我遇到错误阅读,没有从'int'转换为'TrashCan'的功能样式的匹配转换。
这是标头中的声明:
class TrashCan
{
friend TrashCan operator +( TrashCan& left,
TrashCan& right);
public:
TrashCan();
int size=0;
int item=0;
void setSize(int);
void addItem();
这是我的实现:
TrashCan operator +(const TrashCan& left,
const TrashCan& right) {
TrashCan t= TrashCan( left.size + right.size );
return( t );
}
这是底部的主要运算符:
int main( ) {
cout << "Welcome to My TrashCan Program!" << endl;
TrashCan myCan;
TrashCan yourCan;
yourCan.setSize( 12 );
myCan.setSize( 12 );
yourCan.addItem( );
yourCan.addItem( );
myCan.addItem( );
myCan.printCan();
yourCan.printCan();
//TrashCan combined = yourCan + myCan;
最佳答案
编辑
您可以这样声明构造函数:TrashCan();
但是您可以这样称呼:TrashCan t= TrashCan( left.size + right.size );
。
您需要第二个构造函数,例如TrashCan(int nsize) : size(nsize) { }
。
godel9已经在答案中添加了答案,但这是一个工作代码示例:
#include <iostream>
class TrashCan {
// Your declaration did not match your definition
// Need to put const here
friend TrashCan operator +(const TrashCan& left,
const TrashCan& right);
public:
TrashCan(int nsize) : size(nsize) { }
~TrashCan() { }
int size;
};
TrashCan operator +(const TrashCan& left,
const TrashCan& right)
{
TrashCan t= TrashCan( left.size + right.size );
return( t );
}
int main()
{
TrashCan tc1(10);
TrashCan tc2(20);
std::cout << (tc1 + tc2).size;
// outputs 30
return 0;
}
关于c++ - C++运算符重载错误,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/19759748/