Question

我被困在这个标准化部分。我正在使用字符串并逐个读取数据。我一直空白。该程序编译。任何关于下一步该做什么的提示都会很棒。我如何完成以下5个步骤？第3步和第4步工作正常。

使用逐个字符的I / O读取文本文件并执行以下规范化任务的程序：

用8个空格替换所有制表符
用小写字母替换所有大写字母
所有@符号将替换为“at”。
所有=标志将被一系列19 =标志所取代。
当您找到星号时，您将打印一系列星号。星号后面的字符表示要打印的星号数。使用以下字符的ASCII值。星号数是ASCII值减去32加1.星号后面的字符仅用作计数器，而不是数据字符。

#include <iostream>
#include <fstream>
#include <string>
#include <cstring>
using namespace std;

int main() {
    ifstream fin;
    ofstream fout;
    char fname[256];
    char ofname[256];
    char norm[256] = ".normal";
    int count = 0;
    //Opening input and output file
    cout << "What file do you want to be normalized? \n";
    cin >> fname;
    cout << "\n";
    fin.open(fname);
    if (fin.fail()) {
        cout << "Error opening input file! \n";
        return 0;
    }
    strcpy( ofname, fname);
    strcat( ofname, norm);
    fout.open(ofname);
    if (fout.fail()) {
    cout << "Error opening output file! \n";
    return 0;
    }
    cout << "Your output file name is: " << ofname << "\n";
    //Normalization begins here
    char data;
    while (fin.get(data)) {
        if (data == "\t") { //***
            fout << "        ";
        }// else if (isupper(data)) { //***
           // fout << tolower(data);  //***

        else if (data == "@") {
            fout << "at";
        } else if (data == "=") {
            fout << "===================";
        } else if (data == "*") {
            fout << "some shit";
        }
}
    fin.close();
    fout.close();
    return 0;
}

[/代码]

Answer 1

你走在正确的轨道上。我没有逐行解读，而是在下面添加了评论。您面临的主要挑战是尝试阅读string data;，问题的目的似乎需要char data;同样在阅读character-by-character时，您需要包含流修饰符 noskipws以确保您不会跳过空格字符。有很多很多方法可以做到这一点。这只是与您正在采取的方法进行比较的一个示例：

#include <iostream>
#include <fstream>
#include <string>
#include <cstring>
using namespace std;

int main () {
    ifstream fin;
    ofstream fout;
    char fname[256];
    char ofname[256];
    char norm[256] = ".normal";
    char eq[] = "===================";          // set a convenient 19 char sting of '='


    //Opening input and output file
    cout << endl << " Enter name of file to be normalized: ";
    cin >> fname;
    cout << endl;
    fin.open (fname);
    if (fin.fail ()) {
        cout << "Error opening input file! \n";
        return 0;
    }
    strcpy (ofname, fname);
    strcat (ofname, norm);
    fout.open (ofname);
    if (fout.fail ()) {
        cout << "Error opening output file! \n";
        return 0;
    }
    cout << endl << " Your output file name is: " << ofname << endl << endl;

    //Normalization begins here
    char data;                                  // declare data as 'char' not 'string'
    fin >> noskipws >> data;                    // read each char (including whitespace)
    while (!fin.eof ()) {
        switch (data)
        {
            case '\t' :                         // replace 'tab' by '8 chars'
                fout << "        ";
                break;
            case '@'  :                         // replace '@' by 'at'
                fout << "at";
                break;
            case '='  :                         // replace '=' by series of 19 '='
                fout << eq;
                break;
            case '*'  :                         // replace '*n' by series of (ascii n - 31) '*'
                // fin >> count;
                fin >> data;                    // read next value
                if (fin.eof ())                 // test if eof set
                    break;
                    for (int it=0; it < data - 31; it++)    // output calculate number of asterisks
                        fout << '*';
                break;
            default:                            // use default case to proccess all data and
                if (isupper (data)) {           // test upper/lower-case.
                    char lc = tolower (data);
                fout << lc;
                } else {
                    fout << data;
                }
        }
        fin >> noskipws >> data;               // read the next character
    }
    fin.close ();                              // close files & return
    fout.close ();
    return 0;
}

测试输入：

$ cat dat/test.dat
A program that reads a text:
    Tab '       ' ->  8 spaces.
    U-C letters -> l-c letters.
    All @ -> "at".
    All = signs -> a series of 19 = signs.
    All 'asterisks''n' like '*6'. -> n series of asterisks
    (Where Number of Asterisks is ASCII value minus 32 plus 1).

<强>输出：

$ cat dat/test.dat.normal
a program that reads a text:
    tab '        ' ->  8 spaces.
    u-c letters -> l-c letters.
    all at -> "at".
    all =================== signs -> a series of 19 =================== signs.
    all 'asterisks''n' like '***********************'. -> n series of asterisks
    (where number of asterisks is ascii value minus 32 plus 1).

从c ++中的文本文件中读取数据并更改值

1 个答案: