Question

我是C ++的初学者，特别是数据结构。这个项目我们正在使用截断系列（类似的东西）。我得到这个奇怪的编译错误，我不知道它告诉我什么。在我创建复制构造函数之前程序运行正常，这可能是罪魁祸首。我很难做一个，但我不确定这是不是我应该这样做。

错误看起来像这样：

/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/ostream:1077:1: note: 
      candidate template ignored: could not match
      'basic_string<type-parameter-0-0, type-parameter-0-1, type-parameter-0-2>'
      against 'Series'
operator<<(basic_ostream<_CharT, _Traits>& __os,
^
/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/ostream:1094:1: note: 
      candidate template ignored: could not match
      'shared_ptr<type-parameter-0-2>' against 'Series'
operator<<(basic_ostream<_CharT, _Traits>& __os, shared_ptr<_Yp> const& __p)
^
/Library/Developer/CommandLineTools/usr/bin/../include/c++/v1/ostream:1101:1: note: 
      candidate template ignored: could not match 'bitset<_Size>' against
      'Series'
operator<<(basic_ostream<_CharT, _Traits>& __os, const bitset<_Size>& __x)

代码：

#include <iostream>
#include <vector>

using namespace std;

class Series {
    private:
            size_t degree;
            vector <double> coefs;

    public:
            Series():degree(0){ }

            Series(size_t d): degree(d){
                    for(int i = 0; i < (degree+1); i++){
                            coefs.push_back(0.0);
                    }
            }

            Series(double term){
                    coefs[0] = term;
            }

            Series(size_t d2,vector <double> newcoeffs): Series(d2) {
                            for (int i = 1; i < (d2+1); i++) {
                                    coefs.at(i) = newcoeffs.at(i-1);
                             }
            }

            Series(const Series & rhs) {
                            degree = rhs.degree;
                            vector <double> coefs;
                            for (int i = 1; i < (degree+1); i++) {
                                    coefs.at(i) = rhs.coefs.at(i);
                            }
            }    

            ~Series() {
                    coefs.clear();
            }

            void print(ostream & out) const{
                    if (degree == 0) {
                            return;
                    }
                    for (int i = 1; i < degree+1; i++)
                            out << coefs[i] << "x^" << i << " + ";
                    out << coefs[0];
            }

};
ostream & operator <<(ostream & out, const Series & s){
                    s.print(out);
                    return out;
}

int main(){
    vector <double> v {2.1,3.5,6.2};
    vector <double> z {1.1,2.3,4.0};
    Series two(3,z);
    Series one(two);

    cout << one << end;
    cout << two << endl;
    return 0;
}

Answer 1

这只是您main

中的拼写错误

cout << one << end;

应该是

cout << one << endl;

您收到的无用错误消息是因为end是std命名空间中已存在的函数的名称。这是using namespace std;经常为advised against的原因的一个很好的例子。

Answer 2

与您的问题没有直接关系，但您的代码会导致std::out_of_range异常。这是因为您使用at，威尔是访问std::vector内部值的函数，而不是在其中添加值。

for (int i = 1; i < (degree+1); i++)
    // SEGFAULT !
    coefs.at(i) = rhs.coefs.at(i);

您必须使用std::vector::push_back在容器中添加元素。

for (int i = 1; i < (degree+1); i++)
    coefs.push_back(rhs.coefs.at(i));

现代C ++甚至更短：

for (auto elem : rhs)
    coefs.push_back(elem);

复制构造函数C ++编译错误

2 个答案: