返回局部变量或临时变量的地址警告 - 编程论坛

#2

rjsp2015-05-25 14:08

DateTime* DateTime::operator&(DateTime input)
{
   return &input;
}
看你代码，无从下手，因为不知道你想干什么。（你需要多看书，多思考，两者都缺）
DateTime* DateTime::operator&()
{
    return this;
}
const DateTime* DateTime::operator&() const
{
    return this;
}
当然，虽然可以编译通过，但还是不懂你想干什么

#3

OscarWu2015-05-25 14:55

谢谢回复。我想完成下面一些操作：
   typedef struct {
      long        mjd;
      double      fracOfDay;
   }  MJD;
   class  DateTime
   {
      friend ostream &operator<<( ostream &output, DateTime &dt );

      public:
         // constructors
         DateTime();
         DateTime( Time time );

         // destructor
         ~DateTime();

         // initializers
         void  SetTime( Time time );

         // selectors
         GPSTime      GetTime();
         // manipulators
         const DateTime &operator=(const DateTime &DT2);
     DateTime* operator&(DateTime input);
         DateTime operator + ( const double days );
         double   operator - ( const DateTime &DT2 );
         bool    operator == ( const DateTime &DT2 );
         bool    operator != ( const DateTime &DT2 );
         bool    operator >  ( const DateTime &DT2 );
         bool    operator >= ( const DateTime &DT2 );
         bool    operator <  ( const DateTime &DT2 );
         bool    operator <= ( const DateTime &DT2 );
      private:
         long        mjd;
         double      fractionOfDay;
   };

DateTime* DateTime::operator&(DateTime input)
{
   return &input;
}

// const return avoids: (a1 = a2 ) = a3
const DateTime &DateTime::operator=(const DateTime &DT2)
{
   if( &DT2 != this ) // avoids self assignment
   {
     mjd = DT2.mjd;
     fractionOfDay = DT2.fractionOfDay;
   }
   return *this;
}

#4

wmf20142015-05-25 22:30

问题出在 return &input;input是一个不确定的局部变量，用它作为指针可能是一个无有效值的指向，你如果改成下述代码可能没有警告：
DateTime* DateTime::operator&(DateTime input)
{
   DateTime *p;
   p=new DateTime;
   *p=input;
   return p;
}

#5

OscarWu2015-05-26 13:05

谢谢，没有警告了。