如何从仅使用空格作为分隔符的文件中获取要写入的二维数组
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【中文标题】如何从仅使用空格作为分隔符的文件中获取要写入的二维数组【英文标题】:How to get a 2D array to write from a file with only spaces used as separators 【发布时间】:2018-12-08 06:38:16 【问题描述】:我正在尝试获取一个文件并将其放入 3 个不同的数组中。其中两个数组是一维数组,另一个是二维数组。文本文件如下
Bill Hansley 1 1 1 1 1
Todd Howard 2 3 1 0 0
Sam Duke 0 1 1 0 0
Danny Martin 1 0 2 0 1
我正在尝试获取这个文本文件,并将名字插入到名为 firstNames[] 的数组中,然后将另一个用于姓氏的数组称为 lastNames[],最后将数字插入到数组中称为 Productsorders[][]。我的代码如下。
bool loadOrderFile(string orderFN,
string firstNames[], string lastNames[],
int productsOrders[MAX_ORDERS][MAX_PRODS],
int &namesCount, int &prodCount, string &menuName)
ifstream File;
File.open(orderFN.c_str());
if (File.is_open())
cout << "Order file opened..." << endl;
int i = 0;
getline(File, menuName);
(File >> prodCount);
while (File)
File.get();
(File >> firstNames[i]);
(File >> lastNames[i]);
(File >> productsOrders[i][i]);
(File >> productsOrders[i + 1][i + 1]);
(File >> productsOrders[i + 2][i + 2]);
(File >> productsOrders[i + 3][i + 3]);
(File >> productsOrders[i + 4][i + 4]);
(i++);
cout << "Menu name: " << menuName << endl;
cout << "Product Count: " << prodCount << endl;
cout << "There were " << (prodCount - 1) << " orders read in." << endl;
for (int i = 0; i < 10; i++)
cout << productsOrders[i][i] << endl;
for (int i = 0; i < 10; i++)
cout << firstNames[i] << lastNames[i] << endl;
return true;
名称数组似乎可以正常工作,因为它们按应有的方式输出名称,但二维数组输出
1
2
0
1
0
2
0
1
0
0
应该是什么时候
1 1 1 1 1
2 3 1 0 0
0 1 1 0 0
1 0 2 0 1
我将不胜感激。
【问题讨论】:
您从文件内容中显示的信息;内容之前有标题信息吗?还是从您显示的文件开始?我看到你有一个getline()
调用menuName
,然后从文件中提取prodCount
,最后在你的while 循环中你有一个fstream::get()
调用,然后你开始填充你的数组。只是想更好地理解您的文件结构以尝试正确解析它。
【参考方案1】:
你的问题是你在这里没有正确处理你的二维数组。
例如,在3x3
二维数组中,您有两个索引[a][b]
,该数组的二维表示如下所示:
[0][0] [0][1] [0][2]
[1][0] [1][1] [1][2]
[2][0] [2][1] [2][2]
所以当你输出例如:
for (int i = 0; i < 10; i++)
cout << productsOrders[i][i] << endl;
你可以看到你只会得到一条穿过数组的对角线,而不是所有的项目 ([0][0], [1][1], [2][2]
)。要打印整个数组,您需要使用两个循环。
您的输入存在类似问题,您同时增加了两个索引。
【讨论】:
【参考方案2】:在解析文本文件时诚实;我会根据其独特的职责将逻辑分成单独的功能。
你的代码看起来像这样:
#include <vector>
#include <string>
#include <sstream>
#include <iostream>
#include <fstream>
#include <exception>
// Structure of your data type...
struct Order
std::string firstName;
std::string lastName;
std::vector<int> productOrders; // maybe use vector instead of array...
// any other variable(s) or container(s) you may need...
;
// Simple ostream operator<< overload to print your Order Struct
// in a nice readable format.
std::ostream& operator<<(std::ostream& os, const Order& order );
// Function to split a string based on a single character delimiter
std::vector<std::string> splitString( const std::string& s, char delimiter );
// Function that will get each line text from a file and stores it into a vector
// of strings. Then closes the file handle once the entire file has been read.
void getAllLinesFromFile(const char* filename, std::vector<std::string>& output);
// This function will parse a single line of text or string that is contained in
// the vector of strings. The declaration of this can vary; if you have other
// information such as header information before the actual data structures
// you would have to modify this function's declaration-definition to accommodate
// for those variables.
void parseLine( const std::vector<std::string>& fileContents, std::vector<Order>& orders );
// Simple and clean looking main function.
int main()
try
std::vector<std::string> fileContents;
getAllLinesFromFile( "filename.txt", fileContents );
std::vector<Order> orders;
parseLine( fileContents, orders );
for ( auto& o : orders )
std::cout << o << '\n';
catch( std::runtime_error& e )
std::cerr << e.what() << std::endl;
return EXIT_FAILURE;
return EXIT_SUCCESS;
std::ostream& operator<<(std::ostream& os, const Order& order )
os << order.firstName << " " << order.lastName << '\n';
for (auto& p : order.productOrders)
os << p << " ";
os << '\n';
return os;
std::vector<std::string> splitString( const std::string& s, char delimiter )
std::vector<std::string> tokens;
std::string token;
std::istringstream tokenStream( s );
while( std::getline( tokenStream, token, delimiter ) )
tokens.push_back( token );
return tokens;
void getAllLinesFromFile(const char* filename, std::vector<std::string>& output)
std::ifstream file(filename);
if (!file)
std::stringstream stream;
stream << "failed to open file " << filename << '\n';
throw std::runtime_error(stream.str());
else
std::cout << "File " << filename << " opened successfully.\n";
std::string line;
while (std::getline(file, line))
if (line.size() > 0)
output.push_back(line);
file.close();
void parseLine( const std::vector<std::string>& fileContents, std::vector<Order>& orders )
// Here is where you would do the logic to parse the vector of strings
// this function may vary based on your file structure. If there is any
// header information you would have to extract that first from the
// vector's index.
// Once you get to the index in the vector that describes your data structure
// it is hear that you would want to call `splitString()` using the current
// current index of that vector and the space character as your delimiter.
// This will create a vector of strings that are now considered to be tokens.
// On each pass of the loop for each line of contents you will want to
// create an instance of the Order Structure, then use that to populate
// the vector of Orders that was passed in by reference.
// Once all of the contents are done being parsed the function will exit
// and your vector of Orders will have the appropriate data.
在解析文本文件时,拥有这种结构会更容易调试。我相信从文件中检索所有数据并将其存储到某个容器中会更优雅;最容易使用的是string
,或者一些stream buffer
,例如stringstream
。完成所有内容后关闭文件。从文本文件中读取所有内容并将其存储到字符串容器中之后。这是您想要解析strings
或streams
并检查信息是否有效的地方。多次打开和关闭文件句柄效率低下且速度慢,而且可能会出错。我认为从文件中获取内容、保存、关闭并完成文件,然后继续实际工作会更容易。
【讨论】:
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