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Text File  |  1992-05-06  |  2.4 KB  |  79 lines
  1. a      = 
  2. b      = 1000000001
  3. c      = 10101010101010101010
  4. d      = 00110011001100110011
  5. e      = 11110000111100001111
  6. f = b  = 1000000001
  7. g = ~e = 00001111000011110000
  8. h = d  = 00110011001100110011
  9. bits in e:
  10. 0 1 2 3 8 9 10 11 16 17 18 19 
  11. clear bits in g (reverse order):
  12. 19 18 17 16 11 10 9 8 3 2 1 0 
  13. ~c     = 01010101010101010101
  14. c & d  = 00100010001000100010
  15. c | d  = 10111011101110111011
  16. c - d  = 10001000100010001000
  17. c ^ d  = 10011001100110011001
  18. c + d  = 1010101010101010101000110011001100110011
  19. c <<2  = 0010101010101010101010
  20. c >>2  = 101010101010101010
  21. f &= c = 1000000000
  22. f |= d = 10110011001100110011
  23. f -= e = 10000000000000000000
  24. f ^= c = 00101010101010101010
  25. f += b = 001010101010101010101000000001
  26. f <<=5 = 00000001010101010101010101000000001
  27. f >>=10= 0101010101010101000000001
  28. l = 101010101010101010100011001100110011001110101010101010101010
  29. BitPattern pat = 0011XXXX0011XXXX0011
  30. pat.pattern    = 00110011001100110011
  31. pat.mask       = 11110000111100001111
  32. l.index(pat)   = 20
  33. l.index(pat,-1)= 20
  34. l.before(pat)  = 10101010101010101010
  35. l.at(pat)      = 00110011001100110011
  36. l.after(pat)   = 10101010101010101010
  37. b.set(0)       :1000000001
  38. b.set(65):
  39. 100000000100000000000000000000000000000000000000000000000000000001
  40. b.clear(2):
  41. 100000000100000000000000000000000000000000000000000000000000000001
  42. b.invert(11,20):
  43. 100000000100111111111000000000000000000000000000000000000000000001
  44. b.set(21,30):
  45. 100000000100111111111111111111100000000000000000000000000000000001
  46. b.clear(31, 40):
  47. 100000000100111111111111111111100000000000000000000000000000000001
  48. b.set(0)       :10110011001100110011
  49. b.set(65):
  50. 101100110011001100110000000000000000000000000000000000000000000001
  51. b.clear(2):
  52. 100100110011001100110000000000000000000000000000000000000000000001
  53. b.invert(11,20):
  54. 100100110010110011001000000000000000000000000000000000000000000001
  55. b.set(21,30):
  56. 100100110010110011001111111111100000000000000000000000000000000001
  57. b.clear(31, 40):
  58. 100100110010110011001111111111100000000000000000000000000000000001
  59. k = 0101
  60. c.before(k) = 1
  61. c.at(k)     = 0101
  62. c.after(k)  = 010101010101010
  63. c.after(k)=k :101010101
  64. c.before(k)=k:010101010101
  65. reverse(k)           = 1010
  66. k.left_trim(0)       : 101
  67. k.right_trim(1)      : 10
  68. k = 0110
  69. c.before(k) = 0
  70. c.at(k)     = 0110
  71. c.after(k)  = 011001100110011
  72. c.after(k)=k :001100110
  73. c.before(k)=k:011001100110
  74. reverse(k)           = 0110
  75. k.left_trim(0)       : 110
  76. k.right_trim(1)      : 110
  77.  
  78. End of test.
  79.