#include "misc.h"
+/*
+ * Usage notes:
+ * * Do not call the DIVMOD_WORD macro with expressions such as array
+ * subscripts, as some implementations object to this (see below).
+ * * Note that none of the division methods below will cope if the
+ * quotient won't fit into BIGNUM_INT_BITS. Callers should be careful
+ * to avoid this case.
+ * If this condition occurs, in the case of the x86 DIV instruction,
+ * an overflow exception will occur, which (according to a correspondent)
+ * will manifest on Windows as something like
+ * 0xC0000095: Integer overflow
+ * The C variant won't give the right answer, either.
+ */
+
#if defined __GNUC__ && defined __i386__
typedef unsigned long BignumInt;
typedef unsigned long long BignumDblInt;
#define BIGNUM_TOP_BIT 0x80000000UL
#define BIGNUM_INT_BITS 32
#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
-typedef struct {
- unsigned __int32 quot;
- unsigned __int32 remd;
-} msvc_quorem;
-static __declspec(naked) msvc_quorem __stdcall msvc_divmod(
- unsigned __int32 hi,
- unsigned __int32 lo,
- unsigned __int32 w)
-{
- __asm {
- mov edx, dword ptr [esp+4]
- mov eax, dword ptr [esp+8]
- div dword ptr [esp+12]
- ret 12
- };
-}
+/* Note: MASM interprets array subscripts in the macro arguments as
+ * assembler syntax, which gives the wrong answer. Don't supply them.
+ * <http://msdn2.microsoft.com/en-us/library/bf1dw62z.aspx> */
#define DIVMOD_WORD(q, r, hi, lo, w) do { \
- const msvc_quorem qr = msvc_divmod((hi), (lo), (w)); \
- (q) = qr.quot; (r) = qr.remd; \
-} while (0)
+ __asm mov edx, hi \
+ __asm mov eax, lo \
+ __asm div w \
+ __asm mov r, edx \
+ __asm mov q, eax \
+} while(0)
#else
typedef unsigned short BignumInt;
typedef unsigned long BignumDblInt;
*/
q = BIGNUM_INT_MASK;
} else {
- DIVMOD_WORD(q, r, h, a[i], m0);
+ /* Macro doesn't want an array subscript expression passed
+ * into it (see definition), so use a temporary. */
+ BignumInt tmplo = a[i];
+ DIVMOD_WORD(q, r, h, tmplo, m0);
/* Refine our estimate of q by looking at
h:a[i]:a[i+1] / m0:m1 */
* round up (rounding down might make it less than x again).
* Therefore if we multiply the bit count by 28/93, rounding
* up, we will have enough digits.
+ *
+ * i=0 (i.e., x=0) is an irritating special case.
*/
i = bignum_bitcount(x);
- ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
+ if (!i)
+ ndigits = 1; /* x = 0 */
+ else
+ ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
ndigits++; /* allow for trailing \0 */
ret = snewn(ndigits, char);