在c ++中一次从输入文件中读取一个字

时间:2013-10-24 05:24:57

标签: c++

我正在尝试将输入文件中的一组单词读入3个不同的字符串数组。文件中的单词用'#'分隔。我的代码由于某种原因运行了两次,第一个数组是空白的,而且是单词。请帮助,我的循环肯定是错的,我必须忽略一些东西。请让我知道我做错了什么。感谢

 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;
}

2 个答案:

答案 0 :(得分:1)

此代码和示例文本文件中有几个小错误:

1-示例文件最后应该有一个#,否则最后一个循环将永远运行。

2 -

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

总是评估为false,因为delimitCount将为零。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,以便它变为:

while (getHardWord != "#")

3-如果你在最后关闭文件,你的while条件应该是检查文件是否打开,所以代替:

while (inFile)

使用

while (inFile.is_open())

我使用这些更改测试了您的代码并且运行正常:

// 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;
}

答案 1 :(得分:0)

你有没有理由不使用getline?如果我正确理解你的代码,可以通过在该函数中使用#作为分隔符来简化它。

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.