Using a variable for a filename

This is a discussion on Using a variable for a filename within the C++ Programming forums, part of the General Programming Boards category; i seem to be having trouble using a variable as a file name for an ifstream. i am making a ...

  1. #1
    Registered User
    Join Date
    Apr 2007
    Posts
    13

    Using a variable for a filename

    i seem to be having trouble using a variable as a file name for an ifstream.
    i am making a program that is quite simple. all it does is ask for a filename, reads the file and displays it to the screen. my source code is
    Code:
    #include <cstdlib>
    #include <iostream>
    #include <string>
    #include <fstream>
    
    using namespace std;
    
    int main(int argc, char *argv[])
    {
        string file;
        char letter;
        ifstream in_stream;
        cout << "Please enter the name of \n"
                "the file you wish to open. \n"
                "Please use only text files. \n";
        cin >> file;
        in_stream.open(file);
        if (in_stream.fail())
        {
                    cout << "File Failed To Open";
                    system("pause");
                    exit(1);
                    }
        do 
        {
                    in_stream.get(letter);
                    cout.put(letter);
                    }
        while(! in_stream.eof());
        cout << endl;
        system("pause");
    }
    the underlined line is where i receive errors. I'm sure that I should be able to do this. Im following the syntax out of a book. Can someone tell me whats wrong?

  2. #2
    Captain - Lover of the C
    Join Date
    May 2005
    Posts
    341
    ifstream::open takes a character array as the first parameter, not a string. Either read in the filename as a char array or convert it to one.
    Last edited by Brad0407; 12-14-2007 at 11:25 PM. Reason: disable smilies
    Don't quote me on that... ...seriously

  3. #3
    Registered User
    Join Date
    Jun 2007
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    41
    which can be done via

    Code:
    in_stream.open(file.c_str());

  4. #4
    and the hat of sweating
    Join Date
    Aug 2007
    Location
    Toronto, ON
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    3,545
    Or even better, don't create the ifstream until you're actually ready to use it and then use the constructor instead of open()
    Code:
        string file;
        cout << "Please enter the name of \n"
                "the file you wish to open. \n"
                "Please use only text files. \n";
        cin >> file;
        ifstream in_stream( file.c_str() );
        if ( !in_stream )

  5. #5
    Registered User
    Join Date
    Apr 2007
    Posts
    13
    thank you. my program is fine now. I changed the problem part to
    Code:
        in_stream.open(file.c_str());

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