ref: 0d86fe4b748e72fb8491a7e8bea7bb0a923b9f79
parent: 6fb890e0ea20a3c366ffd2de45d26a0c1c454dd6
author: Simon Tatham <anakin@pobox.com>
date: Sun Apr 16 04:44:33 EDT 2023
Move obfuscator tests into obfusc.c. I just found these self-tests lying around in mines.c under an #ifdef that nobody ever enables. Let's put them somewhere more sensible! We already have a separate tool for working with the obfuscation system in a puzzle-independent way, and it seems reasonable to put them in there.
--- a/auxiliary/obfusc.c
+++ b/auxiliary/obfusc.c
@@ -30,6 +30,75 @@
#include "puzzles.h"
+static bool self_tests(void)
+{+ bool ok = true;
+
+ #define FAILED (ok = false, "failed")
+
+ /*
+ * A few simple test vectors for the obfuscator.
+ *
+ * First test: the 28-bit stream 1234567. This divides up
+ * into 1234 and 567[0]. The SHA of 56 70 30 (appending
+ * "0") is 15ce8ab946640340bbb99f3f48fd2c45d1a31d30. Thus,
+ * we XOR the 16-bit string 15CE into the input 1234 to get
+ * 07FA. Next, we SHA that with "0": the SHA of 07 FA 30 is
+ * 3370135c5e3da4fed937adc004a79533962b6391. So we XOR the
+ * 12-bit string 337 into the input 567 to get 650. Thus
+ * our output is 07FA650.
+ */
+ {+ unsigned char bmp1[] = "\x12\x34\x56\x70";
+ obfuscate_bitmap(bmp1, 28, false);
+ printf("test 1 encode: %s\n",+ memcmp(bmp1, "\x07\xfa\x65\x00", 4) ? FAILED : "passed");
+ obfuscate_bitmap(bmp1, 28, true);
+ printf("test 1 decode: %s\n",+ memcmp(bmp1, "\x12\x34\x56\x70", 4) ? FAILED : "passed");
+ }
+ /*
+ * Second test: a long string to make sure we switch from
+ * one SHA to the next correctly. My input string this time
+ * is simply fifty bytes of zeroes.
+ */
+ {+ unsigned char bmp2[50];
+ unsigned char bmp2a[50];
+ memset(bmp2, 0, 50);
+ memset(bmp2a, 0, 50);
+ obfuscate_bitmap(bmp2, 50 * 8, false);
+ /*
+ * SHA of twenty-five zero bytes plus "0" is
+ * b202c07b990c01f6ff2d544707f60e506019b671. SHA of
+ * twenty-five zero bytes plus "1" is
+ * fcb1d8b5a2f6b592fe6780b36aa9d65dd7aa6db9. Thus our
+ * first half becomes
+ * b202c07b990c01f6ff2d544707f60e506019b671fcb1d8b5a2.
+ *
+ * SHA of that lot plus "0" is
+ * 10b0af913db85d37ca27f52a9f78bba3a80030db. SHA of the
+ * same string plus "1" is
+ * 3d01d8df78e76d382b8106f480135a1bc751d725. So the
+ * second half becomes
+ * 10b0af913db85d37ca27f52a9f78bba3a80030db3d01d8df78.
+ */
+ printf("test 2 encode: %s\n",+ memcmp(bmp2, "\xb2\x02\xc0\x7b\x99\x0c\x01\xf6\xff\x2d\x54"
+ "\x47\x07\xf6\x0e\x50\x60\x19\xb6\x71\xfc\xb1\xd8"
+ "\xb5\xa2\x10\xb0\xaf\x91\x3d\xb8\x5d\x37\xca\x27"
+ "\xf5\x2a\x9f\x78\xbb\xa3\xa8\x00\x30\xdb\x3d\x01"
+ "\xd8\xdf\x78", 50) ? FAILED : "passed");
+ obfuscate_bitmap(bmp2, 50 * 8, true);
+ printf("test 2 decode: %s\n",+ memcmp(bmp2, bmp2a, 50) ? FAILED : "passed");
+ }
+
+ #undef FAILED
+
+ return ok;
+}
+
int main(int argc, char **argv)
{ enum { BINARY, DEFAULT, HEX } outputmode = DEFAULT;@@ -36,7 +105,7 @@
char *inhex = NULL;
unsigned char *data;
int datalen;
- enum { UNKNOWN, DECODE, ENCODE } mode = UNKNOWN;+ enum { UNKNOWN, DECODE, ENCODE, SELFTEST } mode = UNKNOWN;bool doing_opts = true;
while (--argc > 0) {@@ -56,6 +125,9 @@
case 'd':
mode = DECODE;
break;
+ case 't':
+ mode = SELFTEST;
+ break;
case 'b':
outputmode = BINARY;
break;
@@ -81,7 +153,12 @@
if (mode == UNKNOWN) {fprintf(stderr, "usage: obfusc < -e | -d > [ -b | -h ] [hex data]\n");
+ fprintf(stderr, " or: obfusc -t to run self-tests\n");
return 0;
+ }
+
+ if (mode == SELFTEST) {+ return self_tests() ? 0 : 1;
}
if (outputmode == DEFAULT)
--- a/mines.c
+++ b/mines.c
@@ -1902,72 +1902,7 @@
static bool *new_mine_layout(int w, int h, int n, int x, int y, bool unique,
random_state *rs, char **game_desc)
{- bool *grid;
-
-#ifdef TEST_OBFUSCATION
- static int tested_obfuscation = false;
- if (!tested_obfuscation) {- /*
- * A few simple test vectors for the obfuscator.
- *
- * First test: the 28-bit stream 1234567. This divides up
- * into 1234 and 567[0]. The SHA of 56 70 30 (appending
- * "0") is 15ce8ab946640340bbb99f3f48fd2c45d1a31d30. Thus,
- * we XOR the 16-bit string 15CE into the input 1234 to get
- * 07FA. Next, we SHA that with "0": the SHA of 07 FA 30 is
- * 3370135c5e3da4fed937adc004a79533962b6391. So we XOR the
- * 12-bit string 337 into the input 567 to get 650. Thus
- * our output is 07FA650.
- */
- {- unsigned char bmp1[] = "\x12\x34\x56\x70";
- obfuscate_bitmap(bmp1, 28, false);
- printf("test 1 encode: %s\n",- memcmp(bmp1, "\x07\xfa\x65\x00", 4) ? "failed" : "passed");
- obfuscate_bitmap(bmp1, 28, true);
- printf("test 1 decode: %s\n",- memcmp(bmp1, "\x12\x34\x56\x70", 4) ? "failed" : "passed");
- }
- /*
- * Second test: a long string to make sure we switch from
- * one SHA to the next correctly. My input string this time
- * is simply fifty bytes of zeroes.
- */
- {- unsigned char bmp2[50];
- unsigned char bmp2a[50];
- memset(bmp2, 0, 50);
- memset(bmp2a, 0, 50);
- obfuscate_bitmap(bmp2, 50 * 8, false);
- /*
- * SHA of twenty-five zero bytes plus "0" is
- * b202c07b990c01f6ff2d544707f60e506019b671. SHA of
- * twenty-five zero bytes plus "1" is
- * fcb1d8b5a2f6b592fe6780b36aa9d65dd7aa6db9. Thus our
- * first half becomes
- * b202c07b990c01f6ff2d544707f60e506019b671fcb1d8b5a2.
- *
- * SHA of that lot plus "0" is
- * 10b0af913db85d37ca27f52a9f78bba3a80030db. SHA of the
- * same string plus "1" is
- * 3d01d8df78e76d382b8106f480135a1bc751d725. So the
- * second half becomes
- * 10b0af913db85d37ca27f52a9f78bba3a80030db3d01d8df78.
- */
- printf("test 2 encode: %s\n",- memcmp(bmp2, "\xb2\x02\xc0\x7b\x99\x0c\x01\xf6\xff\x2d\x54"
- "\x47\x07\xf6\x0e\x50\x60\x19\xb6\x71\xfc\xb1\xd8"
- "\xb5\xa2\x10\xb0\xaf\x91\x3d\xb8\x5d\x37\xca\x27"
- "\xf5\x2a\x9f\x78\xbb\xa3\xa8\x00\x30\xdb\x3d\x01"
- "\xd8\xdf\x78", 50) ? "failed" : "passed");
- obfuscate_bitmap(bmp2, 50 * 8, true);
- printf("test 2 decode: %s\n",- memcmp(bmp2, bmp2a, 50) ? "failed" : "passed");
- }
- }
-#endif
-
- grid = minegen(w, h, n, x, y, unique, rs);
+ bool *grid = minegen(w, h, n, x, y, unique, rs);
if (game_desc)
*game_desc = describe_layout(grid, w * h, x, y, true);