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
我需要为浮点数回答同样的问题，在浮点数中位屏蔽和移位看起来没有希望。我确定的方法适用于有符号和无符号，整数和浮点数。即使没有更大的数据类型可以用于中间计算，它也可以工作。对于所有这些类型，它不是最有效的，但因为它确实适用于所有类型，所以值得使用。
有符号溢出测试，加减法:
Obtain the constants that represent the largest and smallest possible values for the type, MAXVALUE and MINVALUE. Compute and compare the signs of the operands. a. If either value is zero, then neither addition nor subtraction can overflow. Skip remaining tests. b. If the signs are opposite, then addition cannot overflow. Skip remaining tests. c. If the signs are the same, then subtraction cannot overflow. Skip remaining tests. Test for positive overflow of MAXVALUE. a. If both signs are positive and MAXVALUE - A < B, then addition will overflow. b. If the sign of B is negative and MAXVALUE - A < -B, then subtraction will overflow. Test for negative overflow of MINVALUE. a. If both signs are negative and MINVALUE - A > B, then addition will overflow. b. If the sign of A is negative and MINVALUE - A > B, then subtraction will overflow. Otherwise, no overflow.
签名溢出测试，乘法和除法:
Obtain the constants that represent the largest and smallest possible values for the type, MAXVALUE and MINVALUE. Compute and compare the magnitudes (absolute values) of the operands to one. (Below, assume A and B are these magnitudes, not the signed originals.) a. If either value is zero, multiplication cannot overflow, and division will yield zero or an infinity. b. If either value is one, multiplication and division cannot overflow. c. If the magnitude of one operand is below one and of the other is greater than one, multiplication cannot overflow. d. If the magnitudes are both less than one, division cannot overflow. Test for positive overflow of MAXVALUE. a. If both operands are greater than one and MAXVALUE / A < B, then multiplication will overflow. b. If B is less than one and MAXVALUE * B < A, then division will overflow. Otherwise, no overflow.
注意:MINVALUE的最小溢出由3处理，因为我们取的是绝对值。然而,如果 ABS(MINVALUE) > MAXVALUE，那么我们将会有一些罕见的假阳性。
下溢测试类似，但涉及EPSILON(大于零的最小正数)。
其他回答
另一种使用汇编语言的解决方案是外部过程。下面是在Linux x64下使用g++和fasm进行无符号整数乘法的示例。
这个过程将两个无符号整数参数相乘(32位)(根据amd64的规范(第3.2.3节参数传递)。
如果类为INTEGER，则使用序列%rdi、%rsi、%rdx、%rcx、%r8和%r9的下一个可用寄存器
(edi和esi寄存器在我的代码))，并返回结果或0，如果发生溢出。
format ELF64
section '.text' executable
public u_mul
u_mul:
MOV eax, edi
mul esi
jnc u_mul_ret
xor eax, eax
u_mul_ret:
ret
测试:
extern "C" unsigned int u_mul(const unsigned int a, const unsigned int b);
int main() {
printf("%u\n", u_mul(4000000000,2)); // 0
printf("%u\n", u_mul(UINT_MAX/2,2)); // OK
return 0;
}
将程序链接到asm对象文件。在我的例子中，在Qt Creator中将它添加到一个.pro文件中的LIBS中。
我需要为浮点数回答同样的问题，在浮点数中位屏蔽和移位看起来没有希望。我确定的方法适用于有符号和无符号，整数和浮点数。即使没有更大的数据类型可以用于中间计算，它也可以工作。对于所有这些类型，它不是最有效的，但因为它确实适用于所有类型，所以值得使用。
有符号溢出测试，加减法:
Obtain the constants that represent the largest and smallest possible values for the type, MAXVALUE and MINVALUE. Compute and compare the signs of the operands. a. If either value is zero, then neither addition nor subtraction can overflow. Skip remaining tests. b. If the signs are opposite, then addition cannot overflow. Skip remaining tests. c. If the signs are the same, then subtraction cannot overflow. Skip remaining tests. Test for positive overflow of MAXVALUE. a. If both signs are positive and MAXVALUE - A < B, then addition will overflow. b. If the sign of B is negative and MAXVALUE - A < -B, then subtraction will overflow. Test for negative overflow of MINVALUE. a. If both signs are negative and MINVALUE - A > B, then addition will overflow. b. If the sign of A is negative and MINVALUE - A > B, then subtraction will overflow. Otherwise, no overflow.
签名溢出测试，乘法和除法:
Obtain the constants that represent the largest and smallest possible values for the type, MAXVALUE and MINVALUE. Compute and compare the magnitudes (absolute values) of the operands to one. (Below, assume A and B are these magnitudes, not the signed originals.) a. If either value is zero, multiplication cannot overflow, and division will yield zero or an infinity. b. If either value is one, multiplication and division cannot overflow. c. If the magnitude of one operand is below one and of the other is greater than one, multiplication cannot overflow. d. If the magnitudes are both less than one, division cannot overflow. Test for positive overflow of MAXVALUE. a. If both operands are greater than one and MAXVALUE / A < B, then multiplication will overflow. b. If B is less than one and MAXVALUE * B < A, then division will overflow. Otherwise, no overflow.
注意:MINVALUE的最小溢出由3处理，因为我们取的是绝对值。然而,如果 ABS(MINVALUE) > MAXVALUE，那么我们将会有一些罕见的假阳性。
下溢测试类似，但涉及EPSILON(大于零的最小正数)。
一种简单的方法是重写所有操作符(特别是+和*)，并在执行操作之前检查是否有溢出。
为了扩展Head Geek的答案，有一种更快的方法来执行addition_is_safe;
bool addition_is_safe(unsigned int a, unsigned int b)
{
unsigned int L_Mask = std::numeric_limits<unsigned int>::max();
L_Mask >>= 1;
L_Mask = ~L_Mask;
a &= L_Mask;
b &= L_Mask;
return ( a == 0 || b == 0 );
}
这使用了机器架构安全，64位和32位无符号整数仍然可以正常工作。基本上，我创建了一个掩码，它将屏蔽除最重要的位外的所有内容。然后，对两个整数进行掩码，如果其中任何一个没有设置该位，则加法是安全的。
如果在某个构造函数中预初始化掩码，这将更快，因为它永远不会改变。
mozilla::CheckedInt<T>为整数类型T提供溢出检查的整数数学(使用clang和gcc上可用的编译器intrinsic)。该代码是在MPL 2.0下编写的，并且依赖于三个(integertypetrait .h, Attributes.h和Compiler.h)其他仅针对标头的非标准库标头以及mozilla特定的断言机制。如果导入代码，可能需要替换断言机制。
