Function with input but no output?

[cozySam]

Hello!

This function is supposed to have input but no output. I really want to understand how this works. Can someone help me fill in the gaps?

#include <iostream>

using namespace std;

void Hello(char* name)
{
    cout << "Hello " << name << "!";
}

int main()
{
    Hello("John");
    cin.get();
    return 0;
}

So the void means the function has no return. Hello is the name of the function. I'm not sure about that part in the parenthesis.

As far as I know char is a character type and it can store a single character (letter), and char* indicates a character pointer which is a type of variable that can reference memory addresses of other variables.

What type is name? Is it a char variable? Is it a char pointer? Maybe char vector? This I don't understand. I think it's a char pointer. But I can't see how it all fits in the big picture. Where or how is "John" stored? Is it stored at all?

The Hello("John") is a function call, passing in "John" as a string literal and the only argument. As I said this is where I get stuck. What happens to "John"? Is it stored in name? Before it's printed out?

I'm having hard time with this one. I hope someone can help me understand this.

[VictorN]

Hello!

This function is supposed to have input but no output. I really want to understand how this works. Can someone help me fill in the gaps?

#include <iostream>

using namespace std;

void Hello(char* name)
{
    cout << "Hello " << name << "!";
}

int main()
{
    Hello("John");
    cin.get();
    return 0;
}

What type is name? Is it a char variable? Is it a char pointer? Maybe char vector? This I don't understand. I think it's a char pointer.

As you can see from the function definition the type of name is char* - it is a char pointer. Or you can treat it as a character array.

[2kaud]

char* indicates a character pointer which is a type of variable that can reference memory addresses of other variables.

char* is a pointer (ie contains the memory address of) to a section of memory that contains data of type char (usually 1 byte).

What happens to "John"? Is it stored in name? Before it's printed out?

John is a string literal which the compiler stores in memory at some address chosen by the compiler. In Hello(), the parameter name is a memory address and is set by the compiler to the start of the memory address where the compiler stored the characters John. This is called a c-style string and in memory the chars are terminated by the null character (\0) to indicate the end of the string - often referred to as a null-terminated string.

cout has name as a parameter for insertion to the stream. As name is of type char*, the characters stored in memory pointed to by name are sent to cout until a null terminator (\0) is found.

[cozySam]

Alright, so name is a char pointer. And where is it pointing to? Where is the char that cahr* points to?

Why is there no char declaration?

[VictorN]

Alright, so name is a char pointer. And where is it pointing to? Where is the char that cahr* points to?

In the memory beginning with the address name. And there may be more than one char. All of them are between the address name and the address where the NULL character is.

Why is there no char declaration?

Sorry. I don't understand what you mean here...

[cozySam]

In the memory beginning with the address name. And there may be more than one char. All of them are between the address name and the address where the NULL character is.

Sorry. I don't understand what you mean here...

Let me quote 2kaud here, maybe you will see what I mean.

[cozySam]

char* is a pointer (ie contains the memory address of) to a section of memory that contains data of type char (usually 1 byte).

John is a string literal which the compiler stores in memory at some address chosen by the compiler.

This is also the confusing part for me. How does the compiler store the literal "John" in memory? As a char vector? So this is why I don't have to declare a new char vector e.g. char nameVector[]; and assign its address to the char pointer name using the & (address of) operator, e.g. name = &nameVector. The compiler does this automatically in the background?

In Hello(), the parameter name is a memory address

So name is a pointer to the memory location where the compiler auto-created the char vector (nameVector) and stored each individual letter of "John" in?

and is set by the compiler to the start of the memory address where the compiler stored the characters John.

This is a range of addresses. If "John" is stored as a vector then it will be stored as "J"+"o"+"h"+"n" which is concatenated to "John"? And char pointer name points to the first address in this range. (Also called the offset?)

This is called a c-style string and in memory the chars are terminated by the null character (\0) to indicate the end of the string - often referred to as a null-terminated string.

Meaning that the pure C language does not support strings the same way C++ or Java does? It breaks it up into individual characters (e.g. "J"+"o"+"h"+"n") and stores each character as a char in a char vector?

cout has name as a parameter for insertion to the stream.

A parameter?... if it can take parameters, does that mean that cout is a function or a method? I have not studied object orientation yet, but judging by what I know so far things like cout are part of the standard library and they are often objects of classes, and methods is the name for functions that an object can have inside a class. (Correct me if I'm wrong.)

As name is of type char*, the characters stored in memory pointed to by name are sent to cout until a null terminator (\0) is found.

The null terminator is automatically appended to the char vector (nameVector) by the compiler?

[2kaud]

Please don't use the phrase char vector. In c++ this means something specific in a different context.

To store the literal John in memory, the compiler generates code that asks the OS for the address of memory that can be used to stored 5 characters (J o h n \0). The code then copies these characters sequentially into memory starting at the memory address provided by the OS.

The language c was developed before c++. null-terminated strings are part of the c language. c++ originally was based upon c so these null-terminated (c-style strings) can be used with c++. c++ also has a class called string which is the preferred way of using strings in c++. In c++ just referring to string would usually mean referring to the c++ string class. That's why when null-terminated strings are used in c++ they are often referred to as c-style strings (with a type char*).

Consider cout to be an instance of a class. The operator << is overloaded for the class to mean insertion into the class - insertion into the cout class instance means display. Consider

cout << 123;

This displays 123. This is actually

cout.operator<<(123);

where 123 is the parameter for the class function operator<<.

[2kaud]

How are you learning c++? Do you have a previous programming background?

[cozySam]

Alright, so this is what I think might be going on.

This is what the address range for the char vector nameVector might be and what the values might be.

Address     Value
=================
0x0f31      J
0x0f32      o
0x0f33      h
0x0f34      n
0x0f35      NUL

This is what the variable values might be.

Variable    Type    Address     Value
======================================
name        char*   0x0fb20     0x0f31
nameVector  char    0x0f31      John

As you can see the value of the char pointer name is the starting address of the nameVector. It references the first element in the vector.

(I realize now that I may be using the term vector wrong here. You might want to replace that with array.)

[GCDEF]

In C, there really is no such thing as a string variable type. What you're dealing with is the way the C language handles strings. I'd recommend reading up on how they work. This looks like as good a start as any

http://www.programiz.com/c-programming/c-strings

[2kaud]

IMO you're getting too involved with trying to understand some of the low level detail when with c++ it's the high level stuff that you should really be trying to understand. The c++ way for the program in post #1 could be

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

void Hello(const string& name)
{
	cout << "Hello " << name << "!\n";
}

int main()
{
	Hello("John");
	cin.get();
	return 0;
}

and not a pointer in sight! From where did the program in post #1 originate - if from an early part of a c++ course then I take issue with how they are teaching c++.

[cozySam]

How are you learning c++? Do you have a previous programming background?

I have learned some programming fundamentals on my own and I took a basic Python course before I started with this C++ course.

From where did the program in post #1 originate - if from an early part of a c++ course then I take issue with how they are teaching c++.

Yes, it's from the C++ course I'm taking. It's from the course book to be more precise, which is mandatory for the course by the way.

How do you think I feel? I am relatively new to programming in general, and completely new to C++. I don't even have the basic knowledge to recognize the bad practices they are using. Bad teaching practices I should say. All I have is frustration!

I now understand that my frustration stems not only from my inability to solve a computational problem using C++ but from the bad course plan and the bad book I'm reading. It's not because I'm dumb and don't have the will to learn, it's because it's taught to me in a wrong way. Frankly, if they are drilling down low level stuff like pointers and C style strings into my brain in a course titled C++ for beginners then I might as well go take a pure C programming course. That way I will at least have a decent understanding of what I just read, and take it with me as I move on to C++.

The fact that it's a distance learning course doesn't help either. All I have is basically this stupid "mandatory" course book. I might have gotten better off reading some online tutorial than reading this book. But as it was said to be mandatory, because the course is based around the book, I thought I might as well stick to that. Besides, I'm almost done with it, so I don't want to jump to a web tutorial now and neither finish the book nor the tutorial before the course is over.

Another thing that upsets me about this book is that the publisher is not making the code examples available for free download on their website. So every code example I find in the book I have to type in manually to test it. That's so time consuming! They only make the code examples available as downloads along with other so called teaching material for the special teacher's edition of the book. So that the teacher, who presumably already knows C++, can practice and learn even more C++ I guess! While I sit my *** off here and wear out my fingers while learning almost nothing. Honestly, most of the stuff I have read in this book I already knew. If it's anything I have learned here is that there are bad C++ books, and bad C++ teachers. (As if I didn't know that already! This just confirms it.)

The guy who wrote this book is probably a very good and experienced programmer, but as a teacher or a book author... he sucks! I think he just wrote this book for some extra income. I have a very good reason to believe that. But I won't go any further on that. I have already said too much. I don't want this thread to turn into a book review, or a course review. There are simply good books and bad books, good courses and bad courses. This book and this course belongs to the latter category. Let's leave it at that.

The c++ way for the program in post #1 could be

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

void Hello(const string& name)
{
	cout << "Hello " << name << "!\n";
}

int main()
{
	Hello("John");
	cin.get();
	return 0;
}

This code example looks much cleaner.

I know I should probably Google this, but what is the keyword "const" used for? Something to do with constants I'm guessing? I have seen it used in at least 3 different code examples in the book I'm reading and not once has it been explained (unlike the other keywords which have been explained in some detail), and I have almost read the entire book now (unlikely it will be explained at the end of the book).

The & is the "address of" operator? I read it can mean several things in different contexts. I have seen it used as "address of" for pointers and I have seen the double && as "and" operator in conditionals. I'm not sure how to read it in this case.

[VictorN]

I know I should probably Google this, but what is the keyword "const" used for? Something to do with constants I'm guessing? I have seen it used in at least 3 different code examples in the book I'm reading and not once has it been explained (unlike the other keywords which have been explained in some detail), and I have almost read the entire book now (unlikely it will be explained at the end of the book).

http://stackoverflow.com/questions/3...type-specifier

The & is the "address of" operator? I read it can mean several things in different contexts. I have seen it used as "address of" for pointers and I have seen the double && as "and" operator in conditionals. I'm not sure how to read it in this case.

http://programmers.stackexchange.com...ific-functions
http://stackoverflow.com/questions/6...ion-signatures

[2kaud]

const basically means constant ie that the value can't be changed. Consider

int a = 4;
++a;

a is now 5
but

const int a = 4;
++a;

gives an error as a is now a constant and can't have it's value changed. It's good programming practice to make variables const whenever possible to avoid programming errors (values being changed when they shouldn't be).

& has several different meanings depending upon how its used. It can mean bitwise and, address of or reference. In this example it means reference. For functions, arguments can be passed either by value or by reference (there is also r-val reference but that's for advanced c++!). Without the & specified, the parameter is passed by value. This means that the specified variable in the function receives a copy of the value. Any changes to the variable in the function are not passed back to the calling function. If & is specified, then the parameter is passed by reference. This means that the specified variable in the function points to the same data as in the calling function and any changes to the variable are reflected back in the calling function. Consider

#include <iostream>
using namespace std;

void testv(int a)
{
    ++a;
}

void testr(int& a)
{
    ++a;
}

int main()
{
    int x = 5;
    
    cout << "before testv x is " << x << endl;
    testv(x);
    cout << "after testv x is " << x << endl;

    testr(x);
    cout << "affer testr x is " << x << endl;
}

Because pass by value copies the variable, for some data types this can have a significant performance overhead. So the data is passed by reference even when it is not required or desired to change the value used in the calling function. So const is used so that if the value in the function is tried to be changed an error is generated.

It's from the course book to be more precise, which is mandatory for the course by the way

What book is this?

For an on-line course in c++ have a look at
http://www.learncpp.com/