Trying to decide which programming language I want to learn

[FactChecker]

I have issue building the solution. VS gives me error.

Please don't ever ask this without telling us what the complete error statement, including the line number, if given.

1) Why I have to declare int main()? What is int for?

Because main often returns a completion code. 0 = normal satisfactory completion. Anything else indicates a different kind of termination.

2) Why in the first book, it just declare int main(). But in this book it has parameter like int main(int nNumberofArg, char* pszArgs[])?

This is the standard way that the main program can be passed information when it is called on the command line or in scripts. Like if you want to tell main the name of an input file and an output file. nNumberofArg tells main how many arguments were on the command line. All the arguments are stored in the array pszArgs. The first one is special (if it is not null). It gives the name of the executable that is running and contains main. (See https://en.cppreference.com/w/cpp/language/main_function )

 

[yungman]

Please don't ever ask this without telling us what the complete error statement, including the line number, if given.Because main often returns a completion code. 0 = normal satisfactory completion. Anything else indicates a different kind of termination.This is the standard way that the main program can be passed information when it is called on the command line or in scripts. Like if you want to tell main the name of an input file and an output file. nNumberofArg tells main how many arguments were on the command line. All the arguments are stored in the array pszArgs. The first one is special (if it is not null). It gives the name of the executable that is running and contains main. (See https://en.cppreference.com/w/cpp/language/main_function )

Sorry, I added into the other post already.

 

[jtbell]

std::string myThing[20] that tell people myThing[20] is a 20 element string.

No.

std::string myThing; gives you a single C++-style string which is initially "empty". It's a dynamic object which can contain any number of characters depending on what you do with it later. For example, if you later write myThing = "automobile";, it now contains ten characters.

std::string myThing[20]; gives you a 20-element C-style array of std::string objects. You could then write

myThing[0] = "Hello";
myThing[1] = "my";
myThing[2] = "name";
myThing[3] = "is";
myThing[4] = "yungman";

However, I consider such a beast to be a Frankenstein's monster, mixing a C++-style construct with a C-style construct. When I taught C++, I always used std::vector instead of C-style arrays:

std::vector<std::string> myThing(20);
myThing[0] = "Hello";
myThing[1] = "my";
// etc.

After I stopped teaching C++, the C++11 standard (I think that was when it was) introduced C++-style fixed length arrays: std::array<std::string, 20> myThing;.

 

[yungman]

No.

std::string myThing; gives you a single C++-style string which is initially "empty". It's a dynamic object which can contain any number of characters depending on what you do with it later. For example, if you later write myThing = "automobile";, it now contains ten characters.

std::string myThing[20]; gives you a 20-element C-style array of std::string objects. You could then write

myThing[0] = "Hello";
myThing[1] = "my";
myThing[2] = "name";
myThing[3] = "is";
myThing[4] = "yungman";

However, I consider such a beast to be a Frankenstein's monster, mixing a C++-style construct with a C-style construct. When I taught C++, I always used std::vector instead of C-style arrays:

std::vector<std::string> myThing(20);
myThing[0] = "Hello";
myThing[1] = "my";
// etc.

After I stopped teaching C++, the C++11 standard (I think that was when it was) introduced C++-style fixed length arrays: std::array<std::string, 20> myThing;.

I thought myThing is a 20 character string, that is each element is only ONE charater. like if myThing [] = "Hello"
myThing[0] = 'H'; myThing[1] = 'e'; myThing[2] = 'l'; etc. But you are saying myThing[0] = "Hello" a whole word. Which one is correct?

 

[yungman]

I am learning loops, the goto, while, do-while all doing the same thing. Do I have to learn all or them or I just stick with one and be done with it? too many options here.

 

[FactChecker]

There are reasons that goto is frowned on. Some standards forbid it. Bad programmers have put so many gotos in their code that they become spider webs. If you use it, make sure there is no other way and keep it as simple as possible. Avoid it if possible.

 

[yungman]

There are reasons that goto is frowned on. Some standards forbid it. Bad programmers have put so many gotos in their code that they become spider webs. If you use it, make sure there is no other way and keep it as simple as possible. Avoid it if possible.

So "while" it is! too many options in C++.

