Confused with error using ofstream

[rEvolution27]

I'm a c++ newbie here, trying out some stuff and when I try to compile
this:

void create() {
char name;
cout << "Creating a new timetable /n Please type a name for this
timetable";
cin >name;
ofstream editFile;
editFile.open (name, ios::out | ios::app);
editFile << name << endl;
}

I get this:
invalid conversion from `char' to `const char*'

I kinda understand why i'm getting this since when I change:
editFile.open (name, ios::out | ios::app);
To:
editFile.open ("Timetable.txt ", ios::out | ios::app); , it compiles
But how do I get it do do what I want?

[John Harrison]

rE**********@gm ail.com wrote:

I'm a c++ newbie here, trying out some stuff and when I try to compile
this:

void create() {
char name;
cout << "Creating a new timetable /n Please type a name for this
timetable";
cin >name;
ofstream editFile;
editFile.open (name, ios::out | ios::app);
editFile << name << endl;
}

I get this:
invalid conversion from `char' to `const char*'

I kinda understand why i'm getting this since when I change:
editFile.open (name, ios::out | ios::app);
To:
editFile.open ("Timetable.txt ", ios::out | ios::app); , it compiles
But how do I get it do do what I want?

char is a single char, a file name is (obviously) multiple chars.

Put it another way in C++ a char and a string of chars are not the same
thing.

The best way in C++ to do what you want is to use the string class. The
next best way is to use a dynamic array of chars, the worst way is to
use a static arrays of chars.

Here's some example code using the string class

#include <string>
#include <fstream>
using namespace std;

void create() {
string name;
cout << "Creating a new timetable \n Please type a name for this
timetable ";
cin >name;
ofstream editFile(name.c _str());
editFile << name << endl;

You need the c_str() method to convert a C++ string into what ofstream
requires.

char, arrays of chars, dynamic allocation of arrays, strings etc. are a
big topic which should be covered in great detail in your favourite C++
book.

john

[Alan Johnson]

rE**********@gm ail.com wrote:

I'm a c++ newbie here, trying out some stuff and when I try to compile
this:

void create() {
char name;

In the above line you define 'name' as a char. That is, a single
character, 1 byte. Probably not what you want.

cout << "Creating a new timetable /n Please type a name for this
timetable";
cin >name;

Because, as previously mentioned, name is a char, you read exactly 1
character here.

ofstream editFile;
editFile.open (name, ios::out | ios::app);

open expects a POINTER to the first character in an null-terminated
array of characters. If you don't understand pointers and their
relationship to arrays, now is the appropriate time to drop what you are
doing and go read about it.

editFile << name << endl;
}

I get this:
invalid conversion from `char' to `const char*'

Given the above explanation, this message might make sense now. 'name'
is a char, but open wants a pointer to a char.

>
I kinda understand why i'm getting this since when I change:
editFile.open (name, ios::out | ios::app);
To:
editFile.open ("Timetable.txt ", ios::out | ios::app); , it compiles

When you use a string literal (e.g. "Timetable.txt" ), you create a
null-terminated array of characters. And arrays can be implicitly
converted to a pointer to their first element (again, go read about
pointers and arrays if that is new to you), therefore open is getting
what it wants, a pointer to a character.

But how do I get it do do what I want?

A number of ways. First, the not particularly safe, but easier to
understand way:

// I chose the size 80 arbitrarily. This is unsafe.
char name[80] ;
cin >name ;

This will read from standard input until it sees white space. If there
are more than 79 characters (the null-terminator takes 1 character, read
about C-style strings if this is new to you) available then you have
your classic buffer-overflow problem.

A second, more complicated solution, would be to dynamically allocate an
array, and resize it each time it runs out of space. This is not
trivial to do correctly, so I won't even bother. Let's move on to the
third solution.

Use std::string.

#include <string>

....

std::string name ;
cin >name ;

....

editFile.open(n ame.c_str(), ios::out | ios::app) ;
std::string does any necessary memory allocating and reallocating for
you. You can get pointer to a null-terminated array of characters using
the c_str method, as demonstrated above.

--
Alan Johnson

[rEvolution27]

Ok, thanks.
For some reason I thought 'char name' would create a character array
that matches the size of what's put into it. I guess i'll use strings
then.

[rEvolution27]

How would I make it so that the string can accept more than one word?

I've got this so far...
void create() {
string name;
cout << "Creating a new timetable"<< endl <<"Type a name for this
timetable" << endl;
cin >name;
ofstream editFile;
editFile.open ("timetable.txt ", ios::out | ios::trunc);
editFile << name << endl;
}

But when I open timetable.txt, I only see the first word of what I
typed for name.

[Ian Collins]

rE**********@gm ail.com wrote:

How would I make it so that the string can accept more than one word?

I've got this so far...
void create() {
string name;
cout << "Creating a new timetable"<< endl <<"Type a name for this
timetable" << endl;
cin >name;
ofstream editFile;
editFile.open ("timetable.txt ", ios::out | ios::trunc);
editFile << name << endl;
}

But when I open timetable.txt, I only see the first word of what I
typed for name.

That's how std::istream& operator( std::istream& std::string& ) works.
If you want the full line, use getline.

--
Ian Collins.

[rEvolution27]

On Mar 3, 6:04 pm, Ian Collins <[e-mail] mwrote:

rEvolutio...@gm ail.com wrote:

How would I make it so that the string can accept more than one word?

I've got this so far...
void create() {
string name;
cout << "Creating a new timetable"<< endl <<"Type a name for this
timetable" << endl;
cin >name;
ofstream editFile;
editFile.open ("timetable.txt ", ios::out | ios::trunc);
editFile << name << endl;
}

But when I open timetable.txt, I only see the first word of what I
typed for name.

That's how std::istream& operator( std::istream& std::string& ) works.
If you want the full line, use getline.

--
Ian Collins.

You lost me there....

[Jim Langston]

<rE**********@g mail.comwrote in message
news:11******** **************@ h3g2000cwc.goog legroups.com...

On Mar 3, 6:04 pm, Ian Collins <[e-mail] mwrote:

>rEvolutio...@g mail.com wrote:

How would I make it so that the string can accept more than one word?

I've got this so far...
void create() {
string name;
cout << "Creating a new timetable"<< endl <<"Type a name for this
timetable" << endl;
cin >name;
ofstream editFile;
editFile.open ("timetable.txt ", ios::out | ios::trunc);
editFile << name << endl;
}

But when I open timetable.txt, I only see the first word of what I
typed for name.

That's how std::istream& operator( std::istream& std::string& ) works.
If you want the full line, use getline.

--
Ian Collins.

You lost me there....

std::getline( std::cin, name );

[rEvolution27]

The following compiles, but the getline doesn't seem to work... am I
missing something here?

void create() {
string name;
cout << "Creating a new timetable"<< endl <<"Type a name for this
timetable" << endl;
getline(cin,nam e);
ofstream editFile;
editFile.open ("timetable.txt ", ios::out | ios::trunc);
editFile << name << endl;
system ("pause");

[John Harrison]

rE**********@gm ail.com wrote:

The following compiles, but the getline doesn't seem to work... am I
missing something here?

void create() {
string name;
cout << "Creating a new timetable"<< endl <<"Type a name for this
timetable" << endl;
getline(cin,nam e);
ofstream editFile;
editFile.open ("timetable.txt ", ios::out | ios::trunc);
editFile << name << endl;
system ("pause");

Can't tell from the code you've posted, but possibly you've fallen into
the well known gotcha described here

http://www.parashift.com/c++-faq-lit....html#faq-15.6

In other words getline is working, it's just it is reading the newline
that was left behind by some previous input.

john