uudecode this string: H,3DR-S<-"C(X,C<U#0HT.3(V-@T*-C`X,C8-"C(X,S8V#0HR,#<Q-0``
uudecode this string: H,3DR-S<-"C(X,C<U#0HT.3(V-@T*-C`X,C8-"C(X,S8V#0HR,#<Q-0``
When you “uuencode†a file a header line is first written (begin …), then one or more encoded lines, and finally an “end†line. In this case you’ve been given a single encoded line:   H,3DR-S<-"C(X,C<U#0HT.3(V-@T*-C`X,C8-"C(X,S8V#0HR,#<Q-0`` Because the very first character (an “Hâ€) encodes the expected number of decoded bytes in that line (the value is computed by taking its ASCII code minus 32), we can work out that H means:   ASCII("H") = 72, and 72 – 32 = 40. That is, when decoded the rest of the line will yield 40 bytes (characters). The remainder of the line (56 characters) is broken into 14 groups of 4 characters each. (Uuencode always maps 3 bytes → 4 printable characters – with the final group padded if needed; here 14×3 = 42 bytes are “produced†but only the first 40 are considered.) Each character in a group represents a 6–bit value. (Standard uuencode subtracts 32 from the ASCII code; note that a backtick “`†is used in place of a space to represent 0.) Below is a step–by–step (manual) decoding of each of the 14 groups. In each group we label the four characters a, b, c, d. Their numerical value is computed as:   value = (ASCII code of character) – 32, except that because uuencode works modulo 64 a “`†(whose code 96 gives 96 – 32 = 64) is taken as 0 (since 64 mod 64 = 0). Then the 3 decoded bytes come from:   byte1 = (a << 2) | (b >> 4)   byte2 = ((b & 0x0F) << 4) | (c >> 2)   byte3 = ((c & 0x03) << 6) | d For our input the groups are: Group Characters  Values (a,b,c,d)  Calculation  Decoded bytes (in decimal – then ASCII) 1.  “ , 3 D R †  a = “,†→ 44–32 = 12   b = “3†→ 51–32 = 19   c = “D†→ 68–32 = 36   d = “R†→ 82–32 = 50   byte1 = (12<<2) | (19>>4) = 48 | 1 = 49  (‘1’)   byte2 = ((19&15)<<4) | (36>>2) = (3<<4) | 9 = 48+9 = 57  (‘9’)   byte3 = ((36&3)<<6) | 50 = (0<<6) | 50 = 50  (‘2’)   → “192†2.  “- S < -†  a = “-†→ 45–32 = 13   b = “S†→ 83–32 = 51   c = “<†→ 60–32 = 28   d = “-†→ 13   byte1 = (13<<2) | (51>>4) = 52 | 3 = 55  (‘7’)   byte2 = ((51&15)<<4) | (28>>2) = (3<<4) | 7 = 48+7 = 55  (‘7’)   byte3 = ((28&3)<<6) | 13 = (0<<6) | 13 = 13  (ASCII 13, a carriageâ€return)   → “77†+ CR 3.  “" C ( X†  a = “"†→ 34–32 = 2   b = “C†→ 67–32 = 35   c = “(†→ 40–32 = 8   d = “X†→ 88–32 = 56   byte1 = (2<<2) | (35>>4) = 8 | 2 = 10  (ASCII 10, a line–feed)   byte2 = ((35&15)<<4) | (8>>2) = (3<<4) | 2 = 48+2 = 50  (‘2’)   byte3 = ((8&3)<<6) | 56 = (0<<6) | 56 = 56  (‘8’)   → LF, “28†4.  “, C < U†  a = “,†→ 12   b = “C†→ 35   c = “<†→ 28   d = “U†→ 85–32 = 53   byte1 = (12<<2) | (35>>4) = 48 | 2 = 50  (‘2’)   byte2 = ((35&15)<<4) | (28>>2) = (3<<4) | 7 = 48+7 = 55  (‘7’)   byte3 = ((28&3)<<6) | 53 = (0<<6) | 53 = 53  (‘5’)   → “275†5.  “# 0 H T†  a = “#†→ 35–32 = 3   b = “0†→ 48–32 = 16   c = “H†→ 72–32 = 40   d = “T†→ 84–32 = 52   byte1 = (3<<2) | (16>>4) = 12 | 1 = 13  (CR)   byte2 = ((16&15)<<4) | (40>>2) = (0<<4) | 10 = 10  (LF)   byte3 = ((40&3)<<6) | 52 = (0<<6) | 52 = 52  (‘4’)   → CR, LF, “4†6.  “. 3 ( V†  a = “.†→ 46–32 = 14   b = “3†→ 19   c = “(†→ 8   d = “V†→ 86–32 = 54   byte1 = (14<<2) | (19>>4) = 56 | 1 = 57  (‘9’)   byte2 = ((19&15)<<4) | (8>>2) = (3<<4) | 2 = 48+2 = 50  (‘2’)   byte3 = ((8&3)<<6) | 54 = (0<<6) | 54 = 54  (‘6’)   → “926†7.  “- @ T *†  a = “-†→ 13   b = “@†→ 64–32 = 32   c = “T†→ 52   d = “*†→ 42–32 = 10   byte1 = (13<<2) | (32>>4) = 52 | 2 = 54  (‘6’)   byte2 = ((32&15)<<4) | (52>>2) = (0<<4) | 13 = 13  (CR)   byte3 = ((52&3)<<6) | 10 = (0<<6) | 10 = 10  (LF)   → “6†+ CR + LF 8.  “- C ` X†  a = “-†→ 13   b = “C†→ 35   c = “`†→ (backtick, so 96–32 = 64 which is taken as 0)   d = “X†→ 56   byte1 = (13<<2) | (35>>4) = 52 | 2 = 54  (‘6’)   byte2 = ((35&15)<<4) | (0>>2) = (3<<4) | 0 = 48  (‘0’)   byte3 = ((0&3)<<6) | 56 = 0 | 56 = 56  (‘8’)   → “608†9.  “, C 8 -†  a = “,†→ 12   b = “C†→ 35   c = “8†→ 56–32 = 24   d = “-†→ 13   byte1 = (12<<2) | (35>>4) = 48 | 2 = 50  (‘2’)   byte2 = ((35&15)<<4) | (24>>2) = (3<<4) | 6 = 48+6 = 54  (‘6’)   byte3 = ((24&3)<<6) | 13 = (0<<6) | 13 = 13  (CR)   → “26†+ CR 10.  “" C ( X†  (Exactly the same as group 3.)   → LF, “28†11.  “, S 8 V†  a = “,†→ 12   b = “S†→ 51   c = “8†→ 24   d = “V†→ 86–32 = 54   byte1 = (12<<2) | (51>>4) = 48 | 3 = 51  (‘3’)   byte2 = ((51&15)<<4) | (24>>2) = (3<<4) | 6 = 48+6 = 54  (‘6’)   byte3 = ((24&3)<<6) | 54 = (0<<6) | 54 = 54  (‘6’)   → “366†12.  “# 0 H R†  a = “#†→ 3   b = “0†→ 16   c = “H†→ 40   d = “R†→ 50   byte1 = (3<<2) | (16>>4) = 12 | 1 = 13  (CR)   byte2 = ((16&15)<<4) | (40>>2) = 0 | 10 = 10  (LF)   byte3 = ((40&3)<<6) | 50 = 0 | 50 = 50  (‘2’)   → CR, LF, “2†13.  “, # < Q†  a = “,†→ 12   b = “#†→ 3   c = “<†→ 28   d = “Q†→ 81–32 = 49   byte1 = (12<<2) | (3>>4) = 48 | 0 = 48  (‘0’)   byte2 = ((3&15)<<4) | (28>>2) = (3<<4) | 7 = 48+7 = 55  (‘7’)   byte3 = ((28&3)<<6) | 49 = (0<<6) | 49 = 49  (‘1’)   → “071†14.  “ - 0 ` `†  a = “-†→ 13   b = “0†→ 16   c = “`†→ 0   d = “`†→ 0   byte1 = (13<<2) | (16>>4) = 52 | 1 = 53  (‘5’)   byte2 = ((16&15)<<4) | (0>>2) = 0   byte3 = ((0&3)<<6) | 0 = 0   Since the line’s header told us to output only 40 bytes and we already have 39 from groups 1–13, only the first byte from group 14 is used (the extra decoded bytes are simply padding).   → “5†Now, if we write out all the (printable) decoded bytes we get a 40–byte file. When you interpret the CR (ASCII 13) and LF (ASCII 10) in the usual “CRâ€LF†combination as a newline, you get a text file consisting of six 5–digit numbers. Reconstructing in order, the decoded text is:   Line1: “19277†  Line2: “28275†  Line3: “49266†  Line4: “60826†  Line5: “28366†  Line6: “20715†Thus, uudecoding the given string yields:   19277   28275   49266   60826   28366   20715 You can check this yourself with a uuencode/uuencode–decoder tool by “wrapping†the data with a “begin …†header and “end†trailer.
