【发布时间】：2015-03-14 10:08:54
【问题描述】：
当我在 operator* 函数中返回对象 temp 时，它不会被复制到 main 中的 p3。
operator* 函数中的 cout 语句返回正确的值，但 main 中的 p3 只有垃圾。
另外，我在 main 的 return 0 语句中得到一个 _block_type_is_valid(phead->nblockuse)。
这里是代码。
#pragma once
#include<iostream>
using namespace std;
class Polynomial
{
private:
int *coefficients;
int length;
public:
inline int getLength() const {return length;}
Polynomial(int length);
Polynomial(const Polynomial &p);
Polynomial():coefficients(nullptr){}
~Polynomial(void);
extern friend Polynomial operator*(const Polynomial &p1,const Polynomial &p2);
inline int operator[](int n) const { return coefficients[n];}
inline int& operator[](int n) { return coefficients[n];}
Polynomial& operator=(Polynomial &p);
friend void swap(Polynomial& first, Polynomial& second);
void resize(int x);
friend istream& operator>>(istream &is, Polynomial &p);
friend ostream& operator<<(ostream &os, Polynomial &p);
};
这里是多项式.cpp
#include "Polynomial.h"
Polynomial::Polynomial(int length){
this->length = length;
coefficients = new int[length];
for(int i = 0; i < length; i++){
coefficients[i] = 0;
}
}
Polynomial::~Polynomial(void){
cout<<"Deleting: "<<coefficients;
delete[] coefficients;
}
/*
Polynomial Polynomial::operator*(Polynomial p){
Polynomial temp(length + p.getLength());
for(int i = 0; i < length; i++){
for(int j = 0; j < length; j++){
temp[i+j] += coefficients[i] * p[j];
}
}
cout<<temp;
return temp;
}*/
Polynomial operator*(const Polynomial &p1,const Polynomial &p2){
Polynomial temp(p1.getLength() + p2.getLength());
for(int i = 0; i < p1.getLength(); i++){
for(int j = 0; j < p2.getLength(); j++){
temp[i+j] += p1[i] * p2[j];
}
}
cout<<temp;
return temp;
}
void Polynomial::resize(int x){
delete[] coefficients;
coefficients = new int[x];
}
void swap(Polynomial& first,Polynomial& second){
int tempLength = first.getLength();
int *temp = new int[tempLength];
for(int i = 0; i < first.getLength(); i++)
temp[i] = first[i];
first.resize(second.getLength());
for(int i = 0; i < first.getLength(); i++)
first[i] = second[i];
second.resize(tempLength);
for(int i = 0; i < first.getLength(); i++)
second[i] = temp[i];
delete[]temp;
}
Polynomial& Polynomial::operator=(Polynomial &p){
swap(*this,p);
return *this;
}
Polynomial::Polynomial(const Polynomial &p){
//if(coefficients) delete [] coefficients;
coefficients = new int[p.getLength()];
for(int i = 0; i < p.getLength(); i++)
coefficients[i] = p[i];
}
istream& operator>>(istream &is,Polynomial &p){
cout<<"Enter length: ";
is>>p.length;
p.coefficients = new int[p.length];
for(int i = 0; i < p.length; i ++)
is>>p.coefficients[i];
return is;
}
ostream& operator<<(ostream &os,Polynomial &p){
for(int i = 0; i < p.length; i ++)
if(p.coefficients[i])
os<<p.coefficients[i]<<"x^"<<i<<" ";
return os;
}
这里是主要的
#include"Polynomial.h"
#include<iostream>
#include<string>
using namespace std;
int main(){
Polynomial p1,p2,p3;
cin>>p1>>p2;
p3 = (p1 * p2);
cout<<p3[0]<<p3[1]<<"here";
cout<<p3;
return 0;
}
编辑：这是最终更正的代码。我需要做的就是在所有构造函数中用 null 和 length 初始化指针。
#include "Polynomial.h"
Polynomial::Polynomial():
coefficients(nullptr),length(0){}
Polynomial::Polynomial(int length):coefficients(nullptr),length(length){
coefficients = new int[length];
for(int i = 0; i < length; i++){
coefficients[i] = 0;
}
}
Polynomial::~Polynomial(void){
if(coefficients) delete[]coefficients;
}
Polynomial& Polynomial::operator=(const Polynomial &p){
if(coefficients)
delete[]coefficients;
length = p.getLength();
coefficients = new int[length];
for(int i = 0; i < length; i++)
coefficients[i] = p[i];
return *this;
}
Polynomial::Polynomial(const Polynomial &p):
coefficients(nullptr),length(0)
{
length = p.getLength();
coefficients = new int[length];
for(int i = 0; i < length; i++)
coefficients[i] = p[i];
}
Polynomial operator*(const Polynomial &p1,const Polynomial &p2){
Polynomial temp(p1.getLength() + p2.getLength());
for(int i = 0; i < p1.getLength(); i++)
for(int j = 0; j < p2.getLength(); j++)
temp[i+j] += p1[i] * p2[j];
return temp;
}
istream& operator>>(istream &is,Polynomial &p){
cout<<"Enter length: ";
is>>p.length;
p.coefficients = new int[p.length];
for(int i = 0; i < p.length; i ++)
is>>p.coefficients[i];
return is;
}
ostream& operator<<(ostream &os,Polynomial &p){
for(int i = 0; i < p.length; i ++)
if(p.coefficients[i])
os<<p.coefficients[i]<<"x^"<<i<<" ";
return os;
}
【问题讨论】：
-
不确定这是如何编译的，因为
operator=需要一个引用，而(p1 * p2)返回一个不会绑定到非常量引用的临时值。更改operator*和operator=以获取 const refs 并尝试一下。operator*也不应该是成员函数。 -
当您按值返回对象时，例如乘法运算符，会创建一个临时对象，然后将其销毁。你认为这个临时对象中的指针会发生什么？当临时对象被破坏时，它被复制然后删除。那么副本现在指向什么？你需要遵守rule of three。
-
你在谈论垃圾值，但这段代码甚至不应该编译。
error C2679: binary '=' : no operator found which takes a right-hand operand of type 'Polynomial' -
GCC 不会编译这个。
-
我正在使用视觉工作室。我不能在这两个函数中使用 const Polynomial&，正如 Visual Studio 所说，“没有运算符匹配这些操作数，操作数类型是 const Polynomial [int]”，我使用 p[i] 命令。
标签： c++