Question

我是一名 CS 学生，我有一项需要帮助的任务。任务是创建一个vectors_predicate_view 类，它获取一个T 类型参数，一个predictatum 作为模板参数，并且在构造函数中它应该得到2 个std::vector 作为构造函数参数。当范围在视图上时，向量的项目应该根据预测进行排序。如果预测为真，则该项目应放在构造函数的第一个向量中，如果为假，则其他所有内容都应放在第二个向量中。如果范围不在视图中，则所有内容都应设置回其原始状态

所以我的vectors_predicate_view 看起来像这样

#pragma once
#include <vector>

//In this class we add everything to vector1 that gives back true for the predictatum
//everything else goes to vector2
template<typename T, typename Prediktatum>
class vectors_predicate_view
{
private:
    Prediktatum predikt;
    std::vector<T> originalVector1;
    std::vector<T> originalVector2;

public:
    vectors_predicate_view(std::vector<T> &vector1, std::vector<T> &vector2)
    {
        //Putting every element from vector1 into originalVector1
        for (auto element : vector1) 
            originalVector1.push_back(element);
        vector1.clear();    //Emptying vector1

        //Putting every element from vector2 into originalVector2
        for(auto element : vector2)
            originalVector2.push_back(element);
        vector2.clear();    //Emptying vector2

        //We loop through originalVector1
        //if the element gives back true for the predictatum, we add it to vector1
        //else it goes to vector2
        for(auto element : originalVector1)
        {
            if(predikt(element))
                vector1.push_back(element);
            else
                vector2.push_back(element);
        }

        //We loop through originalVector2
        //if the element gives back true for the predictatum, we add it to vector1
        for(auto element : originalVector2)
        {
            if(predikt(element))
                vector1.push_back(element);
        }
    }
};

和主要的不能修改！！！看起来像这样：

#include <iostream>
#include "vecspred.h"
#include <algorithm>
#include <string>
#include <numeric>
#include <functional>
#include "vecspred.h"

struct is_even: std::unary_function<int, bool>
{
  bool operator()( int i ) const
  {
    return 0 == i % 2;
  }
};

struct is_good_language: std::unary_function<std::string, bool>
{
  bool operator()( const std::string& s ) const
  {
    return s == "C++" || s == "C";
  }
};

const int max = 1024;

bool check()
{
  bool c = false;
  std::vector<int> a;
  std::vector<int> b;

  for( int i = 0; i < max; ++i )
  {
    if ( i < max / 2 )
    {
      a.push_back( i );
    }
    else
    {
      b.push_back( i );
    }
  }

  std::vector<std::string> x;
  x.push_back( "Cobol" );
  x.push_back( "Ada" );

  std::vector<std::string> y;
  y.push_back( "Javascript" );
  y.push_back( "C++" );
  y.push_back( "ABAP" );

  if ( !c )
  {
    //vectors_predicate_view is in scope now
    vectors_predicate_view<int, is_even> va( a, b );
    vectors_predicate_view<std::string, is_good_language> vb( x, y );

    c = ( 1 == x.size() && 1 == b[ 0 ] && 2 == a[ 1 ] && "Cobol" == y[ 0 ] );
  } //after this bracket it's out of scope

  //here every vector should be set back to their original state
  if ( !c || "Ada" != x[ 1 ] || "ABAP" != y[ 2 ] || max / 2 != b[ 0 ] )
  {
    return false;
  }

  return c;
}

int main()
{
  std::cout
    << "Your solution is " << (check() ? "" : "not ")
    << "ready for submission.\n";
}

问题是，如果新声明的vectors_predicate_view 对象在范围内，我如何在不修改main.cpp 中的任何内容的情况下检查我的类？

参见此处的示例：

  if ( !c )
  {
    //vectors_predicate_view is in scope now
    vectors_predicate_view<int, is_even> va( a, b );
    vectors_predicate_view<std::string, is_good_language> vb( x, y );

    c = ( 1 == x.size() && 1 == b[ 0 ] && 2 == a[ 1 ] && "Cobol" == y[ 0 ] );
  } //after this bracket it's out of scope

  //here every vector should be set back to their original state
  if ( !c || "Ada" != x[ 1 ] || "ABAP" != y[ 2 ] || max / 2 != b[ 0 ] )
  {
    return false;
  }

vectors_predicate_view 对象在范围内，但在结束大括号之后不是。如果对象不再在范围内，我希望能够将所有向量设置回它们在类中的原始状态。

知道我该怎么做吗？

Answer 1

在您的构造函数中，您可以复制原件，并保留对原件的引用，以便您可以再次修改它们。在析构函数中，您只需将向量设置回其原始值。

template<typename T, typename Prediktatum>
class vectors_predicate_view {
private:
    Prediktatum predikt;
    std::vector<T> originalContents1;
    std::vector<T> originalContents2;

    std::vector<T> &original1; // reference to original vector1
    std::vector<T> &original2; // reference to original vector2

public:
    vectors_predicate_view( std::vector<T> &vector1, std::vector<T> &vector2 ) : original1( vector1 ), original2( vector2 ) {
        //Putting every element from vector1 into originalContents1
        for ( auto element : vector1 )
            originalContents1.push_back( element );
        vector1.clear();    //Emptying vector1

        //Putting every element from vector2 into originalContents2
        for ( auto element : vector2 )
            originalContents2.push_back( element );
        vector2.clear();    //Emptying vector2

        //We loop through originalContents1
        //if the element gives back true for the predictatum, we add it to vector1
        //else it goes to vector2
        for ( auto element : originalContents1 ) {
            if ( predikt( element ) )
                vector1.push_back( element );
            else
                vector2.push_back( element );
        }

        //We loop through originalContents2
        //if the element gives back true for the predictatum, we add it to vector1
        for ( auto element : originalContents2 ) {
            if ( predikt( element ) )
                vector1.push_back( element );
        }
    }

    ~vectors_predicate_view() { // destructor automatically called when object goes out of scope
        original1 = originalContents1; // reset contents of vector1
        original2 = originalContents2; // reset contents of vector2
    }
};

可以对您的代码进行多项改进，但我将保持原样。 original1 和 original2 是对传入的两个向量的引用；因此，对它们的任何更改都会影响这些向量。当对象超出范围时会自动调用析构函数 (~vectors_predicate_view)。

检查对象是否在范围内

1 个答案: