我用c++写了一个程序来寻找ab = C的所有解，其中a, b和C一起使用所有的数字0-9，只使用一次。程序循环遍历a和b的值，并每次对a、b和ab运行数字计数例程，以检查是否满足数字条件。
但是，当ab超出整数限制时，会产生伪解。我最终使用如下代码来检查这个:
unsigned long b, c, c_test;
...
c_test=c*b; // Possible overflow
if (c_test/b != c) {/* There has been an overflow*/}
else c=c_test; // No overflow
是否有更好的方法来测试溢出?我知道有些芯片有一个内部标志，在溢出发生时设置，但我从未见过通过C或c++访问它。
注意，有符号int溢出在C和c++中是未定义的行为，因此您必须在不实际引起它的情况下检测它。对于加法前的有符号整型溢出，请参见在C/ c++中检测有符号溢出。
当前回答
我看到你用的是无符号整数。根据定义，在C中(我不了解c++)，无符号算术不会溢出…所以，至少对C来说，你的观点是没有意义的:)
对于有符号整数，一旦出现溢出，就会发生未定义行为(UB)，程序可以做任何事情(例如:使测试不确定)。
#include <limits.h>
int a = <something>;
int x = <something>;
a += x; /* UB */
if (a < 0) { /* Unreliable test */
/* ... */
}
要创建一个符合要求的程序，您需要在生成溢出之前测试溢出。该方法也可以用于无符号整数:
// For addition
#include <limits.h>
int a = <something>;
int x = <something>;
if (x > 0 && a > INT_MAX - x) // `a + x` would overflow
if (x < 0 && a < INT_MIN - x) // `a + x` would underflow
// For subtraction
#include <limits.h>
int a = <something>;
int x = <something>;
if (x < 0 && a > INT_MAX + x) // `a - x` would overflow
if (x > 0 && a < INT_MIN + x) // `a - x` would underflow
// For multiplication
#include <limits.h>
int a = <something>;
int x = <something>;
// There may be a need to check for -1 for two's complement machines.
// If one number is -1 and another is INT_MIN, multiplying them we get abs(INT_MIN) which is 1 higher than INT_MAX
if (a == -1 && x == INT_MIN) // `a * x` can overflow
if (x == -1 && a == INT_MIN) // `a * x` (or `a / x`) can overflow
// general case
if (x != 0 && a > INT_MAX / x) // `a * x` would overflow
if (x != 0 && a < INT_MIN / x) // `a * x` would underflow
对于除法(INT_MIN和-1特殊情况除外)，不可能超过INT_MIN或INT_MAX。
其他回答
I see that a lot of people answered the question about overflow, but I wanted to address his original problem. He said the problem was to find ab=c such that all digits are used without repeating. Ok, that's not what he asked in this post, but I'm still think that it was necessary to study the upper bound of the problem and conclude that he would never need to calculate or detect an overflow (note: I'm not proficient in math so I did this step by step, but the end result was so simple that this might have a simple formula).
