1

我正在做我的家庭作业,但我卡住了,因为在作业中我们必须要求用户输入文件名,还要输入 wc cc 或 lc(文件的字数、字符数和行数。例如, wc filename.txt. 我想检查文件以查看它是否有效,我理解并且我知道如何比较用户输入以确定要运行的不同类型的函数,但我不明白你怎么能做到一起。有什么想法吗?这就是我目前所拥有的。

#include <iostream>
#include <string>
#include <fstream>

using namespace std;

int main()
{
string line;
string file;

ifstream input; //input file stream
int i;

cout << "Enter a file name" << endl;

while(true){

    cout << ">" ;
    getline(cin,file);

    input.open(file.c_str());

     if (input.fail()) {
        cerr << "ERROR: Failed to open file " << file << endl;
        input.clear();
     }
     else {
        i = 0;
        while (getline(input, line))

            if(line == "wc"){

            cout << "The word count is: " << endl;
    }
        else if(line == "cc"){

            cout << "The character count is: " << endl;
    }
        else if(line == "lc"){

            cout << "The line count is: " << endl;
    }
        else if(line == "exit"){

            return 0;
    }
        else{

            cout << "----NOTE----" << endl;
            cout << "Available Commands: " << endl;
            cout <<"lc \"filename\"" << endl;
            cout <<"cc \"filename\"" << endl;
            cout <<"wc \"filename\"" << endl;
            cout <<"exit" << endl;
    }



     }



}


return 0;
}

void wordCount(){
   //TBD
}

void characterCount(){
    //TBD

}

void lineCount(){
  //TBD

}
4

2 回答 2

0

我想你在问如何验证多个参数:命令和文件。

一个简单的策略是具有如下功能:

#include <fstream> // Note: this is for ifstream below

bool argumentsInvalid(const string& command, const string & command) {
    // Validate the command
    // Note: Not ideal, just being short for demo
    if("wc" != command && "cc" != command && "lc" != command) {
        std::cout << "Invalid command" << std::endl;
        return false;
    }

    // Validate the file
    // Note: This is a cheat that uses the fact that if its valid, its open.
    std::ifstream fileToRead(filename);
    if(!fileToRead) {
        std::cout << "Invalid file: \"" << filename << "\"" << std::endl;
        return false;
    }

    return true;
    // Note: This does rely on the ifstream destructor closing the file and would mean
    // opening the file twice.  Simple to show here, but not ideal real code.
}

如果要在返回错误之前评估所有参数,请在该函数的顶部插入一个标志,例如:

// To be set true if there is an error
bool errorFound = false;

并将条件中的所有回报更改为:

errorFound = true;

最后返回:

return !errorFound;

用法:

....

if(argumentsInvalid(command, filename)) {
    std::cout << "Could not perform command.  Skipping..." << std::endl;
    // exit or continue or whatever
}

// Now do your work

注意:这里的具体有效性测试过于简化。

于 2012-09-11T23:39:16.777 回答
0

您必须在用户输入中找到命令和文件名之间的空格,然后在找到空格的位置拆分字符串。像这样的东西

cout << "Enter a command\n";
string line;
getline(cin, line);
// get the position of the space as an index
size_t space_pos = line.find(' ');
if (space_pos == string::npos)
{
   // user didn't enter a space, so error message and exit
   cout << "illegal command\n";
   exit(1);
}
// split the string at the first space
string cmd = line.substr(0, space_pos);
string file_name = line.substr(space_pos + 1);

这是未经测试的代码。

你可以做得比这更好,例如,如果用户在命令和文件名之间输入两个空格,这将不起作用。但是这种工作很快就会变得非常乏味。由于这是一项任务,我很想继续做更有趣的事情。如果你有时间,你可以随时回来改进。

于 2012-09-07T18:53:40.327 回答