Testing if the previous output contains the string 19277: True
Testing if the previous output contains the string 20715: True
Testing if the previous output contains the string 60826: True
uudecode this string: H,3DR-S<-"C(X,C<U#0HT.3(V-@T*-C`X,C8-"C(X,S8V#0HR,#<Q-0``
uudecode this string: H,3DR-S<-"C(X,C<U#0HT.3(V-@T*-C`X,C8-"C(X,S8V#0HR,#<Q-0``
Below is a stepâ€byâ€step explanation of one way to “uudecode†the input line. (Note that uuencoded data is “selfâ€describing:†the very first character tells you how many bytes are represented on that line. In this case the first character is “H†– since H’s ASCII code is 72 and 72–32 = 40, the line decodes to 40 bytes.) The given encoded string is:   H,3DR-S<-"C(X,C<U#0HT.3(V-@T*-C`X,C8-"C(X,S8V#0HR,#<Q-0`` After the “H†(the length marker) the remaining 56 characters represent 14 groups of 4 encoded characters. Each group of 4 characters decodes into 3 bytes. (In the final group only part of the 3 bytes are “real†because 14×3 = 42 bytes but the header told us that only 40 bytes are in the file.) The basic decoding rule is to subtract 32 from the ASCII code of every character (interpreting the result modulo 64), then for every group of four 6â€bit numbers (call them a, b, c, d) recover three bytes as follows:   byte1 = (a << 2) | (b >> 4)   byte2 = ((b & 0x0F) << 4) | (c >> 2)   byte3 = ((c & 0x03) << 6) | d When you work through the 14 groups you recover these decoded bytes (showing control characters “CR†for carriageâ€return – ASCII 13 – and “LF†for lineâ€feed – ASCII 10): Group Decoded bytes (in characters) 1. “192†2. “77†followed by CR 3. LF, “28†4. “275†5. CR, LF, “4†6. “926†7. “6†then CR, LF 8. “608†9. “26†then CR 10. LF, “28†11. “366†12. CR, LF, “2†13. “071†14. “5†plus two extra bytes [discarded because only 40 bytes are used] If you write the bytes in order (interpreting CR+LF as a newline) the decoded file is made up of six lines (each line being 5 printable characters):   19277   28275   49266   60826   28366   20715 So, the uudecoded output of the given string is exactly these six lines: ------------------------------------------------ 19277 28275 49266 60826 28366 20715 ------------------------------------------------ You can check this result using a standard uudecode tool (by “wrapping†the above into a complete uuencoded file with a “begin …†header, then running uudecode).
Testing if the previous output contains the string 19277: True
Testing if the previous output contains the string 20715: True
Testing if the previous output contains the string 60826: True