重点是问题要求的a b c的上限是98.765.432。不管怎样，先把问题分成琐碎部分和非琐碎部分:
X0 == 1(9、8、7、6、5、4、3、2的所有排列都是解) X1 == x(无解) 0b == 0(不可能解) 1b == 1(无解) Ab, a > 1, b > 1(非平凡)
Now we just need to show that no other solution is possible and only the permutations are valid (and then the code to print them is trivial). We go back to the upper bound. Actually the upper bound is c ≤ 98.765.432. It's the upper bound because it's the largest number with 8 digits (10 digits total minus 1 for each a and b). This upper bound is only for c because the bounds for a and b must be much lower because of the exponential growth, as we can calculate, varying b from 2 to the upper bound:
9938.08^2 == 98765432
462.241^3 == 98765432
99.6899^4 == 98765432
39.7119^5 == 98765432
21.4998^6 == 98765432
13.8703^7 == 98765432
9.98448^8 == 98765432
7.73196^9 == 98765432
6.30174^10 == 98765432
5.33068^11 == 98765432
4.63679^12 == 98765432
4.12069^13 == 98765432
3.72429^14 == 98765432
3.41172^15 == 98765432
3.15982^16 == 98765432
2.95305^17 == 98765432
2.78064^18 == 98765432
2.63493^19 == 98765432
2.51033^20 == 98765432
2.40268^21 == 98765432
2.30883^22 == 98765432
2.22634^23 == 98765432
2.15332^24 == 98765432
2.08826^25 == 98765432
2.02995^26 == 98765432
1.97741^27 == 98765432
注意，例如最后一行:它说1.97^27 ~98M。因此，例如，1^27 == 1和2^27 == 134.217.728，这不是一个解决方案，因为它有9位数字(2 > 1.97，所以它实际上比应该测试的要大)。可以看到，用于测试a和b的组合非常小。对于b == 14，我们需要尝试2和3。对于b == 3，我们从2开始，到462结束。结果均小于~98M。
现在只需测试以上所有的组合，找出不重复任何数字的组合:
['0', '2', '4', '5', '6', '7', '8'] 84^2 = 7056
['1', '2', '3', '4', '5', '8', '9'] 59^2 = 3481
['0', '1', '2', '3', '4', '5', '8', '9'] 59^2 = 3481 (+leading zero)
['1', '2', '3', '5', '8'] 8^3 = 512
['0', '1', '2', '3', '5', '8'] 8^3 = 512 (+leading zero)
['1', '2', '4', '6'] 4^2 = 16
['0', '1', '2', '4', '6'] 4^2 = 16 (+leading zero)
['1', '2', '4', '6'] 2^4 = 16
['0', '1', '2', '4', '6'] 2^4 = 16 (+leading zero)
['1', '2', '8', '9'] 9^2 = 81
['0', '1', '2', '8', '9'] 9^2 = 81 (+leading zero)
['1', '3', '4', '8'] 3^4 = 81
['0', '1', '3', '4', '8'] 3^4 = 81 (+leading zero)
['2', '3', '6', '7', '9'] 3^6 = 729
['0', '2', '3', '6', '7', '9'] 3^6 = 729 (+leading zero)
['2', '3', '8'] 2^3 = 8
['0', '2', '3', '8'] 2^3 = 8 (+leading zero)
['2', '3', '9'] 3^2 = 9
['0', '2', '3', '9'] 3^2 = 9 (+leading zero)
['2', '4', '6', '8'] 8^2 = 64
['0', '2', '4', '6', '8'] 8^2 = 64 (+leading zero)
['2', '4', '7', '9'] 7^2 = 49
['0', '2', '4', '7', '9'] 7^2 = 49 (+leading zero)
没有一个匹配问题(这也可以通过缺少'0'，'1'，…“9”)。
下面是解决该问题的示例代码。还要注意，这是用Python编写的，不是因为它需要任意精确整数(代码不会计算任何大于9800万的数字)，而是因为我们发现测试的数量非常少，所以我们应该使用高级语言来利用其内置的容器和库(还要注意:代码有28行)。
import math
m = 98765432
l = []
for i in xrange(2, 98765432):
inv = 1.0/i
r = m**inv
if (r < 2.0): break
top = int(math.floor(r))
assert(top <= m)
for j in xrange(2, top+1):
s = str(i) + str(j) + str(j**i)
l.append((sorted(s), i, j, j**i))
assert(j**i <= m)
l.sort()
for s, i, j, ji in l:
assert(ji <= m)
ss = sorted(set(s))
if s == ss:
print '%s %d^%d = %d' % (s, i, j, ji)