 

[Mark44]

I thought myThing is a 20 character string, that is each element is only ONE charater. like if myThing [] = "Hello"

No. In the declaration std::string myThing[20];, myThing is a container with 20 elements, each of which is an std::string. Each element is not necessarily a single character.

myThing[0] = 'H'; myThing[1] = 'e'; myThing[2] = 'l'; etc. But you are saying myThing[0] = "Hello" a whole word. Which one is correct?

See above.

I am learning loops, the goto, while, do-while all doing the same thing. Do I have to learn all or them or I just stick with one and be done with it? too many options here.

The goto statement is not a loop -- it's a branch instruction.
There are three types of loops: for (which you didn't mention), while, and do ... while. They don't all do the same thing.
A for loop is usually used when you want the loop to execute a specific number of times.
A while loop is usually used when you don't know how many times the loop should execute. It checks the condition expression first, so the loop body might not execute at all.
A do...while loop is also used when you don't know how many times the loop should execute, but it checks the condition expression after each loop iteration. It will always execute at least once.

There are reasons that goto is frowned on. Some standards forbid it. Bad programmers have put so many gotos in their code that they become spider webs.

Granted, goto is usually frowned on, but there are legitimate reasons to use it, such as to bail out under extraordinary conditions. Some arguments pro, per Steve McConnell's "Code Complete," pp. 348,349.

A well-placed goto can eliminate the need for duplicate code.
The goto is useful in a routine that allocates resources, performs operations on those resource, and then deallocates the resources. With a goto, you can clean up in one section of code.
In some cases, the goto can result in faster and smaller code. Knuth's 1974 article cited a few cases in which the goto produces a legitimate gain.
Two decades' worth of research with gotos has failed to demonstrate their harmfulness. In a survey of the literature, B.A. Sheil concluded that unrealistic test conditions, poor data analysis, and inconclusive result failed to suppor the claim of Shneiderman and others that the number of bugs in code was proportional to the number of gotos (1981).
Finally, the goto was incorporated into the Ada language, the most carefully engineered programming language in history. Ada was developed long after the arguments on both sides of the goto debate had been fully developed.

So "while" it is! too many options in C++.

Suit yourself, but really, you should learn all three loop constructs: for, while, do...while.

 

[sysprog]

@Mark44 http://homepages.inf.ed.ac.uk/rni/papers/realprg.html
Real programmers aren't afraid to use GOTOs.

 

[jtbell]

I thought myThing is a 20 character string, that is each element is only ONE charater. like if myThing [] = "Hello"
myThing[0] = 'H'; myThing[1] = 'e'; myThing[2] = 'l'; etc. But you are saying myThing[0] = "Hello" a whole word. Which one is correct?

With any of these declarations:

std::string myThing[20];
std::vector <std::string> myThing(20);
std::array <std::string, 20> myThing;

myThing is a collection of 20 strings. Each string is a collection of characters. If you want to access individual characters, you need to specify two indexes: which string and which character. Example:

std::string myThing[20];
myThing[0] = "Hello";
myThing[1] = "my";
myThing[2] = "name";
std::cout << "First word is " << myThing[0] << std::endl;
std::cout << "Second letter of third word is " << myThing[2][1] << std::endl;
myThing[0][0] = 'J';
std::cout << "First word is now " << myThing[0] << std::endl;  // guess it, then try it!

 

[yungman]

No. In the declaration std::string myThing[20];, myThing is a container with 20 elements, each of which is an std::string. Each element is not necessarily a single character.
See above.
The goto statement is not a loop -- it's a branch instruction.
There are three types of loops: for (which you didn't mention), while, and do ... while. They don't all do the same thing.
A for loop is usually used when you want the loop to execute a specific number of times.
A while loop is usually used when you don't know how many times the loop should execute. It checks the condition expression first, so the loop body might not execute at all.
A do...while loop is also used when you don't know how many times the loop should execute, but it checks the condition expression after each loop iteration. It will always execute at least once.
Granted, goto is usually frowned on, but there are legitimate reasons to use it, such as to bail out under extraordinary conditions. Some arguments pro, per Steve McConnell's "Code Complete," pp. 348,349.

Suit yourself, but really, you should learn all three loop constructs: for, while, do...while.

