引用绑定到类型值限定符的值 [英] binding of reference to a value of type drops qualifiers

问题描述

我有来自Text Accelerated C ++的以下源代码。当我试图编译源文件时，我得到下面的编译错误列出如下。我是C ++语言的新手，所以您的帮助将不胜感激。

I have the following source from the Text Accelerated C++. When I attempt to compile the source file I get the following compilation error listed below. I'm a novice of the C++ language so your assistance would be appreciated.

#include <algorithm>
#include <iomanip>
#include <ios>
#include <iostream>
#include <string>
#include <vector>
#include <stdexcept>

using namespace std;

struct Student_info {
    string name;
    double midterm, final;
    vector<double> homework;
};

double median(vector<double>);
double grade(const Student_info&);
double grade(double, double, double);
double grade(double, double, const vector<double>&);
istream& read_hw(istream&, vector<double>&);
istream& read(istream&, Student_info&);
bool compare(const Student_info&, Student_info&);

int main() {

    vector<Student_info> students;
    Student_info record;
    string::size_type maxlen = 0;

    while(read(cin, record)) {
        maxlen = max(maxlen, record.name.size());
        students.push_back(record);
    }

    sort(students.begin(), students.end(), compare);

    for(vector<Student_info>::size_type i = 0;
            i != students.size(); ++i) {

        cout << students[i].name << string(maxlen + 1 - students[i].name.size(), ' ');

        try {
            double final_grade = grade(students[i]);
            streamsize prec = cout.precision();
            cout << setprecision(3) << final_grade
                    << setprecision(prec);
        } catch(domain_error& e) {
            cout << e.what();
        }
        cout << endl;
    }
    return 0;
}

double median(vector<double> vec) {
    typedef vector<double>::size_type vec_sz;
    vec_sz size = vec.size();

    if(size == 0)
        throw domain_error("median of an empty vector");

    sort(vec.begin(), vec.end());
    vec_sz mid = size/2;

    return size%2 == 0 ? (vec[mid] + vec[mid - 1])/2 : vec[mid];
}
double grade(const Student_info& s) {
    return grade(s.midterm, s.final, s.homework);
}
double grade(double midterm, double final, double homework) {
    return 0.2*midterm + 0.4*final + 0.4*homework;
}
double grade(double midterm, double final, const vector<double>& hw) {
    if(hw.size() == 0)
        throw domain_error("student has done no homework");
    return grade(midterm, final, median(hw));
}
istream& read_hw(istream& in, vector<double>& hw) {
    if(in) {
        hw.clear();

        double x;
        while(in >> x)
            hw.push_back(x);
        in.clear();
    }

    return in;
}
istream& read(istream& is, Student_info& s) {
    is >> s.name >> s.midterm >> s.final;
    read_hw(is, s.homework);
    return is;
}
bool compare(const Student_info& x, const Student_info& y) {
    return x.name < y.name;
}

将类型为Student_info的引用绑定到类型为Student Student_info 'drops qualifier compute_grades_rev-b

第125行，外部位置：/usr/include/c++/4.2.1/bits/stl_algo.h C / C ++问题

binding of reference to type 'Student_info' to a value of type 'const Student_info' drops qualifiers compute_grades_rev-b
line 125, external location: /usr/include/c++/4.2.1/bits/stl_algo.h C/C++ Problem

这里是来自stl_algo.h的代码：

Here is the code from stl_algo.h:

 template<typename _Tp, typename _Compare>
    inline const _Tp&
    __median(const _Tp& __a, const _Tp& __b, const _Tp& __c, _Compare __comp)
    {
      // concept requirements
      __glibcxx_function_requires(_BinaryFunctionConcept<_Compare,bool,_Tp,_Tp>)
      if (__comp(__a, __b))
    if (__comp(__b, __c))
      return __b;
    else if (__comp(__a, __c))
      return __c;
    else
      return __a;
      else if (__comp(__a, __c))
    return __a;
      else if (__comp(__b, __c))
    return __c;
      else
    return __b;
    }

我更改了比较函数的声明：

I change the declaration of the compare function from:

bool compare(const Student_info&, Student_info&);
bool compare(const Student_info, Student_info);

现在正在编译。

推荐答案

错误表示您不能将非const引用绑定到const对象，因为会在其他编译器中抛弃（ 错误），忽略或忽略 const 限定符。

The error is indicating that you cannot bind a non-const reference to a const-object, as that would drop (discard in other compiler's errors), disregard or ignore the const qualifier.

它尝试指示的是，你可以通过引用修改对象，忽略对象本身 const ，打破const正确性。

What it tries to indicate is that if the operation was allowed you would be able to modify the object through the reference ignoring the fact that the object itself is const, breaking const-correctness.

在您的特定代码中，库中的函数 __ median 使用 __ a ， __b 和 __ c ，并尝试调用 __ comp 函数，你的程序（第一个声明）采用第二个参数的非const引用。为了能够调用 __comp（__ a，__ b）（或在该函数中对 __ comp 的任何其他调用）必须绑定一个对象只能通过 const& 访问第二个参数，它接受一个非const引用。这很可能是一个错字，因为你定义比较下面的两个参数是const引用。

In your particular code, the function __median in the library takes __a, __b, and __c by const reference and tries to call the __comp function, which in your program (first declaration) takes the second argument by non-const reference. To be able to call __comp(__a,__b) (or any other call to __comp in that function) it would have to bind an object accessible only through a const& to the second argument that takes a non-const reference. This is most probably a typo, since you define compare below with both arguments being const references.

更改声明比较之前的主要：

bool compare(const Student_info&, const Student_info&);
//                                ^^^^^

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