我在哪裏使用賦值運算符？

Question

我對類很陌生，雖然我編寫了所有其他代碼，但在我的兩個成員函數結束時，我停滯不前。

這裏是我的頭：

class bignum 
{ 
public: 
// Constructors. 
bignum(); 
bignum(int num_digits); 
bignum(const string &digits); 
bignum(const bignum &other); 

// Destructors. 
~bignum(); 

// Assignment operator. 
bignum &operator=(const bignum &other); 

// Accessors 
int digits() const; 
int as_int() const; 
string as_string() const; 
void print(ostream &out) const; 
bignum add(const bignum &other) const; 
bignum multiply(const bignum &other) const; 
bool equals(const bignum &other) const; 
int PublicNumberTest; 

private: 
// Pointer to a dynamically-allocated array of integers. 

int *digit; 

// Number of digits in the array, not counting leading zeros. 

int ndigits; 
}; 
#endif 

，這裏是我的成員函數：

bignum bignum::multiply(const bignum& other) const{ 
bignum product; 
bignum row; 
int carry = 0; 
int sum = 0; 
int j = 0; 
int *temp_row = new int[]; 
for (int i = 0; i < ndigits-1; i++){ 
    carry = 0; 
    temp_row[i] = 0; 
    for (j; j < other.digits - 1; j++){ 
     sum = digit[i] * other.digit[j] + carry; 
     temp_row[i + j] = sum % 10; 
     carry = sum/10; 
    } 
    if (carry>0) 
     temp_row[i + j] = carry; 
    row = row operator+temp_row //This is what I don't understand. How can I 
     product = product.add(row); //assign the contents of temp_row? 

} 
} 

還有另一種，但它基本上是同樣的問題。我有一個數組，我想複製到我的...課程的內容和地方？我猜？謝謝閱讀。

Answer 1

如果你想使用temp_row[]作爲一組數字對this對象，你可以這樣做：

delete [] digits; 
digits = new_row; 

你或許應該更新ndigits爲好。

Answer 2

2周的方式這樣做的：「舊」的方式，這應該是非常相似的調用，則析構函數拷貝構造函數：這也可以寫成

bignum& bignum::operator=(const bignum& other) 
{ 
    delete[] digit; 
    ndigits = other.ndigits; 
    digit = new int[ndigits]; 
    for (int i = 0; i != ndigits; ++i) { 
     digit[i] = other.digit[i]; 
    } 
    return *this; 
} 

代碼：

bignum& bignum::operator=(const bignum& other) 
{ 
    this->~bignum(); 
    new(this) bignum(other); 
    return *this; 
} 

這種老方式現在不鼓勵，因爲不是例外的安全。

「新」的方式，這是異常安全：您定義不會拋出任何異常的掉期（）成員方法，並用它來實現運營商=（）：

void bignum::swap(bignum& other) noexcept 
{ 
    std::swap(ndigits, other.ndigits); 
    std::swap(digit, other.digit); 
} 

bignum& bignum::operator=(const bignum& other) 
{ 
    bignum tmp(other); 
    this->swap(tmp); 
    return *this; 
} 

Answer 3

而不是鍛鍊身體你多位數乘法，我只留下了幾件事情困擾了，

是否要執行C = A b或A = A b'也就是說，你打算 乘兩個參數，產生一個臨時的，並將其分配給第三個， 或者你打算只乘以一個參數b的對象a？

我的目的是C = A * B

然後你需要分配（新）目標BIGNUM，

使用朋友運算符重載應解決您的需求;這是常見的成語來定義二元運算符（請注意，我並沒有解決您的分配的號碼無誤），

friend bignum operator+(const bignum& a, const bignum& b); //in class 

//you want to calculate A * B --> C 
//you need a temporary place to calculate C, then produce a new C, 
bignum bignum::multiply(const bignum& A, const bignum& B) const 
{ 
    //Your A is this, A.digit[]; 
    //Your B is other, B.digit[]; 
    //Yout C is temp, a buffer for calculations, 
    bignum temp; 

    int carry = 0; 
    int sum = 0; 
    int j = 0; 
    //you don't need this, you have temp.digit[]; 
    for (int i = 0; i < this->ndigits-1; i++) 
    { 
     carry = 0; 
     temp.digit[i] = 0; 
     //what do you want to initialize j to? 
     for (j; j < B.ndigits - 1; j++) 
     { 
      sum = A.digit[i] * B.digit[j] + carry; 
      temp.digit[i + j] = sum % 10; 
      carry = sum/10; 
     } 
     if (carry>0) 
      temp.digit[i + j] = carry; 
    } 
    return bignum(temp); //allocate a new bignum 
} 

對於如何重載二元運算符的例子參見here ...

做你做的方式也有缺點（返回一個常量，不能鏈條超過兩個加法togethere），但基本上你做同樣的事情，但使用(this->)指針，

//you want to calculate A * B --> C 
//you need a temporary place to calculate C, then produce a new C, 
bignum bignum::multiply(const bignum& other) const 
{ 
    //Your A is this, this->digit[]; 
    //Your B is other 
    //Yout C is temp product, a buffer for calculations, 
    bignum temp; 

    int carry = 0; 
    int sum = 0; 
    int j = 0; 
    for (int i = 0; i < this->ndigits-1; i++) 
    { 
     carry = 0; 
     temp.digit[i] = 0; 
     for (j; j < other.ndigits - 1; j++) 
     { 
      sum = this->digit[i] * other.digit[j] + carry; 
      temp.digit[i + j] = sum % 10; 
      carry = sum/10; 
     } 
     if (carry>0) 
      temp.digit[i + j] = carry; 
    } 
    //bignum product; 
    //return *this = temp; //do this, 
    return bignum(temp); //or you should probably return a new object 
} 

你應該return a new object。