C++ programming

allquixotic

Nov 6, 2012 18:23

hi

can you talk?

do you see the chat bar at the bottom?

0

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wireless-networking

dummy answer that question, just be professional and tell him there's no easy way to do that (correct answer) and tell him that he shouldn't try to bypass network policy

me and @r.tanner.f from root access will upvote you and you'll have enough rep

quick before it gets VTC'ed :P

allquixotic

Nov 6, 2012 19:02

argh, you need to get your stackexchange account and your superuser account under the same roof, you need to add your google account to your stackexchange account OR vice versa

try talking now, they look merged

i see your superuser account showing up under jtbadn

i tried giving you write access but it thinks your parent account is meta.stackoverflow

1

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jtbadn

Nov 6, 2012 19:25

hey!

I MADE IT! :)

allquixotic

yay

jtbadn

ok...

now that I'm here... give me a minute...

you will likely hate me for doing this to you

allquixotic

i'm not going to write your code, i will only help with conceptual issues

jtbadn

this is like really basic programming, but I am a dummy and haven't got a clue where to start.

allquixotic

okay

go ahead

jtbadn

Nov 6, 2012 19:29

I need you to help me write the code... so that I understand what the hell I am doing.

Program #1: Write a simple test program to validate passing by value and passing by reference.

Instructions

Initialize a variable and display its value.
Pass the variable by value to a function change and display the value inside the function.
Display the value once the function has returned.
Call the same variable passing by reference to another function change and display the value inside the function.
Display the value once the second function has returned.

allquixotic

uh, in what language?

jtbadn

c++

allquixotic

ok... if the way to do this isn't immediately obvious to you, then you probably don't understand passing by value or passing by reference, or both

jtbadn

I don't understand any of it.

allquixotic

sigh

jtbadn

Nov 6, 2012 19:32

I passed Programming I with a B+ and I have no idea how...

cause I learned NOTHING.

allquixotic

then you probably need to go back and review everything because there has got to be a reason why you aren't retaining the knowledge or really capturing it

jtbadn

Ok, then I should just read the book again from cover to cover I guess.

:)

allquixotic

passing by value / reference may be a new concept to you, but the instructions to the program are pretty clear

do you know how to initialize a variable in C++?

a local variable

jtbadn

int

allquixotic

........

a bit more than that, right?

jtbadn

Nov 6, 2012 19:34

yes

allquixotic

first of all, int is just the type... it doesn't HAVE to be int, it can be anything else

int stands for integer, which is just one type of variable, but we can use int here

initializing a variable in C++ (or C, or Java, or...) really comes down to declaring the type of the variable, and the name of it

the name is how you refer to the variable in other parts of your code, so that the program knows which variable you're referring to

jtbadn

int main()
{
int a;
char b;
return 0;
}

allquixotic

the type is how the compiler knows what kind of data is contained in the variable

ok... right... so that little program you just provided initializes two variables

well, it declares them, but it doesn't initialize them

int a; declares a variable of type int, and name a

like saying "Dog chip;" would declare a variable of type Dog (the variable represents a dog), and name chip (the way that you refer to the variable from here on out is by typing "chip")

do you understand this much?

jtbadn

ok if the value is "Barry" then should I use int Barry; ?

allquixotic

no -- assigning a value to a variable is done with the = operator. So far we've only declared them

you can declare and initialize a variable in the same statement, or you can do it in two statements. the result is the same.

you could write:

int barry = 0;

OR you could write:

int barry;
barry = 0;

those two statements do the same thing

0 is the value you've assigned to the integer, so you use the equals sign just like in math when you say "x = 0" in algebra

jtbadn

Nov 6, 2012 19:39

ok

so...

allquixotic

so the first part of the assignment told you to initialize a variable

jtbadn

Pass the variable by value to a function change and display the value inside the function.

allquixotic

one line of code can do that

jtbadn

if the second value is Thomas...