# Try with non significant zero somewhere
s = ['0'] + s
ss = sorted(set(s))
if s == ss:
print '%s %d^%d = %d (+leading zero)' % (s, i, j, ji)
#include <stdio.h>
#include <stdlib.h>
#define MAX 100
int mltovf(int a, int b)
{
if (a && b) return abs(a) > MAX/abs(b);
else return 0;
}
main()
{
int a, b;
for (a = 0; a <= MAX; a++)
for (b = 0; b < MAX; b++) {
if (mltovf(a, b) != (a*b > MAX))
printf("Bad calculation: a: %d b: %d\n", a, b);
}
}
如果您有一个比您想要测试的数据类型大的数据类型(比如您做了一个32位的添加，而您有一个64位的类型)，那么这将检测是否发生溢出。我的例子是一个8位的添加。但它可以放大。
uint8_t x, y; /* Give these values */
const uint16_t data16 = x + y;
const bool carry = (data16 > 0xFF);
const bool overflow = ((~(x ^ y)) & (x ^ data16) & 0x80);
它基于本页解释的概念:http://www.cs.umd.edu/class/spring2003/cmsc311/Notes/Comb/overflow.html
对于一个32位的例子，0xFF变成0xFFFFFFFF, 0x80变成0x80000000，最后uint16_t变成uint64_t。
注意:这捕获整数加法/减法溢出，我意识到你的问题涉及乘法。在这种情况下，分裂可能是最好的办法。这通常是calloc实现确保参数在相乘以获得最终大小时不会溢出的一种方式。
我看到你用的是无符号整数。根据定义，在C中(我不了解c++)，无符号算术不会溢出…所以，至少对C来说，你的观点是没有意义的:)
对于有符号整数，一旦出现溢出，就会发生未定义行为(UB)，程序可以做任何事情(例如:使测试不确定)。
#include <limits.h>
int a = <something>;
int x = <something>;
a += x; /* UB */
if (a < 0) { /* Unreliable test */
/* ... */
}
要创建一个符合要求的程序，您需要在生成溢出之前测试溢出。该方法也可以用于无符号整数:
// For addition
#include <limits.h>
int a = <something>;
int x = <something>;
if (x > 0 && a > INT_MAX - x) // `a + x` would overflow
if (x < 0 && a < INT_MIN - x) // `a + x` would underflow
// For subtraction
#include <limits.h>
int a = <something>;
int x = <something>;
if (x < 0 && a > INT_MAX + x) // `a - x` would overflow
if (x > 0 && a < INT_MIN + x) // `a - x` would underflow
// For multiplication
#include <limits.h>
int a = <something>;
int x = <something>;
// There may be a need to check for -1 for two's complement machines.
// If one number is -1 and another is INT_MIN, multiplying them we get abs(INT_MIN) which is 1 higher than INT_MAX
if (a == -1 && x == INT_MIN) // `a * x` can overflow
if (x == -1 && a == INT_MIN) // `a * x` (or `a / x`) can overflow
// general case
if (x != 0 && a > INT_MAX / x) // `a * x` would overflow
if (x != 0 && a < INT_MIN / x) // `a * x` would underflow
对于除法(INT_MIN和-1特殊情况除外)，不可能超过INT_MIN或INT_MAX。
有一种方法可以确定一个操作是否可能溢出，使用操作数中最高位的位置和一些基本的二进制数学知识。
对于加法，任何两个操作数的结果(最多)比最大操作数的最高1位多1位。例如:
bool addition_is_safe(uint32_t a, uint32_t b) {
size_t a_bits=highestOneBitPosition(a), b_bits=highestOneBitPosition(b);
return (a_bits<32 && b_bits<32);
}
对于乘法，任何两个操作数的结果(最多)是操作数的位数之和。例如:
bool multiplication_is_safe(uint32_t a, uint32_t b) {
size_t a_bits=highestOneBitPosition(a), b_bits=highestOneBitPosition(b);
return (a_bits+b_bits<=32);
}
类似地，你可以像这样估计a的b次方结果的最大大小:
bool exponentiation_is_safe(uint32_t a, uint32_t b) {
size_t a_bits=highestOneBitPosition(a);
return (a_bits*b<=32);
}
(当然，用比特数代替目标整数。)
我不确定确定数字中最高的1位位置的最快方法，这里有一个蛮力方法:
size_t highestOneBitPosition(uint32_t a) {
size_t bits=0;
while (a!=0) {
++bits;
a>>=1;
};
return bits;
}
它不是完美的，但它能让你在做运算之前知道是否有两个数会溢出。我不知道它是否会比您建议的方式检查结果更快，因为highestOneBitPosition函数中的循环，但它可能会(特别是如果您事先知道操作数中有多少位)。
