Question

我在以下代码中关注STL compilation error。

#include <cstdio>
#include <string>

template <typename T>
class container {
public:
  container(std::string in_key="") {
    m_element_index = 0;
  }
  ~container() {
  }
  // Returns the numbers of elements in the container
  int size() {
    return m_element_index;
  }
  // Assignment operator
  // Assigns a copy of container x as the new content for the container object.
  container& operator= (const container& other) {
    if (this != &other) {
      for ( int idx = 0; idx < other.size(); idx++) {
      }
    }
    return *this;
  }
private:
  int m_element_index;
};

int main ( int argc, char** argv) {
  container<int> v1("my_container");
  container<int> v2("copy_cont");
  v2 = v1;
}

获取以下行的错误

for ( int idx = 0; idx < other.size(); idx++) {

错误是

1>------ Build started: Project: test, Configuration: Debug Win32 ------
1>  test.cpp
1>e:\avinash\test\test\test.cpp(20): error C2662: 'container<T>::size' : cannot convert 'this' pointer from 'const container<T>' to 'container<T> &'
1>          with
1>          [
1>              T=int
1>          ]
1>          Conversion loses qualifiers
1>          e:\avinash\test\test\test.cpp(18) : while compiling class template member function 'container<T> &container<T>::operator =(const container<T> &)'
1>          with
1>          [
1>              T=int
1>          ]
1>          e:\avinash\test\test\test.cpp(30) : see reference to class template instantiation 'container<T>' being compiled
1>          with
1>          [
1>              T=int
1>          ]
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

Answer 1

你需要改变这个：

  int size() {
    return m_element_index;
  }

到此：

  int size() const {
    return m_element_index;
  }

告诉编译器您希望它允许在size()实例上调用const。

Answer 2

在这里，您将const对象传递给赋值运算符：

container& operator= (const container& other) {
    <...>
}

但是，在运营商内部，您正在调用other的{{1}}功能：

size()

为了使用for ( int idx = 0; idx < other.size(); idx++)对象可以使用，函数本身必须声明为const：

const

Answer 3

您需要将size()声明为const方法：

int size() const {
    return m_element_index;
}

因为你的作业运算符

container& operator= (const container& other) { .... }

您正在呼叫other.size()，而other是reference-to-const，这意味着您只能在其上调用const方法。

不能将'this'指针从'const container <t>'转换为'container <t>＆amp;'</t> </t>

3 个答案: