抓住和计算溢出在乘法运算的两个较大的整数
https://stackoverflow.com/questions/1815367
-
06-07-2019 - |
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我在寻找一个有效率(可选标准，优雅的和易于实施的)解决乘以相对较大的数字，并将结果存储到一个或几个整数：
让我们说我有两64位整数，宣布这样的：
uint64_t a = xxx, b = yyy;
当我这样做 a * b, 我怎么检测，如果操作的结果溢出和在这种情况下存储的随身携带的某个地方？
请注意， 我不想用任何大量的图书馆 因为我有约束我的方式储存的数字。
解决方案
<强> 1。检测溢出：
x = a * b;
if (a != 0 && x / a != b) {
// overflow handling
}
编辑：通过0修复分区（谢谢马克！）
<强> 2。计算进位非常复杂。一种方法是将两个操作数分成半字，然后将长乘法应用于一半-words：
uint64_t hi(uint64_t x) {
return x >> 32;
}
uint64_t lo(uint64_t x) {
return ((1L << 32) - 1) & x;
}
void multiply(uint64_t a, uint64_t b) {
// actually uint32_t would do, but the casting is annoying
uint64_t s0, s1, s2, s3;
uint64_t x = lo(a) * lo(b);
s0 = lo(x);
x = hi(a) * lo(b) + hi(x);
s1 = lo(x);
s2 = hi(x);
x = s1 + lo(a) * hi(b);
s1 = lo(x);
x = s2 + hi(a) * hi(b) + hi(x);
s2 = lo(x);
s3 = hi(x);
uint64_t result = s1 << 32 | s0;
uint64_t carry = s3 << 32 | s2;
}
为了看到没有任何部分和本身可以溢出，我们考虑最坏的情况：
x = s2 + hi(a) * hi(b) + hi(x)
让B = 1 << 32。然后我们
x <= (B - 1) + (B - 1)(B - 1) + (B - 1)
<= B*B - 1
< B*B
我相信这会奏效 - 至少它会处理Sjlver的测试用例。除此之外，它是未经测试的（甚至可能无法编译，因为我手头没有C ++编译器了。）
其他提示
这个想法是使用以下事实，这对积分操作是正确的：
a*b > c当且仅当a > c/b
/是这里不可分割的部门。
检查正数溢出的伪代码如下：
if（a <！> gt; max_int64 / b）然后<！> quot; overflow <！> quot; else <！> quot; ok <！> <<; 。
要处理零和负数，您应该添加更多检查。
非负a和b的C代码如下：
if (b > 0 && a > 18446744073709551615 / b) {
// overflow handling
}; else {
c = a * b;
}
注意：
18446744073709551615 == (1<<64)-1
为了计算进位，我们可以使用方法将数字拆分成两个32位数，然后在纸上将它们相乘。我们需要拆分数字以避免溢出。
代码如下：
// split input numbers into 32-bit digits
uint64_t a0 = a & ((1LL<<32)-1);
uint64_t a1 = a >> 32;
uint64_t b0 = b & ((1LL<<32)-1);
uint64_t b1 = b >> 32;
// The following 3 lines of code is to calculate the carry of d1
// (d1 - 32-bit second digit of result, and it can be calculated as d1=d11+d12),
// but to avoid overflow.
// Actually rewriting the following 2 lines:
// uint64_t d1 = (a0 * b0 >> 32) + a1 * b0 + a0 * b1;
// uint64_t c1 = d1 >> 32;
uint64_t d11 = a1 * b0 + (a0 * b0 >> 32);
uint64_t d12 = a0 * b1;
uint64_t c1 = (d11 > 18446744073709551615 - d12) ? 1 : 0;
uint64_t d2 = a1 * b1 + c1;
uint64_t carry = d2; // needed carry stored here
虽然已有若干其它对这个问题的答案，我的几个他们的代码是完全未经测试，并且迄今为止没有一个具有充分比较不同的可能的选择。
出于这一原因，我编写并测试了几种可能实现(最后一个是基于 这个代码 从OpenBSD，讨论上。 在这里，).这里的代码:
/* Multiply with overflow checking, emulating clang's builtin function
*
* __builtin_umull_overflow
*
* This code benchmarks five possible schemes for doing so.
*/
#include <stddef.h>
#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <limits.h>
#ifndef BOOL
#define BOOL int
#endif
// Option 1, check for overflow a wider type
// - Often fastest and the least code, especially on modern compilers
// - When long is a 64-bit int, requires compiler support for 128-bits
// ints (requires GCC >= 3.0 or Clang)
#if LONG_BIT > 32
typedef __uint128_t long_overflow_t ;
#else
typedef uint64_t long_overflow_t;
#endif
BOOL
umull_overflow1(unsigned long lhs, unsigned long rhs, unsigned long* result)
{
long_overflow_t prod = (long_overflow_t)lhs * (long_overflow_t)rhs;
*result = (unsigned long) prod;
return (prod >> LONG_BIT) != 0;
}
// Option 2, perform long multiplication using a smaller type
// - Sometimes the fastest (e.g., when mulitply on longs is a library
// call).
// - Performs at most three multiplies, and sometimes only performs one.
// - Highly portable code; works no matter how many bits unsigned long is
BOOL
umull_overflow2(unsigned long lhs, unsigned long rhs, unsigned long* result)
{
const unsigned long HALFSIZE_MAX = (1ul << LONG_BIT/2) - 1ul;
unsigned long lhs_high = lhs >> LONG_BIT/2;
unsigned long lhs_low = lhs & HALFSIZE_MAX;
unsigned long rhs_high = rhs >> LONG_BIT/2;
unsigned long rhs_low = rhs & HALFSIZE_MAX;
unsigned long bot_bits = lhs_low * rhs_low;
if (!(lhs_high || rhs_high)) {
*result = bot_bits;
return 0;
}
BOOL overflowed = lhs_high && rhs_high;
unsigned long mid_bits1 = lhs_low * rhs_high;
unsigned long mid_bits2 = lhs_high * rhs_low;
*result = bot_bits + ((mid_bits1+mid_bits2) << LONG_BIT/2);
return overflowed || *result < bot_bits
|| (mid_bits1 >> LONG_BIT/2) != 0
|| (mid_bits2 >> LONG_BIT/2) != 0;
}
// Option 3, perform long multiplication using a smaller type (this code is
// very similar to option 2, but calculates overflow using a different but
// equivalent method).
// - Sometimes the fastest (e.g., when mulitply on longs is a library
// call; clang likes this code).
// - Performs at most three multiplies, and sometimes only performs one.
// - Highly portable code; works no matter how many bits unsigned long is
BOOL
umull_overflow3(unsigned long lhs, unsigned long rhs, unsigned long* result)
{
const unsigned long HALFSIZE_MAX = (1ul << LONG_BIT/2) - 1ul;
unsigned long lhs_high = lhs >> LONG_BIT/2;
unsigned long lhs_low = lhs & HALFSIZE_MAX;
unsigned long rhs_high = rhs >> LONG_BIT/2;
unsigned long rhs_low = rhs & HALFSIZE_MAX;
unsigned long lowbits = lhs_low * rhs_low;
if (!(lhs_high || rhs_high)) {
*result = lowbits;
return 0;
}
BOOL overflowed = lhs_high && rhs_high;
unsigned long midbits1 = lhs_low * rhs_high;
unsigned long midbits2 = lhs_high * rhs_low;
unsigned long midbits = midbits1 + midbits2;
overflowed = overflowed || midbits < midbits1 || midbits > HALFSIZE_MAX;
unsigned long product = lowbits + (midbits << LONG_BIT/2);
overflowed = overflowed || product < lowbits;
*result = product;
return overflowed;
}
// Option 4, checks for overflow using division
// - Checks for overflow using division
// - Division is slow, especially if it is a library call
BOOL
umull_overflow4(unsigned long lhs, unsigned long rhs, unsigned long* result)
{
*result = lhs * rhs;
return rhs > 0 && (SIZE_MAX / rhs) < lhs;
}
// Option 5, checks for overflow using division
// - Checks for overflow using division
// - Avoids division when the numbers are "small enough" to trivially
// rule out overflow
// - Division is slow, especially if it is a library call
BOOL
umull_overflow5(unsigned long lhs, unsigned long rhs, unsigned long* result)
{
const unsigned long MUL_NO_OVERFLOW = (1ul << LONG_BIT/2) - 1ul;
*result = lhs * rhs;
return (lhs >= MUL_NO_OVERFLOW || rhs >= MUL_NO_OVERFLOW) &&
rhs > 0 && SIZE_MAX / rhs < lhs;
}
#ifndef umull_overflow
#define umull_overflow2
#endif
/*
* This benchmark code performs a multiply at all bit sizes,
* essentially assuming that sizes are logarithmically distributed.
