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我创建了一个名为Rectangle的类并实现了一个参数化构造函数并创建了 Rectangle 类的三个实例

#include<iostream>

using namespace std;

class Rectangle {
   private:
      int length;
      int bredth;
   public:
      Rectangle(int l=0, int b=0) {
         length=l;
         bredth=b;
      }
};
int main() {
   system("clear");

   Rectangle r1;
   Rectangle r2(10, 5);
   Rectangle r3=Rectangle(10, 5);
   
   return 0;
}

编译此源代码文件时没有问题。

但是在实现复制构造函数之后，我得到了一个错误。

#include<iostream>

using namespace std;

class Rectangle {
   private:
      int length;
      int bredth;
   public:
      Rectangle(int l=0, int b=0) {
         length=l;
         bredth=b;
      }
      Rectangle(Rectangle &r) {
         length=r.length;
         bredth=r.bredth;
      }
};
int main() {
   system("clear");

Rectangle r1;
   Rectangle r2(10, 5);
   Rectangle r3=Rectangle(10, 5);
   Rectangle r4=r3;

   return 0;
}

错误信息

3.cpp:24:14: error: no matching constructor for initialization of 'Rectangle'
   Rectangle r3=Rectangle(10, 5);
             ^  ~~~~~~~~~~~~~~~~
3.cpp:10:7: note: candidate constructor not viable: no known conversion from 'Rectangle' to 'int' for 1st argument                                                                        Rectangle(int l=0, int b=0) {                                                             ^
3.cpp:14:7: note: candidate constructor not viable: expects an lvalue for 1st argument
      Rectangle(Rectangle &r) {
      ^
1 error generated.

为什么在实现复制构造函数之后会发生这种情况，并且只针对显式构造函数调用？

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0 回答 0