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。
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)。
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)。