*/
int main()
{
unsigned long i, j, k;
int count = 0;
unsigned long mult;
unsigned long total = 0;
for (k = 0; k < 0x40000000 / LONG_BIT / LONG_BIT; ++k)
for (i = 0; i != LONG_MAX; i = i*2+1)
for (j = 0; j != LONG_MAX; j = j*2+1) {
count += umull_overflow(i+k, j+k, &mult);
total += mult;
}
printf("%d overflows (total %lu)\n", count, total);
}
这里的结果、测试各种系统编译器和我已经(在这种情况下，所有测试是在OS X，但结果应该是类似于BSD或Linux系统):
+------------------+----------+----------+----------+----------+----------+
| | Option 1 | Option 2 | Option 3 | Option 4 | Option 5 |
| | BigInt | LngMult1 | LngMult2 | Div | OptDiv |
+------------------+----------+----------+----------+----------+----------+
| Clang 3.5 i386 | 1.610 | 3.217 | 3.129 | 4.405 | 4.398 |
| GCC 4.9.0 i386 | 1.488 | 3.469 | 5.853 | 4.704 | 4.712 |
| GCC 4.2.1 i386 | 2.842 | 4.022 | 3.629 | 4.160 | 4.696 |
| GCC 4.2.1 PPC32 | 8.227 | 7.756 | 7.242 | 20.632 | 20.481 |
| GCC 3.3 PPC32 | 5.684 | 9.804 | 11.525 | 21.734 | 22.517 |
+------------------+----------+----------+----------+----------+----------+
| Clang 3.5 x86_64 | 1.584 | 2.472 | 2.449 | 9.246 | 7.280 |
| GCC 4.9 x86_64 | 1.414 | 2.623 | 4.327 | 9.047 | 7.538 |
| GCC 4.2.1 x86_64 | 2.143 | 2.618 | 2.750 | 9.510 | 7.389 |
| GCC 4.2.1 PPC64 | 13.178 | 8.994 | 8.567 | 37.504 | 29.851 |
+------------------+----------+----------+----------+----------+----------+
基于这些结果，我们可以得出一些结论：
- 显然，该司为基础的办法，虽然简单和便携式的，是缓慢的。
- 没有技术是一个明显的胜利者在所有情况。
- 在现代的编译器，使用一个较大的int的方法是最好的,如果你可以用它
- 关于老年编译器，长乘法运算法是最好的
- 令人惊奇的是，海湾合作委员会4.9.0有效的回归过海湾合作委员会4.2.1和海湾合作委员会4.2.1有效的回归过海湾合作委员会3.3
在== 0：
时也适用的版本 x = a * b;
if (a != 0 && x / a != b) {
// overflow handling
}
如果您不仅需要检测溢出而且还需要捕获进位，那么最好将数字分解为32位部分。代码是一场噩梦;以下只是草图：
#include <stdint.h>
uint64_t mul(uint64_t a, uint64_t b) {
uint32_t ah = a >> 32;
uint32_t al = a; // truncates: now a = al + 2**32 * ah
uint32_t bh = b >> 32;
uint32_t bl = b; // truncates: now b = bl + 2**32 * bh
// a * b = 2**64 * ah * bh + 2**32 * (ah * bl + bh * al) + al * bl
uint64_t partial = (uint64_t) al * (uint64_t) bl;
uint64_t mid1 = (uint64_t) ah * (uint64_t) bl;
uint64_t mid2 = (uint64_t) al * (uint64_t) bh;
uint64_t carry = (uint64_t) ah * (uint64_t) bh;
// add high parts of mid1 and mid2 to carry
// add low parts of mid1 and mid2 to partial, carrying
// any carry bits into carry...
