主页 / 问题 / 839844
Accepted
user2173645
Asked:2020-06-09 18:39:08 +0000 UTC2020-06-09 18:39:08 +0000 UTC

将除法规则更改为 0

  • 772

需要改变什么，或者是否可以通过钩子（拦截）来改变划分规则？在我的任务中，将数字除以 0 应该得到 0。我有兴趣更改规定的算法

c++

3 个回答

  1. Best Answer
    #include <iostream>    
using namespace std;
class Int_type {
    int k;
public:
    Int_type(int i = 0) : k(i) {}
    Int_type& operator ++() { ++k; return *this; }
    Int_type operator ++(int) { Int_type t; ++(t.k); return t; }
    Int_type& operator =(const int& i) { k = i; return *this; }
    Int_type& operator /=(const int& i) { k = i ? k / i : 0; return *this;}
    operator  int() const { return k; }
};
inline Int_type operator /(const Int_type& n, const int& i)
{
    Int_type t = n;
    t /= i;
    return t;
}

    您可以使用类对象而不是 int 对象并除以零

    int main()
{   
    Int_type i;
    i = 19;
    ++i;
    cout << i << endl << i/ 2 << endl << i - 2 <<endl << i * 2  << endl 
         << i/0;
    return 0;
}
    • 9

  2. 非常感谢您的回答。然而，由于长时间没有答案，我通过异常过滤器想出了自己的解决方案。应该适用于 32 位应用程序中的所有整数。在这里，当引发异常时，过滤器会检查操作类型并将寄存器中的零替换为一。然后跳转到下一条指令

    #include "stdafx.h"
#include <Windows.h>
#include <iostream>

using namespace std;

LONG WINAPI ExceptionFilter(PEXCEPTION_POINTERS pExceptionInfo)
{
    cout << "Exception code: 0x" << hex << (int)pExceptionInfo->ExceptionRecord->ExceptionCode << " Exception address: 0x" << (DWORD)pExceptionInfo->ContextRecord->Eip << endl;
    if ((int)pExceptionInfo->ExceptionRecord->ExceptionCode == 0xc0000094)
    {
        DWORD presentEIP = (DWORD)pExceptionInfo->ContextRecord->Eip;
        byte EIPvalue[2];
        EIPvalue[0] = *(byte*)presentEIP;
        EIPvalue[1] = *(byte*)(presentEIP + 0x1);
        if (EIPvalue[0] == 0xF7 && EIPvalue[1] == 0xF9)
        {
            pExceptionInfo->ContextRecord->Ecx = 1;
            cout << "Forced ECX to 1 [1]" << endl;
        }
        else if (EIPvalue[0] == 0xF7 && EIPvalue[1] == 0x7D)
        {
            _asm add DWORD PTR SS : [EBP - 8], 1;
            pExceptionInfo->ContextRecord->Eip += 3;
            cout << "Forced ECX to 1 [2]" << endl;
        }
        else return EXCEPTION_CONTINUE_SEARCH;

        return EXCEPTION_CONTINUE_EXECUTION;
    }

    return EXCEPTION_CONTINUE_SEARCH;
}

int f(int x, int y, int
    z) {
    return 2 + x / (2 / (y - z));
}

int main()
{
    AddVectoredExceptionHandler(true, (PVECTORED_EXCEPTION_HANDLER)ExceptionFilter); // Registers our vectored exception handler which is going to catch the exceptions thrown.

    int res = f(2, 3, 3); //F7F9
    cout << "Result: " << dec << res << endl; 

    getchar();

    int zero = 1 * 0;//F77D F8
    int i;
    i = 1 + 5 / zero;
    cout << "Result: " << dec << i << endl;
    getchar();

    return 0;
}
    • 2

  3. 这就是我所做的并且有效：

    #include <iostream>    
using namespace std;

class IntType {
    int k;

public:
    IntType(int i = 0) : k(i) {}
    IntType& operator ++() { ++k; return *this; }
    IntType operator ++(int) { IntType t; ++(t.k); return t; }
    IntType& operator =(const int& i) { k = i; return *this; }
    IntType& operator /=(const int& i) { k = i ? k / i : 0; return *this;}
    operator  int() const { return k; }
};

inline IntType operator /(const IntType& n, const int& i)
{
    IntType t = n;
    t /= i;
    return t;
}
    • 0

相关问题