int main()
{
int barry;
barry = 0;
{

i HAVE THIS SO FAR..

damn caps

allquixotic

ok, now you need to learn what functions are

your int main() function has opening and closing braces to indicate the body of the function

in C++, you can't have functions defined within functions, but you can define more than one function in a file

jtbadn

Nov 6, 2012 19:41

char *sp

sp = "Hello"l

allquixotic

so what you need to do is define a new function that is not named main(), outside of the opening and closing braces of the main() function

jtbadn

;

allquixotic

just hang on, you're declaring a string and I'm trying to teach you about functions.

int main()
{
int barry = 0;
}

You have that program. In that program, you have one function. Its name? main.

Whenever you hear someone say "pass by" (e.g., pass by value, pass by reference), we're talking about passing a value TO a function.

To pass anything, you need to have at least one more function in your program, to pass the variable to.

So instead of a program with just one function -- main() -- you need a second function also.

What does this second function need to do? Well, it needs to accept a variable as a parameter, because you're passing a value (or a reference, in the second case) to that function.

In the end, your program is going to have THREE functions in the completed assignment.

One function is given already -- the main() function.

jtbadn

uhh

allquixotic

The second function will need to work by passing by value.

The third function will need to work by passing by reference.

That's basically the entire assignment.

What you're going to do is, call the other two functions you declared, from the main() function.

Here's another function, in case you still don't get what a function is:

int gain()
{
return 0;
}

What's different about gain() than main()? Well, the name is different, and it has a different implementation.

jtbadn

Nov 6, 2012 19:45

yes, I do understand that.

allquixotic

The implementation of the function is the code that goes between the braces.

So, now you can start to lay out the structure of your completed program:

int main()
{
// declare your variable
// call function2 by value
// call function3 by reference
}

int function2()
{
//change the passed value
}

int function3()
{
//change the passed reference
}

now, all you have to do is replace the comments (starting with //) with your code

and you need to add appropriate parameters to function2() and function3()

jtbadn

ok

allquixotic

So, in C++, what's a value variable versus a reference variable?

Here's a hint: a value variable is one like that int barry = 0; we talked about earlier

you should be able to complete the "pass by value" part of the assignment now, without worrying about "references" at all.

you already declared barry, so you need to pass that value into function2, by calling function2 from main().

jtbadn

right..

like this

include<stdio.h>

int main()
{
int barry;
barry = 0;

}

allquixotic

what you have so far is right, but you're missing a # before the include :)

jtbadn

Nov 6, 2012 19:51

right, i didn't have it highlighted.

allquixotic

you should probably write function2() now. What is the purpose of function2()? It's supposed to accept an integer value as a parameter, then change that value, and print it out.

jtbadn

Variable name has the value "Barry"

calling pass by value
Parameter now has the value "Thomas"

pass by value returned
Variable name has the value "Barry"

calling pass by references
Parameter now has the value "James"

call by reference returned
Variable name has the value "James"

This is the example that's given... so I think it best to follow it.

so function 2 would be Thomas?

allquixotic

oh ok, they're using strings

it doesn't matter what name you give to the function

you aren't printing out the name of the function in the output

jtbadn

function2 is fine.

allquixotic

so, instead of int we're going to use char*, which is a string type variable

change your int to a char* in the main() function

jtbadn

Nov 6, 2012 19:53

#include<stdio.h>

int main()
{
char* barry;
barry = 0;

allquixotic

char* barry = "Barry";

jtbadn

ok

allquixotic

you have to assign it a value -- and 0 isn't a valid value for a string

actually it is, because it's a NULL pointer, but I don't want to confuse you

for now just think that when you use a char* type variable (a string), you need to assign it a string, which is in quotes "like this"

jtbadn

#include<stdio.h>

int main()
{
char* barry = "Barry";

// call function2 by value
// call function3 by reference
}

allquixotic

right, you can stop working on main() now, and we'll come back to it later

jtbadn

Nov 6, 2012 19:55

now you wanted me to write function 2

allquixotic

focus on function2 for now

function2 needs to accept a parameter in order to use that variable, barry, from main(), in function2.