}
问题不仅仅是部分产品，而是任何总和可能溢出的事实。
如果我必须真正做到这一点，我会用本地汇编语言编写扩展乘法例程。也就是说，例如，将两个64位整数相乘得到128-位结果，存储在两个64位寄存器中。所有合理的硬件都在单个本地乘法指令中提供此功能<！>＃8212;它不仅可以从<！> nbsp; C进行访问。
这是极少数情况下，其中最优雅且易于编程的解决方案实际上是使用汇编语言。但它肯定不便携： - （
我这几天一直在处理这个问题而且我不得不说，我看到人们说最好的方式来知道是否存在溢出是分开结果的次数给我留下了深刻的印象，那就是完全没有效率和不必要的。这个功能的要点是它必须尽可能快。
溢出检测有两种选择：
1 <！>＃186; - 如果可能，创建两倍于乘数的结果变量，例如：
struct INT32struct {INT16 high, low;};
typedef union
{
struct INT32struct s;
INT32 ll;
} INT32union;
INT16 mulFunction(INT16 a, INT16 b)
{
INT32union result.ll = a * b; //32Bits result
if(result.s.high > 0)
Overflow();
return (result.s.low)
}
您将立即知道是否存在溢出，并且代码是最快的，而无需在机器代码中编写代码。根据编译器，可以在机器代码中改进此代码。
2 <！>＃186; - 无法创建两倍于乘数变量的结果变量： 然后你应该玩if条件来确定最佳路径。继续举例：
INT32 mulFunction(INT32 a, INT32 b)
{
INT32union s_a.ll = abs(a);
INT32union s_b.ll = abs(b); //32Bits result
INT32union result;
if(s_a.s.hi > 0 && s_b.s.hi > 0)
{
Overflow();
}
else if (s_a.s.hi > 0)
{
INT32union res1.ll = s_a.s.hi * s_b.s.lo;
INT32union res2.ll = s_a.s.lo * s_b.s.lo;
if (res1.hi == 0)
{
result.s.lo = res1.s.lo + res2.s.hi;
if (result.s.hi == 0)
{
result.s.ll = result.s.lo << 16 + res2.s.lo;
if ((a.s.hi >> 15) ^ (b.s.hi >> 15) == 1)
{
result.s.ll = -result.s.ll;
}
return result.s.ll
}else
{
Overflow();
}
}else
{
Overflow();
}
}else if (s_b.s.hi > 0)
{
//Same code changing a with b
}else
{
return (s_a.lo * s_b.lo);
}
}
我希望这段代码可以帮助你提供一个非常有效的程序，我希望代码是清晰的，如果不是，我会放一些代码。
最好的问候。也许解决这个问题的最好方法是有一个函数，它将两个UInt64相乘并产生一对UInt64，UInt128结果的上半部分和下半部分。这是解决方案，包括一个以十六进制显示结果的函数。我想你可能更喜欢C ++解决方案，但我有一个工作的Swift-Solution，它显示了如何管理问题：
func hex128 (_ hi: UInt64, _ lo: UInt64) -> String
{
var s: String = String(format: "%08X", hi >> 32)
+ String(format: "%08X", hi & 0xFFFFFFFF)
+ String(format: "%08X", lo >> 32)
+ String(format: "%08X", lo & 0xFFFFFFFF)
return (s)
}
func mul64to128 (_ multiplier: UInt64, _ multiplicand : UInt64)
-> (result_hi: UInt64, result_lo: UInt64)
{
let x: UInt64 = multiplier
let x_lo: UInt64 = (x & 0xffffffff)
let x_hi: UInt64 = x >> 32
let y: UInt64 = multiplicand
let y_lo: UInt64 = (y & 0xffffffff)
let y_hi: UInt64 = y >> 32
let mul_lo: UInt64 = (x_lo * y_lo)
let mul_hi: UInt64 = (x_hi * y_lo) + (mul_lo >> 32)
let mul_carry: UInt64 = (x_lo * y_hi) + (mul_hi & 0xffffffff)
let result_hi: UInt64 = (x_hi * y_hi) + (mul_hi >> 32) + (mul_carry >> 32)
let result_lo: UInt64 = (mul_carry << 32) + (mul_lo & 0xffffffff)
return (result_hi, result_lo)
}
以下是验证函数是否有效的示例：
var c: UInt64 = 0
var d: UInt64 = 0
(c, d) = mul64to128(0x1234567890123456, 0x9876543210987654)
// 0AD77D742CE3C72E45FD10D81D28D038 is the result of the above example
print(hex128(c, d))
(c, d) = mul64to128(0xFFFFFFFFFFFFFFFF, 0xFFFFFFFFFFFFFFFF)
