在C ++中一次从输入文件读取一个单词

Question

我正在尝试从输入文件中将一组单词读入3个不同的字符串数组。 文件中的单词用“＃”分隔。 由于某种原因，我的代码运行了两次，第一个数组为空，单词为。 请帮助，我的循环肯定是错误的，我必须忽略一些东西。 请让我知道我在做什么错。 谢谢

 Sample input file (input.txt)

complicated
insinuated
complex
juggernaut
#
blah 
...
...
#
...



#include <iostream>
#include <fstream>
#include <string>
#include "conio.h"

using namespace std;

int main() {

ifstream inFile("dictionary.txt");

// Check for error
if (inFile.fail()){
    cout << "Error Opening File." << endl;
    exit(1);
}

string hard[27], medium[29], easy[33];
string getHardWord, getMedWord, getEasyWord; 
int hardCount = 0, medCount = 0, easyCount = 0, delimitCount; // n is number of #
  // delimiters and count is array position

// Read the dictionary file until the end of file
while (inFile){
    inFile >> getHardWord;

    while ((getHardWord != "#") && (delimitCount = 0)){
        hard[hardCount] = getHardWord;
        hardCount++;
        inFile >> getHardWord;
    }

    delimitCount++;
    cout << delimitCount << endl;

    for (int iii = 0; iii < 27; iii++){
        cout << hard[iii] << endl;
    }

    cout << endl;

    inFile >> getMedWord;

    while ((getMedWord != "#") && (delimitCount = 1)){
        medium[medCount] = getMedWord;
        medCount++;
        inFile >> getMedWord;
    }

    delimitCount++;
    cout << delimitCount << endl;

    for (int jjj = 0; jjj < 27; jjj++){
        cout << medium[jjj] << endl;
    }

    cout << endl;

    inFile >> getEasyWord;

    while ((getEasyWord != "#") && (delimitCount = 2)){
        easy[easyCount] = getEasyWord;
        easyCount++;
        inFile >> getEasyWord;
    }

    delimitCount++;
    cout << delimitCount << endl;

    for (int kkk = 0; kkk < 27; kkk++){
        cout << easy[kkk] << endl;
    }

    inFile.close();
}

_getch();

return 0;
}

Answer 1

此代码和示例文本文件中存在一些小错误：

1-示例文件的末尾应带有＃，否则最后一个循环将永远运行。

2-

while ((getHardWord != "#") && (delimitCount = 0))

由于delimitCount将为零，因此将始终为false。 您应该在其声明中将delimitCount初始化为零，并在此循环中删除赋值，以使其变为：

while (getHardWord != "#")

3-如果要在最后关闭文件，则while条件应该是检查文件是否打开，所以不要：

while (inFile)

采用

while (inFile.is_open())

我通过这些更改对您的代码进行了测试，效果很好：

// Test.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
#include <fstream>
#include <string>
#include "conio.h"

using namespace std;

int main() {

ifstream inFile("dictionary.txt");

// Check for error
if (inFile.fail()){
    cout << "Error Opening File." << endl;
    exit(1);
}

string hard[27], medium[29], easy[33];
string getHardWord, getMedWord, getEasyWord; 
int hardCount = 0, medCount = 0, easyCount = 0, delimitCount=0; // n is number of #
  // delimiters and count is array position

// Read the dictionary file until the end of file
while (inFile.is_open()){
    inFile >> getHardWord;

    while ((getHardWord != "#")){
        hard[hardCount] = getHardWord;
        hardCount++;
        inFile >> getHardWord;
    }

    cout<<hard<<endl;

    cout<<endl;


    delimitCount++;
    cout << delimitCount << endl;

    for (int iii = 0; iii < 27; iii++){
        cout << hard[iii] << endl;
    }

    cout << endl;

    inFile >> getMedWord;

    while ((getMedWord != "#") && (delimitCount = 1)){
        medium[medCount] = getMedWord;
        medCount++;
        inFile >> getMedWord;
    }

    delimitCount++;
    cout << delimitCount << endl;

    for (int jjj = 0; jjj < 27; jjj++){
        cout << medium[jjj] << endl;
    }

    cout << endl;

    inFile >> getEasyWord;

    while ((getEasyWord != "#") && (delimitCount = 2)){
        easy[easyCount] = getEasyWord;
        easyCount++;
        inFile >> getEasyWord;
    }

    delimitCount++;
    cout << delimitCount << endl;

    for (int kkk = 0; kkk < 27; kkk++){
        cout << easy[kkk] << endl;
    }

    inFile.close();

}

_getch();

return 0;
}

Answer 2

您不使用getline是有原因的吗？ 如果我正确理解您的代码，则可以通过仅使用＃作为该函数中的定界符来简化该代码。

for (int i = 0; i < 27; i++) //It looks like there are 27 words per type, if that's wrong change this
{
    getline(inFile, getHardWord, '#');
    hard[i] = getHardWord;
}
//And so on for the other difficulties.