运算符重载(+)、(++)、([])、(=)和(*)
运算符重载(+)、(++)、([])、(=)和(*)
提问于 2014-10-17 08:03:47
我正在尝试重载类中的一些运算符。
要加载的运算符：(+)、(=)、(++)、(*)和([])
所以基本上我有一个类A和一个数组，我想使用operator []返回vindex
我想使用运算符+来添加相应的数组元素
我试过了，但是我好像错过了什么。
我在main例程中有一个测试用例，每当我运行我的代码时，我都会得到以下错误：
(78): error C2679: binary '=' : no operator found which takes a right-hand operand of type 'int' (or there is no acceptable conversion)
(67): could be 'A &A::operator =(const A &)'
(82): error C2679: binary '=' : no operator found which takes a right-hand (67): could be 'A &A::operator =(const A &)'
(87): warning C4620: no postfix form of 'operator ++' found for type 'A', using prefix form
(6) : see declaration of 'A'我的代码：
#include <iostream>
using namespace std;
#define NUM 10
class A
{
int v[NUM];
public:
A() { }
A(int x[]) {
for (int i = 0; i<NUM; i++)
v[i] = x[i];
}
// (1) operator: [ index ], which returns v[index]
int &operator [](int i)
{
return v[i];
}
// (2) operator: +, which adds the corresponding array elements
int operator+(const A& b)
{
for (int i = 0; i < NUM; i++)
{
return v[i] = v[i] + v[i + 1];
}
}
// (3) operator: *, which multiplies the corresponding array Elements
int operator*(const A& b)
{
for (int i = 0; i < NUM; i++)
{
return v[i] = v[i] * v[i + 1];
}
}
// (4) operator: ++, which adds each array element by 1
int operator++()
{
for (int i = 0; i < NUM; i++)
{
return v[i++];
}
}
// (5) operator: =, which assigns all the array elements from one object to another
/* void operator=(const A& b)
{
} */
// a print function to print out all the array elements
void print();
};
int main()
{
int x[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
A b1(x), b2, b3, b4, b5;
cout << "b1[5]: " << b1[5] << endl;
b2 = b1;
cout << "b2: ";
b2.print();
b3 = b1 + b2;
cout << "b3: ";
b3.print();
b4 = b1*b2;
cout << "b4: ";
b4.print();
b5 = b4;
b5++;
cout << "b5: ";
b5.print();
}我更新了我的代码，现在我只有一个错误(比以前更好)：
错误在第76行：'A::A(const A &)‘：无法将参数1从'int 10’转换为'const A &‘
我看到它不能将x从int转换为const A&这是有道理的。那我应该把operator[]改成整型吗？
#include <iostream>
using namespace std;
#define NUM 10
class A
{
int v[NUM];
public:
A() {
for (int i = 0; i < NUM; i++)
v[i] = 0;
}
/* operator ([]) */
int &operator[](int i)
{
return v[i];
}
/* operator (+) */
A operator+(const A& b)
{
A result;
for (int i = 0; i < NUM; i++)
{
result.v[i] = v[i] + b.v[i];
}
}
/* operator (*) */
A operator*(const A& b)
{
A result;
for (int i = 0; i < NUM; i++)
{
result.v[i] = v[i] * b.v[i];
}
}
/* operator (=) */
A operator=(const A& b)
{
A result;
for (int i = 0; i < NUM; i++)
{
result.v[i] = b.v[i];
}
}
/* operator (++) */
A operator++(int)
{
A result;
for (int i = 0; i < NUM; i++)
{
result.v[i]++;
}
}
void print();
};
int main()
{
int x[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
A b1(x), b2, b3, b4, b5;
cout << "b1[5]: " << b1[5] << endl;
b2 = b1;
cout << "b2: ";
b2.print();
b3 = b1 + b2;
cout << "b3: ";
b3.print();
b4 = b1*b2;
cout << "b4: ";
b4.print();
b5 = b4;
b5++;
cout << "b5: ";
b5.print();
}回答 1
Stack Overflow用户
发布于 2014-10-17 08:08:59
operator+的返回类型(以及其他一些运算符)和早期的“return”调用是错误的。下面是如何修复它的方法。
A() {
for (int i = 0; i < NUM; i++) {
v[i] = 0;
}
}
A operator+(const A& b) {
A result; // result is initialized with 0s.
for (int i = 0; i < NUM; i++) {
result.v[i] = v[i] + b.v[i];
}
return result;
}另一种方法是，不更改构造函数：
A operator+(const A& b) {
A result = *this;
for (int i = 0; i < NUM; i++) {
result.v[i] += b.v[i];
}
return result;
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/26415873
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