// FFFFFFFFFFFFFFFE0000000000000001 is the result of the above example
print(hex128(c, d))
这是检测两个无符号整数的乘法是否溢出的技巧。
我们观察到，如果我们将N位宽的二进制数与M位宽的二进制数相乘，则乘积不会超过N + M位。
例如，如果要求我们将三位数乘以二十九位数，我们就知道不会溢出三十二位。
#include <stdlib.h>
#include <stdio.h>
int might_be_mul_oflow(unsigned long a, unsigned long b)
{
if (!a || !b)
return 0;
a = a | (a >> 1) | (a >> 2) | (a >> 4) | (a >> 8) | (a >> 16) | (a >> 32);
b = b | (b >> 1) | (b >> 2) | (b >> 4) | (b >> 8) | (b >> 16) | (b >> 32);
for (;;) {
unsigned long na = a << 1;
if (na <= a)
break;
a = na;
}
return (a & b) ? 1 : 0;
}
int main(int argc, char **argv)
{
unsigned long a, b;
char *endptr;
if (argc < 3) {
printf("supply two unsigned long integers in C form\n");
return EXIT_FAILURE;
}
a = strtoul(argv[1], &endptr, 0);
if (*endptr != 0) {
printf("%s is garbage\n", argv[1]);
return EXIT_FAILURE;
}
b = strtoul(argv[2], &endptr, 0);
if (*endptr != 0) {
printf("%s is garbage\n", argv[2]);
return EXIT_FAILURE;
}
if (might_be_mul_oflow(a, b))
printf("might be multiplication overflow\n");
{
unsigned long c = a * b;
printf("%lu * %lu = %lu\n", a, b, c);
if (a != 0 && c / a != b)
printf("confirmed multiplication overflow\n");
}
return 0;
}
一些测试:(在64位系统上）：
$ ./uflow 0x3 0x3FFFFFFFFFFFFFFF 3 * 4611686018427387903 = 13835058055282163709 $ ./uflow 0x7 0x3FFFFFFFFFFFFFFF might be multiplication overflow 7 * 4611686018427387903 = 13835058055282163705 confirmed multiplication overflow $ ./uflow 0x4 0x3FFFFFFFFFFFFFFF might be multiplication overflow 4 * 4611686018427387903 = 18446744073709551612 $ ./uflow 0x5 0x3FFFFFFFFFFFFFFF might be multiplication overflow 5 * 4611686018427387903 = 4611686018427387899 confirmed multiplication overflow
might_be_mul_oflow中的步骤几乎肯定比仅仅进行分区测试要慢，至少在桌面工作站，服务器和移动设备中使用的主流处理器上。在没有良好分工支持的芯片上，它可能很有用。
我觉得还有另一种方法可以做到这种早期拒绝测试。
-
我们从一对数字
arng和brng开始，这些数字被初始化为0x7FFF...FFFF和1。 -
如果
a <= arng且b <= brng我们可以断定没有溢出。 -
否则，我们将
0x3FFF...FFFF移至右侧，并将3移至左侧，向<=>添加一位，以便它们<=>且<=>。 -
如果<=>为零，则完成;否则在2处重复。
该功能现在看起来像：
int might_be_mul_oflow(unsigned long a, unsigned long b)
{
if (!a || !b)
return 0;
{
unsigned long arng = ULONG_MAX >> 1;
unsigned long brng = 1;
while (arng != 0) {
if (a <= arng && b <= brng)
return 0;
arng >>= 1;
brng <<= 1;
brng |= 1;
}
return 1;
}
}
如果你只是想检测溢出，那么如何转换为double，进行乘法以及if
| X | LT <！>; 2 ^ 53，转换为int64
| X | LT <！>; 2 ^ 63，使用int64进行乘法
否则会产生你想要的任何错误？
这似乎有效：
int64_t safemult(int64_t a, int64_t b) {
double dx;
dx = (double)a * (double)b;
if ( fabs(dx) < (double)9007199254740992 )
return (int64_t)dx;
if ( (double)INT64_MAX < fabs(dx) )
return INT64_MAX;
return a*b;
}
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