Опишите функцию умножения двух целых обработайте ошибку переполнения сверху overflow

Although there have been several other answers to this question, I several of them have code that is completely untested, and thus far no one has adequately compared the different possible options.

For that reason, I wrote and tested several possible implementations (the last one is based on this code from OpenBSD, discussed on Reddit here). Here’s the code:

/* 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);
}

Here are the results, testing with various compilers and systems I have (in this case, all testing was done on OS X, but results should be similar on BSD or Linux systems):

+------------------+----------+----------+----------+----------+----------+
|                  | 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 |
+------------------+----------+----------+----------+----------+----------+

Based on these results, we can draw a few conclusions:

• Clearly, the division-based approach, although simple and portable, is slow.
• No technique is a clear winner in all cases.
• On modern compilers, the use-a-larger-int approach is best, if you can use it
• On older compilers, the long-multiplication approach is best
• Surprisingly, GCC 4.9.0 has performance regressions over GCC 4.2.1, and GCC 4.2.1 has performance regressions over GCC 3.3

answered Oct 12, 2014 at 0:33

The GNU Portability Library (Gnulib) contains a module intprops, which has macros that efficiently test whether arithmetic operations would overflow.

For example, if an overflow in multiplication would occur, INT_MULTIPLY_OVERFLOW (a, b) would yield 1.

answered Oct 11, 2019 at 20:58

Here is a trick for detecting whether multiplication of two unsigned integers overflows.

We make the observation that if we multiply an N-bit-wide binary number with an M-bit-wide binary number, the product does not have more than N + M bits.

For instance, if we are asked to multiply a three-bit number with a twenty-nine bit number, we know that this doesn’t overflow thirty-two bits.

#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 formn");
    return EXIT_FAILURE;
  }

  a = strtoul(argv[1], &endptr, 0);

  if (*endptr != 0) {
    printf("%s is garbagen", argv[1]);
    return EXIT_FAILURE;
  }

  b = strtoul(argv[2], &endptr, 0);

  if (*endptr != 0) {
    printf("%s is garbagen", argv[2]);
    return EXIT_FAILURE;
  }

  if (might_be_mul_oflow(a, b))
    printf("might be multiplication overflown");

  {
    unsigned long c = a * b;
    printf("%lu * %lu = %lun", a, b, c);
    if (a != 0 && c / a != b)
      printf("confirmed multiplication overflown");
  }

  return 0;
}

A smattering of tests: (on 64 bit system):

$ ./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

The steps in might_be_mul_oflow are almost certainly slower than just doing the division test, at least on mainstream processors used in desktop workstations, servers and mobile devices. On chips without good division support, it could be useful.

It occurs to me that there is another way to do this early rejection test.

1. We start with a pair of numbers arng and brng which are initialized to 0x7FFF...FFFF and 1.

2. If a <= arng and b <= brng we can conclude that there is no overflow.

3. Otherwise, we shift arng to the right, and shift brng to the left, adding one bit to brng, so that they are 0x3FFF...FFFF and 3.

4. If arng is zero, finish; otherwise repeat at 2.

The function now looks like:

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;
  }
}

answered Jun 12, 2017 at 18:42

When your using e.g. 64 bits variables, implement ‘number of significant bits’ with nsb(var) = { 64 — clz(var); }.

clz(var) = count leading zeros in var, a builtin command for GCC and Clang, or probably available with inline assembly for your CPU.

Now use the fact that nsb(a * b) <= nsb(a) + nsb(b) to check for overflow. When smaller, it is always 1 smaller.

Ref GCC: Built-in Function: int __builtin_clz (unsigned int x)
Returns the number of leading 0-bits in x, starting at the most significant bit position. If x is 0, the result is undefined.

answered Jan 11, 2021 at 7:31

I was thinking about this today and stumbled upon this question, my thoughts led me to this result. TLDR, while I find it «elegant» in that it only uses a few lines of code (could easily be a one liner), and has some mild math that simplifies to something relatively simple conceptually, this is mostly «interesting» and I haven’t tested it.

