C++ 从具有不同变量类型的文件逐行读取
对于我正在做的一个项目,我们必须用“House”对象填充一个队列。“House”对象从名为“data.dat”的文件中获取信息。文件的每一行都是进入house对象的另一个对象。首先我取一个字符*作为地址,然后是一个int,另一个int,第三个int,然后是另一个char*。我们不会大声使用字符串来获取char*变量,我相信这就是我遇到问题的地方。每次编译时,它都告诉我有一个分段错误。这是我的queue.cpp区域,我非常确定错误在其中C++ 从具有不同变量类型的文件逐行读取,c++,file,file-io,fault,C++,File,File Io,Fault,对于我正在做的一个项目,我们必须用“House”对象填充一个队列。“House”对象从名为“data.dat”的文件中获取信息。文件的每一行都是进入house对象的另一个对象。首先我取一个字符*作为地址,然后是一个int,另一个int,第三个int,然后是另一个char*。我们不会大声使用字符串来获取char*变量,我相信这就是我遇到问题的地方。每次编译时,它都告诉我有一个分段错误。这是我的queue.cpp区域,我非常确定错误在其中 #include"queue.h" #include<
#include"queue.h"
#include<iostream>
#include<fstream>
#include<istream>
Queue::Queue(const char *filename){
head = NULL;
tail = NULL;
std::ifstream infile(filename);
char * address = NULL;
int footage = 0;
int bedrooms = 0;
int bathrooms = 0;
char * features = NULL;
while(!infile.eof()){
while(infile.get() != '\n'){
//std::cout << infile.get();
infile.get(address[i]);
}
infile >> footage >> bedrooms >> bathrooms;
while(infile.get() != '\n'){
infile.get(features[i]);
}
enqueue(House(address, footage, bedrooms, bathrooms, features));
}
infile.close();
#包括“queue.h”
#包括
#包括
#包括
队列::队列(常量字符*文件名){
head=NULL;
tail=NULL;
std::ifstream infle(文件名);
字符*地址=空;
整数进尺=0;
内部卧室=0;
内部浴室=0;
char*features=NULL;
而(!infle.eof()){
而(infle.get()!='\n'){
//标准::cout>镜头>>卧室>>浴室;
而(infle.get()!='\n'){
获取(特征[i]);
}
排队(房屋(地址、录像、卧室、浴室、特色);
}
infle.close();
}
以下是房屋对象头文件:
House();
House(char * ad, int fo, int be, int ba, char * fe);
char * getAddress();
int getFootage();
int getBedrooms();
int getBathrooms();
char * getFeatures();
void setAddress(char * ad);
void setFootage(int fo);
void setBedrooms(int be);
void setBathrooms(int ba);
void setFeatures(char * fe);
friend std::ostream& operator<<(std::ostream& out, House& house);
private:
char * address;
int footage;
int bedrooms;
int bathrooms;
char * features;
House();
房屋(char*ad、int-fo、int-be、int-ba、char*fe);
char*getAddress();
int gettaines();
int getbeddrooms();
int getBathrooms();
char*getFeatures();
无效设置地址(字符*ad);
无效设置进尺(int-fo);
空立根柱面(int be);
休息室(内部ba);
无效设置特征(字符*fe);
friend std::ostream&operator您需要首先初始化特性
和地址
,方法是使用新建
,或者将其创建为一个具有一定长度的字符数组。
按照这种方式,您试图写入尚未分配的内存-因此缓冲区溢出。您需要首先初始化功能
和地址
,方法是使用新建
,或将其创建为具有一定长度的字符数组。
按照这种方式,您试图写入尚未分配的内存-因此缓冲区溢出。为了好玩,这里有一个已清理的版本
- 它主要用
std::string
替换char*
(因为我们在做C++)
- 它使整个事情变得自给自足
- 它使用正确的输入验证(不要在(!infle.eof())
时使用,请检查提取运算符)
我没有实现您的队列:)
#include <iostream>
#include <fstream>
#include <sstream>
struct House {
House()
: address(), footage(0), bedrooms(0), bathrooms(0), features()
{ }
House(std::string ad, int fo, int be, int ba, std::string fe)
: address(ad), footage(fo), bedrooms(be), bathrooms(ba), features(fe)
{ }
std::string getAddress() const { return address; }
int getFootage() const { return footage; }
int getBedrooms() const { return bedrooms; }
int getBathrooms() const { return bathrooms; }
std::string getFeatures() const { return features; }
void setAddress(std::string ad) { address = ad; }
void setFootage(int fo) { footage = fo; }
void setBedrooms(int be) { bedrooms = be; }
void setBathrooms(int ba) { bathrooms = ba; }
void setFeatures(std::string fe) { features = fe; }
friend std::ostream &operator<<(std::ostream &out, House const &house) {
return out << "Address: " << house.getAddress() << '\n'
<< "Footage: " << house.getFootage() << '\n'
<< "Bed rooms: " << house.getBedrooms() << '\n'
<< "Bath rooms: " << house.getBathrooms() << '\n'
<< "Features: " << house.getFeatures() << '\n';
}
private:
std::string address;
int footage;
int bedrooms;
int bathrooms;
std::string features;
};
struct Queue {
Queue(std::string filename);
struct Node {
House value;
Node* next;
};
Node *head, *tail;
void enqueue(House const& h) {
// TODO
std::cout << h << "\n";
}
};
Queue::Queue(std::string filename) : head(nullptr), tail(nullptr) {
std::ifstream infile(filename);
std::string address;
int footage = 0;
int bedrooms = 0;
int bathrooms = 0;
std::string features;
std::string line; // lines
while (getline(infile, address) && getline(infile, line)) {
std::istringstream iss(line);
if (iss >> footage >> bedrooms >> bathrooms && getline(iss, features)) {
enqueue(House(address, footage, bedrooms, bathrooms, features));
}
}
}
int main()
{
Queue q("data.dat");
}
它打印输出:
Address: Blv. Dreams Abroken 24, 78377d XG, ClassyCode
Footage: 2
Bed rooms: 4
Bath rooms: 1
Features: pool sauna porch indoor-parking
Address: Oyi. Qernzf Noebxra 24, 78377q KT, PynfflPbqr
Footage: 3
Bed rooms: 8
Bath rooms: 2
Features: cbby fnhan cbepu vaqbbe-cnexvat
为了好玩,这里有一个清理版本
- 它主要用
std::string
替换char*
(因为我们在做C++)
- 它使整个事情变得自给自足
- 它使用正确的输入验证(不要在(!infle.eof())
时使用,请检查提取运算符)
我没有实现您的队列:)
#include <iostream>
#include <fstream>
#include <sstream>
struct House {
House()
: address(), footage(0), bedrooms(0), bathrooms(0), features()
{ }
House(std::string ad, int fo, int be, int ba, std::string fe)
: address(ad), footage(fo), bedrooms(be), bathrooms(ba), features(fe)
{ }
std::string getAddress() const { return address; }
int getFootage() const { return footage; }
int getBedrooms() const { return bedrooms; }
int getBathrooms() const { return bathrooms; }
std::string getFeatures() const { return features; }
void setAddress(std::string ad) { address = ad; }
void setFootage(int fo) { footage = fo; }
void setBedrooms(int be) { bedrooms = be; }
void setBathrooms(int ba) { bathrooms = ba; }
void setFeatures(std::string fe) { features = fe; }
friend std::ostream &operator<<(std::ostream &out, House const &house) {
return out << "Address: " << house.getAddress() << '\n'
<< "Footage: " << house.getFootage() << '\n'
<< "Bed rooms: " << house.getBedrooms() << '\n'
<< "Bath rooms: " << house.getBathrooms() << '\n'
<< "Features: " << house.getFeatures() << '\n';
}
private:
std::string address;
int footage;
int bedrooms;
int bathrooms;
std::string features;
};
struct Queue {
Queue(std::string filename);
struct Node {
House value;
Node* next;
};
Node *head, *tail;
void enqueue(House const& h) {
// TODO
std::cout << h << "\n";
}
};
Queue::Queue(std::string filename) : head(nullptr), tail(nullptr) {
std::ifstream infile(filename);
std::string address;
int footage = 0;
int bedrooms = 0;
int bathrooms = 0;
std::string features;
std::string line; // lines
while (getline(infile, address) && getline(infile, line)) {
std::istringstream iss(line);
if (iss >> footage >> bedrooms >> bathrooms && getline(iss, features)) {
enqueue(House(address, footage, bedrooms, bathrooms, features));
}
}
}
int main()
{
Queue q("data.dat");
}
它打印输出:
Address: Blv. Dreams Abroken 24, 78377d XG, ClassyCode
Footage: 2
Bed rooms: 4
Bath rooms: 1
Features: pool sauna porch indoor-parking
Address: Oyi. Qernzf Noebxra 24, 78377q KT, PynfflPbqr
Footage: 3
Bed rooms: 8
Bath rooms: 2
Features: cbby fnhan cbepu vaqbbe-cnexvat
data.dat看起来像…?data.dat看起来像。。。?