when you declare function2, you need to tell the program what type of parameter it accepts.

jtbadn

ok

allquixotic

you know how we have the open and closed parentheses after the name of the function?

jtbadn

do I need to initialize Thomas like this then,...

int function2()
{
char* thomas = "Thomas";

//change the passed value
}

allquixotic

within those parentheses is where you put your formal parameter declaration.

jtbadn

Nov 6, 2012 19:56

Yeah

allquixotic

you have a good idea, and that's pretty close, but instead of having char* thomas in the function body, it needs to be within the parentheses after function2, to indicate that it is a parameter of function2().

THEN, what you can do is assign the string "Thomas" to the parameter itself, within the function body.

jtbadn

int function2(thomas)
{

?

allquixotic

int function2(char* thomas).

jtbadn

oic

allquixotic

without the period...

the thing is, within the body of function2, you can directly refer to thomas (in lowercase, without quotation marks) to read from or write to that passed-in parameter.

jtbadn

Nov 6, 2012 19:59

i understand.

allquixotic

so now, within the body of `function2`, you can do:
thomas = "Thomas";

jtbadn

ok

allquixotic

So here's the thing. Since we are passing by value, when you modify the variable thomas within the body of function2, you are not modifying the value of variable barry within the body of main!

"Pass by value" means exactly this behavior. Whenever you pass the value into a function, that function receives a copy of the value you passed, and can modify it independently.

Let me explain pass by value in terms of real life experience.

On a piece of paper, I write "john". I want to pass this to you by value. To do that, I get another piece of paper and write "john" on it, and hand you the second piece of paper.

If you then erase "john" from your paper with an eraser and write "smith" instead, it won't change the fact that "john" is written on my piece of paper.

jtbadn

and then I can write "athan" at the end of john and change it.

allquixotic

If you pass by reference, it works differently.

jtbadn

Nov 6, 2012 20:03

i see.

allquixotic

If we were passing by reference, then I take a piece of paper, and write "john", and hand you the same paper that I wrote "john" on.

You can then erase "john" and write "smith", and now the paper is changed for both of us -- if I look at the paper you wrote on, it doesn't say "john" anymore. it says "smith"!

jtbadn

understood.

allquixotic

So passing by value makes a copy of the data that you pass to the function. Passing by reference does not copy the data, it's referring to the same data in both functions.

So in function2(), since we passed by value, changing the variable to "Thomas" doesn't make char* barry in main() also say "Thomas".

But in function3() we're going to pass by reference, and when we do that, it will change the value of that char* barry to something else.

jtbadn

so passing reference will change the function that it's been pointed at..

passing function makes a copy

allquixotic

Passing by reference will change the value of the variable in the function that passed it in.

jtbadn

Nov 6, 2012 20:06

passing by value rather.

allquixotic

Passing by value makes a copy.

jtbadn

passing by reference modifies the reference completely.

allquixotic

It modifies the value contained in the variable, not the reference itself.

jtbadn

ok

allquixotic

the reference itself can be changed, but that's not what we're talking about.

so anyway, the assignment asks you to print out some stuff

have you seen cout << something << endl; before?

jtbadn

Nov 6, 2012 20:08

yes

allquixotic

the "something" is where you need to put your strings that you print out in the program, so that you can show proof by running the program that you did the assignment...

jtbadn

cin requires user input, cout prints out something to the screen.

allquixotic

from what I understand, they want you to print out the value of the parameter within the function2 function, and then print out the value of the barry variable back in main().

jtbadn

so in function2...

it should read ..

int function2(char* thomas)
{
thomas = "Thomas";
cout << thomas << endl;

allquixotic

Within main(), before you call function2(), you should cout << "Variable name has the value \"" << barry << "\"" << endl;

What you just wrote is correct, but I think they want you to include some other verbiage, too, I'll help you with that.

Instead of the cout << thomas line, change it to: cout << "Parameter now has the value \"" << thomas << "\"" << endl;

that includes the output that your sample shows

actually, update function2() to have a return type of void, which means it doesn't return anything. Just change int to void.

jtbadn

Nov 6, 2012 20:12

#include<stdio.h>

int main()
{
char* barry = "Barry";
cout << "Variable name has the value \ "" << barry << "\"" << endl;

// call function2 by value
// call function3 by reference
}

int function2(char* thomas)
{
thomas = "Thomas";
cout << "Variable name has the value \ "" << thomas << "\"" << endl;

//change the passed value
}

allquixotic

Great! Almost there. Change the return type of function2(), and then you just have to write function3() and add some more cout statements, and call the functions from main().

The program's starting to take shape though.

jtbadn

change the return type of function2() ?

allquixotic

Yeah. Read up before your paste.

1 min ago, by allquixotic

actually, update function2() to have a return type of void, which means it doesn't return anything. Just change int to void.

jtbadn

void function2(char* thomas)

allquixotic

Yes.

jtbadn

Nov 6, 2012 20:14

ok

allquixotic

Do you know how to call a function and pass a parameter to it?

jtbadn

i don't recall

allquixotic

Instead of that // call function2 by value you need to call the function. It's one line of code. In fact, it looks almost like the function signature of function2 which you already wrote, except for omitting the return type and changing the parameter to an actual parameter.

Give it a try. It's real simple. You just type the name of the function, then an opening parenthesis, then the name of the variable you're passing in, then the closing parenthesis, then a semicolon.

(don't include the type of the variable you're passing in, only its name.)

you want to pass in barry, by the way.. just in case you were confused

jtbadn

function2(barry); ???

allquixotic

YES!!!!

Perfect :)

jtbadn

Nov 6, 2012 20:17

yay

allquixotic

What that line of code does is, it causes the body of function2() to execute, and when you read or write thomas within the body, it's going to initially contain a copy of the value from barry.

But, since we passed by value (remember that?) you aren't changing barry to "Thomas"!

jtbadn

barry remains Barry

thomas is Thomas

allquixotic

Yes

They're two separate variables because we passed by value.

In function3() the exact opposite will be true.

So now, below the line of code you just wrote -- that function2(barry); -- you need to use cout for the next line of program output.

cout << Variable name has the value \"" << barry << "\"" << endl;

And indeed, barry will have the value of "Barry".

Because we passed by value, not by reference.

"Thomas", in fact, will be destroyed when function2() exits.

So I think at this point you're done all of your program up through the point where you have to pass by reference.

Passing by value and printing out the stuff is done

jtbadn

#include<stdio.h>

int main()
{
char* barry = "Barry";

function2(barry); //calling function2 by value

cout << "Variable name has the value \ "" << barry << "\"" << endl;

// call function3 by reference
}

allquixotic

Post your code so far

jtbadn

Nov 6, 2012 20:22

this what you mean?

allquixotic

Yes, you've got main() right so far

jtbadn

here is everything

#include<stdio.h>

int main()
{
char* barry = "Barry";

function2(barry); //calling function2 by value

cout << "Variable name has the value \ "" << barry << "\"" << endl;

// call function3 by reference
}

void function2(char* thomas)
{
thomas = "Thomas";
cout << "Variable name has the value \ "" << thomas << "\"" << endl;

//change the passed value
}

int function3()
{
//change the passed reference
}

allquixotic

Actually, if you want your code to output the same as the instructor's example, you need to add a cout before you call function2, in main().

So it'll be something like:
cout << ...
function2(barry);
cout <<... (you already have this line written)

jtbadn

#include<stdio.h>

int main()
{
char* barry = "Barry";

cout << "Variable name has the value \ "" << barry << "\"" << endl;

function2(barry); //calling function2 by value

// call function3 by reference
}

void function2(char* thomas)
{
thomas = "Thomas";
cout << "Variable name has the value \ "" << thomas << "\"" << endl;

//change the passed value
}

int function3()
{
//change the passed reference
}

that is how I originally had it.

I thought you had asked me to change that.

allquixotic

Well you need to display it both BEFORE and AFTER calling function2, to match the instructor's output, right?

The point is to show that the value of barry does not change from "Barry" even though the variable was purportedly "modified" in the function2. Because it's by value, it doesn't change.

jtbadn

Nov 6, 2012 20:24

ok

so you were right..

so this is main..

#include<stdio.h>

int main()
{
char* barry = "Barry";

cout << "Variable name has the value \ "" << barry << "\"" << endl;

function2(barry); //calling function2 by value

cout << "Variable name has the value \ "" << barry << "\"" << endl;

// call function3 by reference

}

allquixotic

that's 100% correct, and the // call function3 by reference is going to be almost exactly the same as the three lines where you called function2 and the couts.

you may as well copy and paste from the first cout through the second cout, and change function2 to function3 in the copied code.

jtbadn

huh

?

allquixotic

#include<stdio.h>

int main()
{
char* barry = "Barry";

cout << "Variable name has the value \ "" << barry << "\"" << endl;
cout << "Calling by value" << endl;
function2(barry);
cout << "Variable name has the value \ "" << barry << "\"" << endl;

cout << "Variable name has the value \ "" << barry << "\"" << endl;
cout << "Calling by reference" << endl;
function3(barry);
cout << "Variable name has the value \ "" << barry << "\"" << endl;

return 0;
}

see what I did?

jtbadn

yeah, made "calling by value" and calling by reference" show up on screen

allquixotic

that should be your entire main function -- I think it's totally done, for both functions... I added in a return 0; at the end... now it's done... you just have to write function3

just copy my code from there, and focus on function3

Maybe this will help you, since it's dealing with exactly this problem:
http://www.learncpp.com/cpp-tutorial/73-passing-arguments-by-reference/

I gotta AFK for a minute so read that webpage, it says it exactly as I would say it... pay special attention when you see & because that's what gives away the fact that the function's parameter is being passed by reference

jtbadn

Nov 6, 2012 20:30

ok will do

i think though, you have one too many cout << "Variable...... " in there though..

allquixotic

you're right :D go ahead and remove the extra one

jtbadn

one right after the other one.

allquixotic

it'd just say the same thing, anyway

jtbadn

void function3(int &barry) // barry is now a reference
{
using namespace std;

cout << "barry =" << barry << endl;
char james = "James";

cout << "Variable name has the value \ "" << James << "\"" << endl;

cout << "Call by reference returned." << endl;

}

Tell me if that's right when you get back...

allquixotic

yay!!!!!!! well, except for using namespace std; you should put that at the top of your program :)

jtbadn

Nov 6, 2012 20:39

oh right

allquixotic

also, change << James << to << james <<

and the type of char james should be char* james

and...............

jtbadn

oh right

allquixotic

you need to modify int &barry to reflect the correct type

jtbadn

void function3(char* &barry) // barry is now a reference
{

cout << "barry =" << barry << endl;
char* james = "James";

cout << "Variable name has the value \ "" << james << "\"" << endl;

cout << "Call by reference returned." << endl;

}

allquixotic

char* &barry

yayyyyyy!

uh...

you don't need to declare variable char *james; you can directly modify the value of barry to "James"

barry = "James";

and guess what? lol? the variable barry in the main() method will change to "James"

because you passed by reference!

jtbadn

Nov 6, 2012 20:41

void function3(char* &barry) // barry is now a reference
{

cout << "barry =" << barry << endl;

barry = "James";

cout << "Variable name has the value \ "" << james << "\"" << endl;

cout << "Call by reference returned." << endl;

}

allquixotic

cout << "Variable name has the value \"" << barry << "\"" << endl;

you got rid of james, remember?

hi @KronoS, feel free to stick around if you want to learn basic C++ ;P

KronoS

just as an FYI we do have this room:

Fake Programmers

The room for elite Super Users who realize they know nothing a...

allquixotic

@jtbadn is a real life friend so I'm helping him personally, and think that more people would just confuse him

jtbadn

#include<stdio.h>

using namespace std;

int main()
{
char* barry = "Barry";

cout << "Variable name has the value \ "" << barry << "\"" << endl;
cout << "Calling by value" << endl;

function2(barry);
cout << "Variable name has the value \ "" << barry << "\"" << endl;

cout << "Calling by reference" << endl;

function3(barry);
cout << "Variable name has the value \ "" << barry << "\"" << endl;

return 0;
}

void function2(char* thomas)
{
thomas = "Thomas";

cout << "Variable name has the value \ "" << thomas << "\"" << endl;

that's the whole thing.

changed it back to barry

allquixotic

you have an extra function2 body at the end of the program

seems copy and pasted

oh wait no

KronoS

Nov 6, 2012 20:44

@allquixotic thanks... I know a little, but can always learn more

allquixotic

NEVERMIND

i didn't click "(see full text)"

jtbadn

ok, so this should work then?

allquixotic

well, it looks pretty darn good so far, i'm not sure if it'll output precisely the same text as the example, but the concept is right

jtbadn

it actually does make a lot more sense now that you've taught me how to do this... the second project should be easier.

allquixotic

I can't see your prof giving you less than a B on this because it really looks like you have the two concepts down

to summarize, you declare a variable in main(), you assign it the value "Barry", you pass it by value to function2(), function2 changes its local copy of the value to "Thomas" (but not the main() version of the variable), then you pass the main() variable by reference to function3(), and function3 changes the value of the variable which is shared between main() and function3(), resulting in the value of "Barry" changing to "James" in main().

KronoS

Nov 6, 2012 20:46

What's the original problem?

allquixotic

@KronoS basically he has to print out some strings and write a program to distinguish between passing by value and passing by reference

KronoS

Got it.

jtbadn

problem with this line...

cout << "Variable name has the value \ "" << barry << "\"" << endl;

allquixotic

there's a space between the first backslash and the quotation mark

jtbadn

oh yes

allquixotic

Nov 6, 2012 20:49

the backslash means "treat the next character as part of the string literal", but space doesn't have any special meaning, so you can't escape it.

you have to put the backslash there to tell the compiler "this quotation mark is part of the string, I don't want to actually end the string literal here"

jtbadn

still getting error

allquixotic

what error?

jtbadn

cout << "Variable name has the value \"" << barry << "\"" << endl; is that right?

It's going to keep showing up cause I copy/pasted

illegal character \

allquixotic

should be, unless there's an error elsewhere

is it erroring on that line?

KronoS

What IDE are you using?

jtbadn

Nov 6, 2012 20:53

still giving error

Right now I'm using Dev-CPP

allquixotic

heh

jtbadn

cause it's a lot easier and quicker than visual studio

and it has its own compiler.

@allquixotic WHAT?

allquixotic

>_>

KronoS

@jtbadn The prob with Visual-Studio is that you have to use the Visual C++ if you don't want to configure anything

however if you use CMAKE you can have multiplatform support.

allquixotic

oy no

don't teach him cmake now lol

KronoS

Nov 6, 2012 20:55

I know... that's why I'm agreeing with him :)

jtbadn

what is wrong with this line?

cout << "Variable name has the value \"" << barry << "\"" << endl;

too many "

allquixotic

nothing? unless there's a space before a \

KronoS

It'd be nice if we got a more specific error message

jtbadn

" or ' ' ???

allquixotic

always "

it's one character. a double quote character

jtbadn

Nov 6, 2012 20:59

9 C:\Users\Joshua\Desktop\main.cpp `cout' undeclared (first use this function)

yes

allquixotic

AH

#include <iostream>

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