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:

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#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");
}

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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?

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.

which can be done via

Code:

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in_stream.open(file.c_str());

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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:

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    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 )

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thank you. my program is fine now. I changed the problem part to

Code:

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    in_stream.open(file.c_str());

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