Question

我正在尝试获取一个文件并将其放入3个不同的数组中。这些阵列中的两个是1D阵列，另一个是2D阵列。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

我正在尝试获取此文本文件，并将其名字插入到名为firstNames []的数组中，然后将另一个用于姓氏的数组称为lastNames []，最后将其插入数字，我希望将它们放在数组中称为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，因为它们可以按原样输出名称，但是2D数组可以输出

应该在什么时候

我将不胜感激。

Answer 1

您的问题是您在此处无法正确处理2D阵列。

例如，在3x3 2D数组中，您有两个索引[a][b]，该数组的2D表示如下所示：

[0][0] [0][1] [0][2]

[1][0] [1][1] [1][2]

[2][0] [2][1] [2][2]

例如，当您输出时：

for (int i = 0; i < 10; i++) {
    cout << productsOrders[i][i] << endl;
}

您会看到您只会在阵列中得到一条对角线，而不是所有项（[0][0], [1][1], [2][2]）。要打印整个数组，您需要使用两个循环。

输入有一个类似的问题，即同时增加两个索引。

Answer 2

诚实地解析文本文件时；我将根据其各自的职责将逻辑分为各个功能。

您的代码应如下所示：

#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.    
}

具有这种结构将使在解析文本文件时更易于调试。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好一些。最容易使用的是string或某些stream buffer，例如stringstream。完成获取所有内容后，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之后。这是您要解析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，并且可能会出错。我认为从文件中获取内容，保存，关闭并完成文件操作，然后再进行实际工作会更容易。

如何使2D数组从仅使用空格作为分隔符的文件中写入

2 个答案: