自動轉換時出錯

Question

class Sample
{
public:
  Sample();
  Sample(int i);
  Sample(Sample& s);
  ~Sample();
};

Sample::Sample()
{
  cout<<"Default constructor called\n";
}

Sample::Sample(int i)
{
  cout<<"1-argument constructor called\n";
}

Sample::Sample(Sample& s)
{
  cout<<"Copy constructor called\n";
}

Sample::~Sample()
{
  cout<<"Destructor called\n";
}

void Fun(Sample s)
{

}

int main()
{
  Sample s1;
  Fun(5);

  return 0;
}

我期望隱式轉換為5。但是，當我編譯以上代碼時，出現以下錯誤：

main.cpp:7:8: error: no matching function for call to ‘Sample::Sample(Sample)’
main.cpp:7:8: note: candidates are:
Sample.h:10:3: note: Sample::Sample(Sample&)
Sample.h:10:3: note:   no known conversion for argument 1 from ‘Sample’ to ‘Sample&’
Sample.h:9:3: note: Sample::Sample(int)
Sample.h:9:3: note:   no known conversion for argument 1 from ‘Sample’ to ‘int’
Sample.h:8:3: note: Sample::Sample()
Sample.h:8:3: note:   candidate expects 0 arguments, 1 provided
Helper.h:6:13: error:   initializing argument 1 of ‘void Fun(Sample)’

問題是什么？ 當我刪除復制構造函數時，以上代碼成功編譯。

提前致謝。

Answer 1

臨時不能綁定到非常量引用。 您的副本構造函數應為：

Sample::Sample(const Sample&)

刪除它會告訴編譯器生成一個簡單的文件，該文件將具有上述簽名。