C++ 大数溢出检测(unsigned long long)

Question

我正在处理大整数(unsigned long long)并且需要防止溢出情况。无论是否确实存在，代码都会抛出异常:

try
{
    unsigned long long y = std::numeric_limits<unsigned long long>::max() - 2;
    unsigned long long z = 1;
    int size = - 1;
    if((y+z) ^ y < 0) //If y+z causes overflow its sign will be changed => y and (y+z) will have opposite signs
        throw std::overflow_error("overflow of y+z");
    //int* myarray= new int[size]; VS Debug Library catches it earlier than catch()
    printf("%d\n", y*(y+z));
}
catch(exception& e)
{
    cout << e.what() << endl;
}

由于它已经是最大的数据类型(64 位)，因此没有提升到更大的空间。

新代码:

try
{
    unsigned long long int y = std::numeric_limits<unsigned long long int>::max() - 2;
    unsigned long long int z = std::numeric_limits<unsigned long long int>::max() / 2;
    unsigned long long delta = std::numeric_limits<unsigned long long int>::max() - y;
    int size = - 1;
    if(z > delta) //If y+z causes overflow its sign will be changed => y and (y+z) will have opposite signs
        throw std::overflow_error("overflow of y+z");
    //int* myarray= new int[size]; VS Debug Library catches it earlier than catch()
    printf("%d\n", (y+z));
}
catch(exception& e)
{
    cout << e.what() << endl;
}

Answer 1

y < 0将始终为假，任何 xor 0 将始终是那个东西(您是否错过了 < 的评估优先级高于 ^ ？)。

因此除非x + y国防部 <the max value> happens to equal 0, you will throw (and probably appears to always throw in practice unless you contrived specific inputs).

Perhaps what you meant to do was something like this: if((std::numeric_limits<unsigned long long>::max() - y) < z) throw ...;