1

我正在从事一个大学项目，我们将在其中将一些 c++ 字符串类实现为 Mystring。我正在研究重载赋值运算符，这是它的当前代码：

Mystring& Mystring::operator=(const Mystring& orig)
{
    if(this != &orig)
    {
        delete ptr_buffer;
        len = orig.len;
        buf_size = orig.buf_size;
        ptr_buffer = orig.ptr_buffer;
    }
    return *this;

}

ptr_buffer，长度。和 buf_size 是 Mystring 类的三个私有变量。这是我要测试的主要程序：

void check (const Mystring s, const string name)
{
    cout << "checking " << name << endl;
    cout << name << " contains " << s << endl;
    cout << name << " capacity() is " << s.capacity() << endl;
    cout << name << " length() is " << s.length() << endl;
    cout << name << " size() is " << s.size() << endl;
    cout << name << " max_size() is " << s.max_size() << endl << endl;
}


int main()
{
    Mystring s1("Hi there!");
    check(s1, "s1");

    Mystring s2("Testing before assignment!");
    check(s2, "s2");
    s2 = s1;
    check(s2, "s2");
    return 0;
}

这就是输出：

checking s1
s1 contains Hi there!
s1 capacity() is 10
s1 length() is 9
s1 size() is 9
s1 max_size() is 1073741820

checking s2
s2 contains Testing before assignment!
s2 capacity() is 27
s2 length() is 26
s2 size() is 26
s2 max_size() is 1073741820

checking s2
s2 contains Hi there!
s2 capacity() is 10
s2 length() is 9
s2 size() is 9
s2 max_size() is 1073741820

free(): double free detected in tcache 2

Process finished with exit code 134 (interrupted by signal 6: SIGABRT)

如您所见，赋值确实可以设置所有成员变量，但是我得到一个非零退出代码和 free(): double free detected in tcache 2 错误。我做错了什么，这个错误是什么意思？我已经确认退出代码来自调用分配，因为当我注释掉 s2 = s1; 时，它以退出代码 0 完成。

4

2 回答 2

3

在这个复制赋值运算符中

Mystring& Mystring::operator=(const Mystring& orig)
{
    if(this != &orig)
    {
        delete ptr_buffer;
        len = orig.len;
        buf_size = orig.buf_size;
        ptr_buffer = orig.ptr_buffer;
    }
    return *this;
}

至少有两个问题。如果指针指向的字符数组ptr_buffer是动态分配的，那么您必须使用运算符delete []而不是delete

delete [] ptr_buffer;

第二个问题是这个赋值之后

ptr_buffer = orig.ptr_buffer;

两个指针指向同一个动态分配的内存。

您需要分配一个新的扩展内存并将分配对象的字符串复制到那里。

运算符至少可以通过以下方式定义

Mystring & Mystring::operator =( const Mystring &orig )
{
    if(this != &orig)
    {
        if ( buf_size != orig.buf_size )
        {
            delete [] ptr_buffer;
            ptr_buffer = new char[orig.buf_size];
            buf_size = orig.buf_size;
        } 
        len = orig.len;
        strcpy( ptr_buffer, orig.ptr_buffer );
        // or if the class does not store strings then
        // memcpy( ptr_buffer, orig.ptr_buffer, len ); 
    }

    return *this;
}
于 2021-10-22T14:49:00.913 回答
2

据推测，Mystring有一个如下所示的析构函数：

Mystring::~Mystring() {
  delete[] ptr_buffer;
}

因此，您不能只获取 from 的所有权而ptr_buffer不用orig其他东西替换它。这在move分配中是可行的，但是由于您处于复制分配中，因此您必须保持orig原样。

这意味着您必须将orig的数据复制到其中，并在必要时分配一个新数据：

Mystring& Mystring::operator=(const Mystring& orig)
{
    if(this != &orig)
    {
        std::size_t min_buf_size = orig.len + 1; // or perhaps orig.buf_size, it depends.

        // no need to allocate a new buffer if the current one is big enough already.
        if(buf_size < min_buf_size) {
          delete[] ptr_buffer;
          buf_size = min_buf_size;
          ptr_buffer = new char[buf_size];
        }

        len = orig.len;

        assert(buf_size >= len + 1);
        std::memcpy(ptr_buffer, orig.ptr_buffer, len + 1);
    }
    return *this;
}
于 2021-10-22T14:48:21.477 回答