C++ 练习

今天下午看书的时候发现一个很有趣的问题。
《C++ Primer Plus》
示例14.1(studentc.h)中,定义了两个非常奇怪的类成员函数

#ifndef STUDENTC_H_
#define STUDENTC_H_

#include <iostream>
#include <string>
#include <valarray>

class Student
{
private:
        typedef std::valarray<double> ArrayDb;
        std::string name;
        ArrayDb scores;
        std::ostream & arr_out(std::ostream & os) const;
public:
        Student() :name("Null Student"), scores() {}
        explicit Student(const std::string & s) :name(s), scores() {}
        explicit Student(int n) :name("Nully"), scores(n) {}
        Student(const std::string & s, int n) :name(s), scores(n) {}
        Student(const std::string & s, const ArrayDb & a) : name(s), scores(a) {}
        Student(const char * str, const double * pd, int n) :name(str), scores(pd,n) {}
        ~Student() {}
        double Average() const;
        const std::string & Name() const;
//[]重载函数1,似乎能匹配所有的调用
        double & operator[](int i);
//[]重载函数2,在有上面的重载函数下,似乎永远匹配不到
        double operator[](int i) const;

        friend std::istream & operator>>(std::istream & is, Student & stu);
        friend std::istream & getline(std::istream & is, Student & stu);
        friend std::ostream & operator<<(std::ostream & os,const Student & stu);
};

#endif

在实现中分别添加两个重载函数的输出,查看调用的情况

#include "studentc.h"
using std::ostream;
using std::endl;
using std::istream;
using std::string;

double Student::Average() const
{
	if (scores.size() > 0)
		return scores.sum() / scores.size();
	else
		return 0;
}

const string & Student::Name() const
{
	return name;
}

double & Student::operator[](int i)
{
	std::cout << "Use double & " << std::endl;
	return scores[i];
}

double Student::operator[](int i) const
{
	std::cout << "Use double " << std::endl;
	return scores[i];
}

ostream & Student::arr_out(ostream & os) const
{
	int i;
	int lim = scores.size();
	if (lim > 0)
	{
		for(i = 0; i < lim; i++)
		{
			os << scores[i] << " ";
			if ( i % 5 == 4)
				os << endl;
		}
		if ( i % 5 != 0)
			os << endl;
	}
	else
		os << " empty array ";
	return os;
}

istream & operator>>(istream & is, Student & stu)
{
	is >> stu.name;
	return is;
}

istream & getline(istream & is, Student & stu)
{
	getline(is,stu.name);
	return is;
}

ostream & operator<<(ostream & os, const Student & stu)
{
	os << "Scores for " << stu.name << ":\n";
	stu.arr_out(os);
	return os;
}
#include "studentc.h"
#include <iostream>
using std::cin;
using std::cout;
using std::endl;

void set(Student & sa, int n);
const int pupils = 3;
const int quizzes = 5;

int main(void)
{
	Student ada[pupils] = { Student(quizzes), Student(quizzes), Student(quizzes) };
	int i;
	for (i = 0; i < pupils; ++i)
		set(ada[i], quizzes);
	cout << "\nStudent List:\n";
	for (i = 0; i < pupils; ++i)
		cout << ada[i].Name() << endl;
	cout << "\nResults:";
	for (i = 0; i < pupils; ++i)
	{
		cout << endl << ada[i];
		cout << "average: " << ada[i].Average() << endl;
	}
	cout << "Done.\n";
	return 0;
}

void set(Student & sa, int n)
{
	cout << "Please enter the student's name: ";
	getline(cin,sa);
	cout << "Please enter " << n << " quiz scores:\n";
	for (int i = 0; i < n; i++)
		cin >> sa[i];
	while (cin.get() != '\n')
		continue;
}
[code title="输出结果"]
Please enter the student's name: Gil Bayts
Please enter 5 quiz scores:
Use double & 
92 94 96 93 95
Use double & 
Use double & 
Use double & 
Use double & 
Please enter the student's name: Pat Roone
Please enter 5 quiz scores:
Use double & 
83 89 72 78 95
Use double & 
Use double & 
Use double & 
Use double & 
Please enter the student's name: Fleur O'Day
Please enter 5 quiz scores:
Use double & 
92 89 96 74 64
Use double & 
Use double & 
Use double & 
Use double & 

Student List:
Gil Bayts
Pat Roone
Fleur O'Day

Results:
Scores for Gil Bayts:
92 94 96 93 95 
average: 94

Scores for Pat Roone:
83 89 72 78 95 
average: 83.4

Scores for Fleur O'Day:
92 89 96 74 64 
average: 83
Done.

可以看到,const版本的operator[]完全没有匹配到。