Help writing function please,

[tonyjeffs]

In the following code, what type is examplefile?
I know that FILE is incorrect!
Thanks
Tony

main ()
{
char infile[]="test.it";
ifstream examplefile (infile);
myFunction (examplefile);
return 0;
}

myfunction(FILE ef)
{
/*do something with ef*/
......
.....
}

[Chris Dams]

Dear Tony,

to*******@aol.com writes:

In the following code, what type is examplefile?
I know that FILE is incorrect! ifstream examplefile (infile);
myFunction (examplefile); myfunction(FILE ef)

Examplefile is of type ifstream. FILE should be ifstream&. I.e., a reference
to an ifstream.

Good luck,
Chris

[Ken Human]

to*******@aol.com wrote:

In the following code, what type is examplefile?
I know that FILE is incorrect!
Thanks
Tony

main ()
{
char infile[]="test.it";
ifstream examplefile (infile);
myFunction (examplefile);
return 0;
}

myfunction(FILE ef)
{
/*do something with ef*/
.....
....
}

examplefile is an object of type ifstream. Think of it this way, when
you declare a variable or object, you type: variable_type variable_name;
Compare it to the line above of where you declare examplefile, you
typed: char infile[] = "test.it"; A variable of type char named infile,
the [] tells us that it's an array, the = says that we're going to
define infile right now, and "test.it" is what we're defining infile as.

Your code should look more like:

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

void myFunction(ifstream&);

int main() {
/*char infile[]="test.it"; this isn't necessary*/
ifstream examplefile("test.it");
myFunction(examplefile);
examplefile.close();
return 0;
}

void myFunction(ifstream& ef) {
/*do stuff to the stream, probably read it*/
while(ef.good()) cout << (char)ef.get();
cout << endl;
}

HTH

[Jonathan Mcdougall]

> In the following code, what type is examplefile?

I know that FILE is incorrect!
You don't seem to understand how and object is defined. You should
back up and learn the basics.
main ()
main() returns an int. Always.

int main()
{
char infile[]="test.it";
ifstream examplefile (infile);
Make sure you included <fstream> and replace this with

std::ifstream examplefile(infile);
myFunction (examplefile);
return 0;
} myfunction(FILE ef)

This is illegal in C++. A function must have a return value. Since
you want myfunction() to operate on an std::ifstream, just use that
type:

void myfunction(std::ifstream &if)

Note that streams cannot be copied so you'll have to use references or
pointers.
Jonathan

[tonyjeffs]

Thanks to all...
I had got close, but I left out the ampersand.
so... "ifstream = the address of the 'if' "
- is that a reasonable way of thinking about it?
The truest comment is Jonathan's:
"You should back up and learn the basics."
I know, I know. I'm itching to successfully write something more
ambitious,as a kind of reward, but I realise I'm not ready; so back to
the textbook next week.
Tony
Jonathan Mcdougall wrote:

In the following code, what type is examplefile?
I know that FILE is incorrect!
You don't seem to understand how and object is defined. You should
back up and learn the basics.

main ()

main() returns an int. Always.

int main()

{
char infile[]="test.it";
ifstream examplefile (infile);

Make sure you included <fstream> and replace this with

std::ifstream examplefile(infile);

myFunction (examplefile);
return 0;
}

myfunction(FILE ef)

This is illegal in C++. A function must have a return value. Since
you want myfunction() to operate on an std::ifstream, just use that
type:

void myfunction(std::ifstream &if)

Note that streams cannot be copied so you'll have to use references

or pointers.
Jonathan

[ben]

> Thanks to all...

I had got close, but I left out the ampersand.
so... "ifstream = the address of the 'if' "
- is that a reasonable way of thinking about it?

More advices here:

1. Don't use "if" for the argument, it is a C++ keyword!
2. ifstream is a type, Jonathan's "if" is an object. A type is not an
object, just like a fruit is not an apple.
3. Unless you are using pointers, you don't have to worry about the address.

Given

void myfunction(std::ifstream &inputFile);

Think in this way:

This is a declaration of function named myfunction which, takes only one
parameter named inputFile which is a reference to an object of type
std::ifstream, and returns nothing.

[tonyjeffs]

Got it.
Hey I've just written a little program to find the size of a bitmap
file, passing a reference to ifstream to a function .
It works, and it's the first complicated (for me) program I've actually
'written' - without copying code wholesale. And whether or not it's
the best way, it's structured the way I planned it to be. *Thanks
again for help with the detail. *
Encouraged by success, I feel I'm actually getting somewhere.
Now I can get back to working through Accelerated C++.
(Maybe tomorrow)

Happy day!
Tony