C ++使用链表，模板和堆栈设计用于大整数加法和减法的类

Question

目前我只实施了添加部分。

在main函数中，我的程序实际上可以分别打印出n2 n3，但它不能打印出下一个“n2 + n3”的情况，因为它会得到运行时错误。

设置断点后，我发现当bigInteger＆n传入case“n2 + n3”时，follow语句不起作用，因此n的内容不会被修改。

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以下是该程序的三段代码。 还有另一个链表头文件，它定义了node，iterator（p，q）和一些成员函数（如insert（）和length（））的用法。

不管怎样，谢谢你的帮助。

class bigInteger
{
private:
int sign;                       // set 0 for positive, 1 for negative
linkedListType<int> digits;     // internal linked-list for storing digits in reverse order

public:
bigInteger();                           // default constructor
bigInteger(const bigInteger& other);    // copy constructor

// Overload constructor
// Use an numerical string to construct this bigInteger
// For negative number, the first char in the string is '-'
// e.g. "-12345"
bigInteger(const string& number);       

// overload the assignment operator
const bigInteger& operator= (const bigInteger& other);

// Return a new bigInteger that is equal to *this + other
// The contents of this and other should not be modified
bigInteger& operator+ (bigInteger& other);

// Return a new bigInteger that is equal to *this - other
// The contents of this and other should not be modified
bigInteger& operator- (bigInteger& other);

// Print a big integer
// Since the digits are stored in reverse order in the internal
// list, you should print the list reversely
// Print "undefined" if the digits list is empty
friend ostream& operator<<(ostream& os, bigInteger& n);
};

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第二个是关于成员函数的实现。

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 bigInteger& bigInteger::operator+ ( bigInteger& other ) { bigInteger resultBI; //saving the answer bigInteger forSwap; //not used bool explicitSign = 0; //.. stack<int> resultStack; // stack saving the answer, later converting to Type BigInteger int result; //local var for addition int carry = 0; //.. linkedListIterator<int> p = digits.begin(); //iterator marking the first node linkedListIterator<int> q = other.digits.begin(); //.. if ( this->digits.length() >= other.digits.length() ) { while ( q != NULL ) { result = ( *p + *q + carry ) % 10; // '*' acts like dereference carry = ( *p + *q + carry ) / 10; ++p; // "++' acts like increment to the link ++q; resultStack.push(result); } while ( p != NULL ) //remaining carry { result = ( *p + carry ) % 10; carry = ( *p + carry ) / 10; ++p; resultStack.push(result); } } if ( this->digits.length() < other.digits.length() ) { while ( p != NULL ) { result = ( *p + *q + carry ) % 10; carry = ( *p + *q + carry ) / 10; ++p; ++q; resultStack.push(result); } while ( q != NULL ) { result = ( *q + carry ) % 10; carry = ( *q + carry ) / 10; ++q; resultStack.push(result); } } if ( carry != 0 ) //push and remaining carry { resultStack.push(carry); } while ( !resultStack.empty() ) //convert the stack to Type bigInteger { resultBI.digits.insert ( resultStack.top() ); resultStack.pop(); } if ( explicitSign == 1 ) // not used resultBI.sign = 1; return resultBI; } 

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最后一部分是主要功能。

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 int main() { //-------- Test Case 1 bigInteger n1; bigInteger n2("987654321"); bigInteger n3("123456789"); cout << "n1 = "; cout << n1 << endl; // undefined cout << "n2 = "; cout << n2 << endl; // 987654321 cout << "n3 = "; cout << n3 << endl; // 123456789 //-------- Test Case 2 cout << "n2 + n3 = "; cout << n2 + n3 << endl; // 1111111110 //run-time error return 0; } 

Answer 1

问题是您通过引用返回结果返回一个局部变量，该变量将在您返回后立即消失。

将运算符定义为按值返回，它应该更好地工作：

bigInteger operator+ (bigInteger& other);   // return by value.  

附加信息：

• 这篇文章的指南很好地解释了这个问题（“按值返回对象”一节）。

• 这里有一篇关于运算符重载的文章 ，该文章由运营商家族解释了采取的方法和要避免的陷阱。