Question

我使用默认构造函数，复制构造函数，赋值运算符和析构函数声明了一个简单的结构。但是，该结构不能用作std :: map的值类型。

以下是代码：

#include <string.h>
#include <iostream>
#include <string>
#include <map>

class Foo;
std::ostream & operator<<(std::ostream & os, const Foo & v);

typedef unsigned char BYTE;

struct Foo {
  char type_;        // char to label type
  size_t num_;       // number of elem, useful if array
  size_t total_;     // total memory
  BYTE * data_;      // content of memory

  Foo(const char * t) : type_('c'), num_(strlen(t)+1), total_(strlen(t)+1), data_(NULL) {
    data_ = new BYTE[total_];
    memcpy(data_, t, total_-1);
    ((char *)data_)[total_-1] = '\0';
  }

  Foo() : type_(), num_(), total_(), data_(NULL) {}

  Foo(const Foo& rhs) : type_(rhs.type_), num_(rhs.num_), total_(rhs.total_), data_(NULL) {
    if (total_) {
    data_ = new BYTE[total_];
    memcpy((char *)data_, (const char *)&rhs.data_, total_);
    }
  }

  Foo & operator=(const Foo& rhs) {
    if (&rhs != this) {
      releaseData();
      type_ = rhs.type_;
      num_ = rhs.num_;
      total_ = rhs.total_;
      data_ = new BYTE[total_];
      memcpy(data_, &rhs.data_, total_);
    }
    return *this;
  }

  ~Foo() {
    releaseData();
  }

private:
  void releaseData() {
    delete [] data_; data_ = NULL;
  }
};

inline std::ostream & operator<<(std::ostream & os, const Foo & v) {
  os << "(type: " << v.type_ << ", num: " << v.num_ << ", total: " << v.total_ << ", data: " << (const char *)v.data_ << ", data addr: " << (void *)v.data_ << ")";
  return os;
}


int main() {
  Foo c("/home/data/");
  std::map<std::string, Foo> store;
  store["abc"] = Foo("/home/data/");

  std::cout << c << std::endl;
  std::cout << store["abc"] << std::endl;
}

使用gcc 4.9.2在Linux上编译代码。第一个打印正确打印出字符串，但第二个没有。

这段代码有什么问题？

Answer 1

您在复制构造函数和赋值运算符中对memcpy()的调用都是错误的。您在两种情况下都指定&rhs.data_作为来源：

memcpy((char *)data_, (const char *)&rhs.data_, total_);
...
memcpy(data_, &rhs.data_, total_);

使用'＆amp;'以这种方式，您正在复制紧跟在内存中data_成员之后的随机字节， NOT data_指向的字节。

由于data_已经是指向要复制的数据的指针，因此您需要删除&并按原样使用rhs.data_（并且不需要类型 - 铸）：

memcpy(data_, rhs.data_, total_);

或者，摆脱所有这些手动逻辑，只需使用std::string或std::vector，让编译器和STL为您处理所有内存管理和数据复制：

struct Foo {
  char type_;        // char to label type
  std::string data_; // content of memory

  Foo(const char * t) : type_('c'), data_(t) {}

  Foo() : type_() {}
};

inline std::ostream & operator<<(std::ostream & os, const Foo & v) {
  os << "(type: " << v.type_ << ", num: " << v.data_.length() << ", total: " << v.data_.capacity() << ", data: " << v.data_.c_str() << ", data addr: " << (void *)v.data_.data() << ")";
  return os;
}

struct Foo {
  char type_;        // char to label type
  std::vector<BYTE> data_; // content of memory

  Foo(const char * t) : type_('c') { std::copy(t, t+(strlen(t)+1), std::back_inserter(data_)); }

  Foo() : type_() {}
};

inline std::ostream & operator<<(std::ostream & os, const Foo & v) {
  os << "(type: " << v.type_ << ", num: " << v.data_.size() << ", total: " << v.data_.capacity() << ", data: " << (const char*) &v.data_[0] << ", data addr: " << (void *)&v.data_[0] << ")";
  return os;
}

C ++ std :: map中对值的要求？

1 个答案: