gcc/testsuite/gdc.test/runnable/pi.d

   1 // PERMUTE_ARGS:
   2 // EXECUTE_ARGS: 1000
   3
   4 import core.stdc.stdio;
   5 import core.stdc.time;
   6
   7 const int LONG_TIME=4000;
   8
   9 byte[] p;
  10 byte[] t;
  11 int q;
  12
  13 int main(char[][] args)
  14 {
  15         time_t startime, endtime;
  16         int i;
  17
  18         if (args.length == 2) {
  19                 sscanf(&args[1][0],"%d",&q);
  20         } else {
  21                 printf("Usage: pi [precision]\n");
  22                 return 1;
  23         }
  24
  25         if (q < 0)
  26         {
  27                 printf("Precision was too low, running with precision of 0.\n");
  28                 q = 0;
  29         }
  30
  31         if (q > LONG_TIME)
  32         {
  33             printf("Be prepared to wait a while...\n");
  34         }
  35
  36         // Compute one more digit than we display to compensate for rounding
  37         q++;
  38
  39         p.length = q + 1;
  40         t.length = q + 1;
  41
  42         /* compute pi */
  43
  44         time(&startime);
  45         arctan(2);
  46         arctan(3);
  47         mul4();
  48         time(&endtime);
  49
  50         // Return to the number of digits we want to display
  51         q--;
  52
  53         /* print pi */
  54
  55         printf("pi = %d.",cast(int)(p[0]));
  56         for (i = 1; i <= q; i++)
  57         printf("%d",cast(int)(p[i]));
  58         printf("\n");
  59         printf("%lld seconds to compute pi with a precision of %d digits.\n", cast(long)(endtime-startime),q);
  60
  61         return 0;
  62 }
  63
  64 void arctan(int s)
  65 {
  66         int n;
  67
  68         t[0] = 1;
  69         div(s); /* t[] = 1/s */
  70         add();
  71         n = 1;
  72         do {
  73                 mul(n);
  74                 div(s * s);
  75                 div(n += 2);
  76                 if (((n-1) / 2) % 2 == 0)
  77                         add();
  78                 else
  79                         sub();
  80         } while (!tiszero());
  81 }
  82
  83 void add()
  84 {
  85         int j;
  86
  87         for (j = q; j >= 0; j--)
  88         {
  89                 if (t[j] + p[j] > 9) {
  90                         p[j] += t[j] - 10;
  91                         p[j-1] += 1;
  92                 } else
  93                         p[j] += t[j];
  94         }
  95 }
  96
  97 void sub()
  98 {
  99         int j;
 100
 101         for (j = q; j >= 0; j--)
 102                 if (p[j] < t[j]) {
 103                         p[j] -= t[j] - 10;
 104                         p[j-1] -= 1;
 105                 } else
 106                         p[j] -= t[j];
 107 }
 108
 109 void mul(int multiplier)
 110 {
 111         int b;
 112         int i;
 113         int carry = 0, digit = 0;
 114
 115         for (i = q; i >= 0; i--) {
 116                 b = (t[i] * multiplier + carry);
 117                 digit = b % 10;
 118                 carry = b / 10;
 119                 t[i] = cast(byte)digit;
 120         }
 121 }
 122
 123 /* t[] /= l */
 124
 125 void div(int divisor)
 126 {
 127         int i, b;
 128         int quotient, remainder = 0;
 129
 130         foreach (ref x; t)
 131         {
 132                 b = (10 * remainder + x);
 133                 quotient = b / divisor;
 134                 remainder = b % divisor;
 135                 x = cast(byte)quotient;
 136         }
 137 }
 138
 139 void div4()
 140 {
 141         int i, c, d = 0;
 142
 143         for (i = 0; i <= q; i++) {
 144                 c = (10 * d + p[i]) / 4;
 145                 d = (10 * d + p[i]) % 4;
 146                 p[i] = cast(byte)c;
 147         }
 148 }
 149
 150 void mul4()
 151 {
 152         int i, c, d;
 153
 154         d = c = 0;
 155
 156         for (i = q; i >= 0; i--) {
 157                 d = (p[i] * 4 + c) % 10;
 158                 c = (p[i] * 4 + c) / 10;
 159                 p[i] = cast(byte)d;
 160         }
 161 }
 162
 163 int tiszero()
 164 {
 165         int k;
 166
 167         for (k = 0; k <= q; k++)
 168                 if (t[k] != 0)
 169                         return false;
 170         return true;
 171 }
 172
 173
 174