Simple user request for filename for output and input

Question

How can I request the user to input the filename that my program needs to read from and have it output the name with .out extension instead?

Example:

char fileName[256];
cout << "What is the file name that should be processed?";
cin >> fileName;

inFile.open(fileName);
outFile.open(fileName);

But I need it to save the file as a filename.out instead of the original document type (IE:.txt)

I've tried this:

char fileName[256];
cout << "What is the file name that should be processed?";
cin >> fileName;

inFile.open(fileName.txt);
outFile.open(fileName.out);

But I get these errors:

c:\users\matt\documents\visual studio 2008\projects\dspi\dspi\dspi.cpp(41) : error C2228: left of '.txt' must have class/struct/union 1> type is 'char [256]'

c:\users\matt\documents\visual studio 2008\projects\dspi\dspi\dspi.cpp(42) : error C2228: left of '.out' must have class/struct/union 1> type is 'char [256]'

score 1 · Answer 1 · answered Sep 11 '10 at 22:25

1

You are using iostreams, implying the use of C++. That in turn means you should probably be using std::string, which has overloaded operators for string concatenation - and the nice side effect of automatic memory management and added security.

#include <string>
// ...
// ...
std::string input_filename;
std::cout << "What is the file name that should be processed?\n";
std::cin >> input_filename;
// ...
infile.open(input_filename + ".txt");

answered Sep 11 '10 at 22:25

Jim Brissom

31,821
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I tried that but got an error. Looked it up and .open doesn't take string as an argument. But I found a line of code that makes it so. inFile.open(inputFile.c_str(), ios::in); outFile.open(outputFile.c_str(), ios::in); – Matt Swezey Sep 12 '10 at 18:56

score 1 · Accepted Answer · 2010-09-12T20:35:43.303

1

To change fileName extension:

string fileName;
cin >> fileName;
string newFileName = fileName.substr(0, fileName.find_last_of('.')) + ".out";

edited Sep 12 '10 at 20:35

answered Sep 11 '10 at 22:30

1

Try testing with a file name like 'x.y.z', or path like: "abc.def\xyz\file.txt". – Jerry Coffin Sep 11 '10 at 22:53
That was supposed to be find_last_of, not find_first_of :) Thanks for pointing out my mistake. – Sep 11 '10 at 22:57

score 0 · Answer 3 · answered Sep 11 '10 at 22:27

0

Writing filename.txt implies that fileName is an object and you want to access it's data member .txt. (Similar argument applies to fileName.out). Instead, use

inFile.open(fileName + ".txt");
outFile.open(fileName + ".out");

answered Sep 11 '10 at 22:27

Michael Mior

28,107
9
89
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Yeah, now that i think about it that makes perfect logical sense! – Matt Swezey Sep 12 '10 at 18:58

Simple user request for filename for output and input

3 Answers3