Neat idea from Gareth: if you put a % on the end of the mine count

[sgt/puzzles] / random.c

diff --git a/random.c b/random.c
index e0a179e..d70dd00 100644 (file)
--- a/random.c
+++ b/random.c
@@ -15,15 +15,6 @@
 
 #include "puzzles.h"
 
-typedef unsigned long uint32;
-
-typedef struct {
-    uint32 h[5];
-    unsigned char block[64];
-    int blkused;
-    uint32 lenhi, lenlo;
-} SHA_State;
-
 /* ----------------------------------------------------------------------
  * Core SHA algorithm: processes 16-word blocks into a message digest.
  */
@@ -108,14 +99,14 @@ static void SHATransform(uint32 * digest, uint32 * block)
  * the end, and pass those blocks to the core SHA algorithm.
  */
 
-static void SHA_Init(SHA_State * s)
+void SHA_Init(SHA_State * s)
 {
     SHA_Core_Init(s->h);
     s->blkused = 0;
     s->lenhi = s->lenlo = 0;
 }
 
-static void SHA_Bytes(SHA_State * s, void *p, int len)
+void SHA_Bytes(SHA_State * s, void *p, int len)
 {
     unsigned char *q = (unsigned char *) p;
     uint32 wordblock[16];
@@ -158,7 +149,7 @@ static void SHA_Bytes(SHA_State * s, void *p, int len)
     }
 }
 
-static void SHA_Final(SHA_State * s, unsigned char *output)
+void SHA_Final(SHA_State * s, unsigned char *output)
 {
     int i;
     int pad;
@@ -196,7 +187,7 @@ static void SHA_Final(SHA_State * s, unsigned char *output)
     }
 }
 
-static void SHA_Simple(void *p, int len, unsigned char *output)
+void SHA_Simple(void *p, int len, unsigned char *output)
 {
     SHA_State s;
 
@@ -231,7 +222,7 @@ random_state *random_init(char *seed, int len)
 
 unsigned long random_bits(random_state *state, int bits)
 {
-    int ret = 0;
+    unsigned long ret = 0;
     int n;
 
     for (n = 0; n < bits; n += 8) {
@@ -251,7 +242,13 @@ unsigned long random_bits(random_state *state, int bits)
        ret = (ret << 8) | state->databuf[state->pos++];
     }
 
-    ret &= (1 << bits) - 1;
+    /*
+     * `(1 << bits) - 1' is not good enough, since if bits==32 on a
+     * 32-bit machine, behaviour is undefined and Intel has a nasty
+     * habit of shifting left by zero instead. We'll shift by
+     * bits-1 and then separately shift by one.
+     */
+    ret &= (1 << (bits-1)) * 2 - 1;
     return ret;
 }