (0 + (c = b + j).toString(16).toUpperCase()).slice(-2) + " " + (c < 32 || i == 127 ? z[c] : String.fromCharCode(c)) + (b==112?"\n":(b < 80 ? " " : "  "));
    if(b==112)j++;
  }
  return a
}
document.write("<pre>" + t() + "</pre>")

---

^{1 - fixed spacing
2 - again fixed spacing}

Answer 14

MATLAB, 363 bytes

Not as small as C but comparable...

h='';s=@sprintf;for i=[4 4 2 2 2 3 3 2]h=s([h 'Dec Hex%*s'],i,'');end
h=[h 10];a='NULDLESOHDC1STXDC2ETXDC3EOTDC4ENQNAKACKSYNBELETBBS CANHT EM LF SUBVT ESCFF FS CR GS SO RS SI US ';for i=0:15for j=0:7k=i+16*j;if j<2b=1+6*i+3*j;h=[h s('%3d %02X %-3s ',k,k,a(b:b+2))];else
h=[h s('%*d %02X %c ',3+(j>5),k,k,k)];end
end
h=[h(1:end-1) 10];end
disp([h(1:end-2) 'DEL']);

Answer 15

Common Lisp (SBCL), 309

(progn(format t"~{~v,0TDec Hex  ~}"#1='(0 11 22 31 40 49 59 69))(dotimes(y 16)(fresh-line)(loop for x from 0 for k in #1#for i =(+(* 16 x)y)do(format t"~v,0T~3d ~2,'0x ~a"k i i(if(graphic-char-p #2=(code-char i))#2#(case i(8"BS")(9"HT")(10"LF")(12"FF")(13"CR")(127"DEL")(t(string-upcase(char-name #2#)))))))))

A script to be run which outputs the following to standard output, ie. the version without a final newline (md5: 41b6ecde6a3a1324be4836871d8354fe).

Dec Hex    Dec Hex    Dec Hex  Dec Hex  Dec Hex  Dec Hex   Dec Hex   Dec Hex  
  0 00 NUL  16 10 DLE  32 20    48 30 0  64 40 @  80 50 P   96 60 `  112 70 p
  1 01 SOH  17 11 DC1  33 21 !  49 31 1  65 41 A  81 51 Q   97 61 a  113 71 q
  2 02 STX  18 12 DC2  34 22 "  50 32 2  66 42 B  82 52 R   98 62 b  114 72 r
  3 03 ETX  19 13 DC3  35 23 #  51 33 3  67 43 C  83 53 S   99 63 c  115 73 s
  4 04 EOT  20 14 DC4  36 24 $  52 34 4  68 44 D  84 54 T  100 64 d  116 74 t
  5 05 ENQ  21 15 NAK  37 25 %  53 35 5  69 45 E  85 55 U  101 65 e  117 75 u
  6 06 ACK  22 16 SYN  38 26 &  54 36 6  70 46 F  86 56 V  102 66 f  118 76 v
  7 07 BEL  23 17 ETB  39 27 '  55 37 7  71 47 G  87 57 W  103 67 g  119 77 w
  8 08 BS   24 18 CAN  40 28 (  56 38 8  72 48 H  88 58 X  104 68 h  120 78 x
  9 09 HT   25 19 EM   41 29 )  57 39 9  73 49 I  89 59 Y  105 69 i  121 79 y
 10 0A LF   26 1A SUB  42 2A *  58 3A :  74 4A J  90 5A Z  106 6A j  122 7A z
 11 0B VT   27 1B ESC  43 2B +  59 3B ;  75 4B K  91 5B [  107 6B k  123 7B {
 12 0C FF   28 1C FS   44 2C ,  60 3C <  76 4C L  92 5C \  108 6C l  124 7C |
 13 0D CR   29 1D GS   45 2D -  61 3D =  77 4D M  93 5D ]  109 6D m  125 7D }
 14 0E SO   30 1E RS   46 2E .  62 3E >  78 4E N  94 5E ^  110 6E n  126 7E ~
 15 0F SI   31 1F US   47 2F /  63 3F ?  79 4F O  95 5F _  111 6F o  127 7F DEL

Details

• (0 11 22 31 40 49 59 69) contains the position of each column in a line, as obtained by inspecting the desired output. There were many corner cases that are simply better handled by hard-coding the exact position for each column and letting the ~T format directive tabulate as needed.

• Since names of characters are implementation-dependent, this code targets only the SBCL interpreter (I tested with CCL, which uses other names). I rely on char-name for most non-graphic characters, but some of them have explicit mappings to the expected output.

Ungolfed

(progn
  (format t"~{~v,0TDec Hex  ~}" '(0 11 22 31 40 49 59 69))
  (dotimes(y 16)
    (fresh-line)
    (loop for x from 0
          for k in '(0 11 22 31 40 49 59 69)
          for i = (+ (* 16 x) y)
          for c = (code-char i)
          do (format t
                     "~v,0T~3d ~2,'0x ~a"
                     k
                     i
                     i
                     (if (graphic-char-p c)
                         c
                         (case i
                           (8"BS")
                           (9"HT")
                           (10"LF")
                           (12"FF")
                           (13"CR")
                           (127"DEL")
                           (t (string-upcase (char-name c)))))))))