If you think of an unsigned integer as being a single digit with radix 2^n where n is the number of bits in the integer, then you can map those numbers to radians around the unit circle, e.g.

radians(x) = x * (2 * pi * rad / 2^n)

When the integer overflows, it is equivalent to wrapping around the circle. So calculating the carry is equivalent to calculating the number of times multiplication would wrap around the circle. To calculate the number of times we wrap around the circle we divide radians(x) by 2pi radians. e.g.

wrap(x) = radians(x) / (2*pi*rad)
        = (x * (2*pi*rad / 2^n)) / (2*pi*rad / 1)
        = (x * (2*pi*rad / 2^n)) * (1 / 2*pi*rad)
        = x * 1 / 2^n
        = x / 2^n

Which simplifies to

wrap(x) = x / 2^n

This makes sense. The number of times a number, for example, 15 with radix 10, wraps around is 15 / 10 = 1.5, or one and a half times. However, we can’t use 2 digits here (assuming we are limited to a single 2^64 digit).

Say we have a * b, with radix R, we can calculate the carry with

Consider that: wrap(a * b) = a * wrap(b)
wrap(a * b) = (a * b) / R
a * wrap(b) = a * (b / R)
a * (b / R) = (a * b) / R

carry = floor(a * wrap(b))

Take for example a = 9 and b = 5, which are factors of 45 (i.e. 9 * 5 = 45).

wrap(5) = 5 / 10 = 0.5
a * wrap(5) = 9 * 0.5 = 4.5
carry = floor(9 * wrap(5)) = floor(4.5) = 4

Note that if the carry was 0, then we would not have had overflow, for example if a = 2, b=2.

In C/C++ (if the compiler and architecture supports it) we have to use long double.

Thus we have:

long double wrap = b / 18446744073709551616.0L; // this is b / 2^64
unsigned long carry = (unsigned long)(a * wrap); // floor(a * wrap(b))
bool overflow = carry > 0;
unsigned long c = a * b;

c here is the lower significant «digit», i.e. in base 10 9 * 9 = 81, carry = 8, and c = 1.

This was interesting to me in theory, so I thought I’d share it, but one major caveat is with the floating point precision in computers. Using long double, there may be rounding errors for some numbers when we calculate the wrap variable depending on how many significant digits your compiler/arch uses for long doubles, I believe it should be 20 more more to be sure. Another issue with this result, is that it may not perform as well as some of the other solutions simply by using floating points and division.

answered Apr 19, 2021 at 20:14

Although there have been several other answers to this question, I several of them have code that is completely untested, and thus far no one has adequately compared the different possible options.

For that reason, I wrote and tested several possible implementations (the last one is based on this code from OpenBSD, discussed on Reddit here). Here’s the code:

/* 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);
}

Here are the results, testing with various compilers and systems I have (in this case, all testing was done on OS X, but results should be similar on BSD or Linux systems):

+------------------+----------+----------+----------+----------+----------+
|                  | 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 |
+------------------+----------+----------+----------+----------+----------+

Based on these results, we can draw a few conclusions:

• Clearly, the division-based approach, although simple and portable, is slow.
• No technique is a clear winner in all cases.
• On modern compilers, the use-a-larger-int approach is best, if you can use it
• On older compilers, the long-multiplication approach is best
• Surprisingly, GCC 4.9.0 has performance regressions over GCC 4.2.1, and GCC 4.2.1 has performance regressions over GCC 3.3

answered Oct 12, 2014 at 0:33

The GNU Portability Library (Gnulib) contains a module intprops, which has macros that efficiently test whether arithmetic operations would overflow.

For example, if an overflow in multiplication would occur, INT_MULTIPLY_OVERFLOW (a, b) would yield 1.

answered Oct 11, 2019 at 20:58

Here is a trick for detecting whether multiplication of two unsigned integers overflows.

We make the observation that if we multiply an N-bit-wide binary number with an M-bit-wide binary number, the product does not have more than N + M bits.

For instance, if we are asked to multiply a three-bit number with a twenty-nine bit number, we know that this doesn’t overflow thirty-two bits.

#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 formn");
    return EXIT_FAILURE;
  }

  a = strtoul(argv[1], &endptr, 0);

  if (*endptr != 0) {
    printf("%s is garbagen", argv[1]);
    return EXIT_FAILURE;
  }

  b = strtoul(argv[2], &endptr, 0);

  if (*endptr != 0) {
    printf("%s is garbagen", argv[2]);
    return EXIT_FAILURE;
  }

  if (might_be_mul_oflow(a, b))
    printf("might be multiplication overflown");

  {
    unsigned long c = a * b;
    printf("%lu * %lu = %lun", a, b, c);
    if (a != 0 && c / a != b)
      printf("confirmed multiplication overflown");
  }

  return 0;
}

A smattering of tests: (on 64 bit system):

$ ./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

The steps in might_be_mul_oflow are almost certainly slower than just doing the division test, at least on mainstream processors used in desktop workstations, servers and mobile devices. On chips without good division support, it could be useful.

It occurs to me that there is another way to do this early rejection test.

1. We start with a pair of numbers arng and brng which are initialized to 0x7FFF...FFFF and 1.

2. If a <= arng and b <= brng we can conclude that there is no overflow.

3. Otherwise, we shift arng to the right, and shift brng to the left, adding one bit to brng, so that they are 0x3FFF...FFFF and 3.

4. If arng is zero, finish; otherwise repeat at 2.

The function now looks like:

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;
  }
}

answered Jun 12, 2017 at 18:42

When your using e.g. 64 bits variables, implement ‘number of significant bits’ with nsb(var) = { 64 — clz(var); }.

clz(var) = count leading zeros in var, a builtin command for GCC and Clang, or probably available with inline assembly for your CPU.

Now use the fact that nsb(a * b) <= nsb(a) + nsb(b) to check for overflow. When smaller, it is always 1 smaller.

Ref GCC: Built-in Function: int __builtin_clz (unsigned int x)
Returns the number of leading 0-bits in x, starting at the most significant bit position. If x is 0, the result is undefined.

answered Jan 11, 2021 at 7:31

I was thinking about this today and stumbled upon this question, my thoughts led me to this result. TLDR, while I find it «elegant» in that it only uses a few lines of code (could easily be a one liner), and has some mild math that simplifies to something relatively simple conceptually, this is mostly «interesting» and I haven’t tested it.

If you think of an unsigned integer as being a single digit with radix 2^n where n is the number of bits in the integer, then you can map those numbers to radians around the unit circle, e.g.

radians(x) = x * (2 * pi * rad / 2^n)

When the integer overflows, it is equivalent to wrapping around the circle. So calculating the carry is equivalent to calculating the number of times multiplication would wrap around the circle. To calculate the number of times we wrap around the circle we divide radians(x) by 2pi radians. e.g.

wrap(x) = radians(x) / (2*pi*rad)
        = (x * (2*pi*rad / 2^n)) / (2*pi*rad / 1)
        = (x * (2*pi*rad / 2^n)) * (1 / 2*pi*rad)
        = x * 1 / 2^n
        = x / 2^n

Which simplifies to

wrap(x) = x / 2^n

This makes sense. The number of times a number, for example, 15 with radix 10, wraps around is 15 / 10 = 1.5, or one and a half times. However, we can’t use 2 digits here (assuming we are limited to a single 2^64 digit).

Say we have a * b, with radix R, we can calculate the carry with

Consider that: wrap(a * b) = a * wrap(b)
wrap(a * b) = (a * b) / R
a * wrap(b) = a * (b / R)
a * (b / R) = (a * b) / R

carry = floor(a * wrap(b))

Take for example a = 9 and b = 5, which are factors of 45 (i.e. 9 * 5 = 45).

wrap(5) = 5 / 10 = 0.5
a * wrap(5) = 9 * 0.5 = 4.5
carry = floor(9 * wrap(5)) = floor(4.5) = 4

Note that if the carry was 0, then we would not have had overflow, for example if a = 2, b=2.

In C/C++ (if the compiler and architecture supports it) we have to use long double.

Thus we have:

long double wrap = b / 18446744073709551616.0L; // this is b / 2^64
unsigned long carry = (unsigned long)(a * wrap); // floor(a * wrap(b))
bool overflow = carry > 0;
unsigned long c = a * b;

c here is the lower significant «digit», i.e. in base 10 9 * 9 = 81, carry = 8, and c = 1.

This was interesting to me in theory, so I thought I’d share it, but one major caveat is with the floating point precision in computers. Using long double, there may be rounding errors for some numbers when we calculate the wrap variable depending on how many significant digits your compiler/arch uses for long doubles, I believe it should be 20 more more to be sure. Another issue with this result, is that it may not perform as well as some of the other solutions simply by using floating points and division.

answered Apr 19, 2021 at 20:14