在C ++中一次從輸入文件讀取一個單詞

Question

我正在嘗試從輸入文件中將一組單詞讀入3個不同的字符串數組。 文件中的單詞用“＃”分隔。 由於某種原因，我的代碼運行了兩次，第一個數組為空，單詞為。 請幫助，我的循環肯定是錯誤的，我必須忽略一些東西。 請讓我知道我在做什么錯。 謝謝

 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

此代碼和示例文本文件中存在一些小錯誤：

1-示例文件的末尾應帶有＃，否則最后一個循環將永遠運行。

2-

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

由於delimitCount將為零，因此將始終為false。 您應該在其聲明中將delimitCount初始化為零，並在此循環中刪除賦值，以使其變為：

while (getHardWord != "#")

3-如果要在最后關閉文件，則while條件應該是檢查文件是否打開，所以不要：

while (inFile)

采用

while (inFile.is_open())

我通過這些更改對您的代碼進行了測試，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是有原因的嗎？ 如果我正確理解您的代碼，則可以通過僅使用＃作為該函數中的定界符來簡化該代碼。

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.