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Add direct tests of division/modulus to testbn.

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I'm about to rewrite the division code, so it'll be useful to have a
way to test it directly, particularly one which exercises difficult
cases such as extreme values of the leading word and remainders just
above and below zero.
This commit is contained in:
Simon Tatham 2015-12-13 14:46:43 +00:00
parent 90c7b1562c
commit 984792e9f4
2 changed files with 74 additions and 0 deletions
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59
sshbn.c
View File

@ -2059,6 +2059,65 @@ int main(int argc, char **argv)
freebn(modulus);
freebn(expected);
freebn(answer);
} else if (!strcmp(buf, "divmod")) {
Bignum n, d, expect_q, expect_r, answer_q, answer_r;
int fail;
if (ptrnum != 4) {
printf("%d: divmod with %d parameters, expected 4\n", line, ptrnum);
exit(1);
}
n = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
d = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
expect_q = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
expect_r = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
answer_q = bigdiv(n, d);
answer_r = bigmod(n, d);
fail = FALSE;
if (bignum_cmp(expect_q, answer_q) != 0) {
char *as = bignum_decimal(n);
char *bs = bignum_decimal(d);
char *cs = bignum_decimal(answer_q);
char *ds = bignum_decimal(expect_q);
printf("%d: fail: %s / %s gave %s expected %s\n",
line, as, bs, cs, ds);
fail = TRUE;
sfree(as);
sfree(bs);
sfree(cs);
sfree(ds);
}
if (bignum_cmp(expect_r, answer_r) != 0) {
char *as = bignum_decimal(n);
char *bs = bignum_decimal(d);
char *cs = bignum_decimal(answer_r);
char *ds = bignum_decimal(expect_r);
printf("%d: fail: %s mod %s gave %s expected %s\n",
line, as, bs, cs, ds);
fail = TRUE;
sfree(as);
sfree(bs);
sfree(cs);
sfree(ds);
}
freebn(n);
freebn(d);
freebn(expect_q);
freebn(expect_r);
freebn(answer_q);
freebn(answer_r);
if (fail)
fails++;
else
passes++;
} else {
printf("%d: unrecognised test keyword: '%s'\n", line, buf);
exit(1);

15
testdata/bignum.py vendored
View File

@ -103,6 +103,21 @@ for i in range(1,4200):
a, b, p = findprod((1<<i)+1, +1, (i, i+1))
print "mul", hexstr(a), hexstr(b), hexstr(p)
# Bare tests of division/modulo.
prefixes = [2**63, int(2**63.5), 2**64-1]
for nsize in range(20, 200):
for dsize in range(20, 200):
for dprefix in prefixes:
d = sqrt(3<<(2*dsize)) + (dprefix<<dsize)
for nprefix in prefixes:
nbase = sqrt(3<<(2*nsize)) + (nprefix<<nsize)
for modulus in sorted({-1, 0, +1, d/2, nbase % d}):
n = nbase - (nbase % d) + modulus
if n < 0:
n += d
assert n >= 0
print "divmod", hexstr(n), hexstr(d), hexstr(n/d), hexstr(n%d)
# Simple tests of modmul.
for ai in range(20, 200, 60):
a = sqrt(3<<(2*ai-1))