超载运算符中的分段错误=

Question

我在为一个FeatureRandomCounts类重载赋值运算符时遇到了一个seg错误,它有_rects作为指针成员指向FeatureCount和大小rhs._dim的数组,其他日期成员是非指针:

FeatureRandomCounts &  FeatureRandomCounts::operator=(const FeatureRandomCounts &rhs)  
{  
  if (_rects) delete [] _rects;  

  *this = rhs;  // segment fault

  _rects = new FeatureCount [rhs._dim];  
  for (int i = 0; i < rhs._dim; i++)  
  {  
    _rects[i]=rhs._rects[i];  
  }  

  return *this;    
}

有人有一些线索吗？感谢致敬!

Answer 1

*this = rhs;

调用operator =(),这是你正在编写的函数.提示无限递归,堆栈溢出,崩溃.

此外,如果您使用std :: vector而不是C样式数组,则可能根本不需要实现operator =().

Answer 2

如上所述,你有无限的递归; 但是,除此之外,这是实现op =的一种万无一失的方法:

struct T {
  T(T const& other);
  T& operator=(T copy) {
    swap(*this, copy);
    return *this;
  }
  friend void swap(T& a, T& b);
};

写一个正确的拷贝ctor和swap,并为你处理异常安全和所有边缘情况!

该拷贝参数按值传递,然后改变.当销毁副本时,将处理当前实例必须销毁的任何资源.这遵循当前的建议并干净地处理自我分配.

#include <algorithm>
#include <iostream>

struct ConcreteExample {
  int* p;
  std::string s;

  ConcreteExample(int n, char const* s) : p(new int(n)), s(s) {}
  ConcreteExample(ConcreteExample const& other)
  : p(new int(*other.p)), s(other.s) {}
  ~ConcreteExample() { delete p; }

  ConcreteExample& operator=(ConcreteExample copy) {
    swap(*this, copy);
    return *this;
  }

  friend void swap(ConcreteExample& a, ConcreteExample& b) {
    using std::swap;
    //using boost::swap; // if available
    swap(a.p, b.p); // uses ADL (when p has a different type), the whole reason
    swap(a.s, b.s); // this 'method' is not really a member (so it can be used
                    // the same way)
  }
};

int main() {
  ConcreteExample a (3, "a"), b (5, "b");
  std::cout << a.s << *a.p << ' ' << b.s << *b.p << '\n';
  a = b;
  std::cout << a.s << *a.p << ' ' << b.s << *b.p << '\n';
  return 0;
}

注意到它与手动管理构件(p)或RAII/SBRM式部件(小号).