latin1.c 1.4 KB

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  1. #include <u.h>
  2. /*
  3. * The code makes two assumptions: strlen(ld) is 1 or 2; latintab[i].ld can be a
  4. * prefix of latintab[j].ld only when j<i.
  5. */
  6. struct cvlist
  7. {
  8. char *ld; /* must be seen before using this conversion */
  9. char *si; /* options for last input characters */
  10. Rune *so; /* the corresponding Rune for each si entry */
  11. } latintab[] = {
  12. #include "../port/latin1.h"
  13. 0, 0, 0
  14. };
  15. /*
  16. * Given 5 characters k[0]..k[4], find the rune or return -1 for failure.
  17. */
  18. long
  19. unicode(Rune *k)
  20. {
  21. long i, c;
  22. k++; /* skip 'X' */
  23. c = 0;
  24. for(i=0; i<4; i++,k++){
  25. c <<= 4;
  26. if('0'<=*k && *k<='9')
  27. c += *k-'0';
  28. else if('a'<=*k && *k<='f')
  29. c += 10 + *k-'a';
  30. else if('A'<=*k && *k<='F')
  31. c += 10 + *k-'A';
  32. else
  33. return -1;
  34. }
  35. return c;
  36. }
  37. /*
  38. * Given n characters k[0]..k[n-1], find the corresponding rune or return -1 for
  39. * failure, or something < -1 if n is too small. In the latter case, the result
  40. * is minus the required n.
  41. */
  42. long
  43. latin1(Rune *k, int n)
  44. {
  45. struct cvlist *l;
  46. int c;
  47. char* p;
  48. if(k[0] == 'X')
  49. if(n>=5)
  50. return unicode(k);
  51. else
  52. return -5;
  53. for(l=latintab; l->ld!=0; l++)
  54. if(k[0] == l->ld[0]){
  55. if(n == 1)
  56. return -2;
  57. if(l->ld[1] == 0)
  58. c = k[1];
  59. else if(l->ld[1] != k[1])
  60. continue;
  61. else if(n == 2)
  62. return -3;
  63. else
  64. c = k[2];
  65. for(p=l->si; *p!=0; p++)
  66. if(*p == c)
  67. return l->so[p - l->si];
  68. return -1;
  69. }
  70. return -1;
  71. }