To me, for is different from the others as you program how many times. The other ones just loop until jumping out. Just use an if statement to to goto start. They all do the same thing at the end.

 

[FactChecker]

@Mark44 http://homepages.inf.ed.ac.uk/rni/papers/realprg.html
Real programmers aren't afraid to use GOTOs.

Well, among sane programmers, GOTOs are fine. But look around at your fellow programmers -- are they sane? Or are they sociopaths? The odds are that some are sociopaths. ;>)

 

[Mark44]

To me, for is different from the others as you program how many times.

All three loop constructs are different, as I explained in post #253.

The other ones just loop until jumping out.

No. The body of a while loop might not execute at all.

Just use an if statement to to goto start. They all do the same thing at the end.

Well, sure, the assembly code that the compiler generates boils down to compare instructions and jump instructions.

too many options in C++.

All procedural programming languages at a higher level than assembly provide the same sorts of loop statements.

 

[yungman]

I have been reviewing int array and char array, and then strings. A lot of my confusion are in this area. I feel I have been asking stupid questions. Also, I am reading back this thread again, memory is failing!

I'll be back.

 

[yungman]

......

char sayHello2 = "Hello";

Both definitions store the letters in the first 5 bytes of memory, followed by an ASCII null character. Both arrays have a length of 5, the number of characters up to the null.
........

This is about post #239
I though this is a 6 elements char array as the last one is '\0'.
I want to confirm these

1) You can declare: char myChar[] = " This" to create a 5 elements char array. But after this, you can ONLY write one element at a time. There's no myChar = "ABCD" one step filling 4 characters.

2) After declaring int myNumbers[] = {1,2,3,4,5}, I cannot read all 5 numbers in one code line, has to be one element at a time eg. cout << myNumbers[1] << endl; to read out "2".

3) There is no command to write all 5 numbers into myNumber[] in one line of code, has to be one number at a time.

Thanks

 

[FactChecker]

C++ and C can be very frustrating for scientific and engineering use. It lacks several features that were in FORTRAN specifically for that use (like reading and writing arrays).

 

[Mark44]

1) You can declare: char myChar[] = " This" to create a 5 elements char array. But after this, you can ONLY write one element at a time. There's no myChar = "ABCD" one step filling 4 characters.

Correct. The reason that you can't assign a new string to it is that myChar is a constant -- it can't be modified. In particular it is a pointer constant, which I don't think you've gotten to yet.

2) After declaring int myNumbers[] = {1,2,3,4,5}, I cannot read all 5 numbers in one code line, has to be one element at a time eg. cout << myNumbers[1] << endl; to read out "2".

Sort of correct, although your use of "read out" is a bit confusing. I would rephrase this to say that you can't output all of the numbers at the same time -- you have to access them one at a time.
I said "sort of correct" because the following is one code line.

cout << myNums[0] << myNums[1] << myNums[2] << myNums[3] << myNums[4] << endl

3) There is no command to write all 5 numbers into myNumber[] in one line of code, has to be one number at a time.

Correct. After the array has been initialized, there's no way to reassign a set of new values other than one at a time.

 

[yungman]

......

char sayHello2[] = "Hello";

Both definitions store the letters in the first 5 bytes of memory, followed by an ASCII null character. Both arrays have a length of 5, the number of characters up to the null.
........

I though this is a 6 elements char array as the last one is '\0'.

Correct. The reason that you can't assign a new string to it is that myChar is a constant -- it can't be modified. In particular it is a pointer constant, which I don't think you've gotten to yet.

I can change the characters in myChar, it is not defined as constant. just has to be one by one.

char myChar[] = "Test";
myChar[0] = 'F'; myChar[1] = 'G'; etc.

If I do cout << myChar; I will get "FGst" as I changed the first two characters. I actually verified with the program.

ThanksI don't mean to challenge you, I just feel this is very important for me to understand this for once. I am NOT learning anything more new until I get through array and strings. I want it to be very clear on this.

 

[Mark44]

From post #239:

......

char sayHello2[] = "Hello";

Both definitions store the letters in the first 5 bytes of memory, followed by an ASCII null character. Both arrays have a length of 5, the number of characters up to the null.
........

I though this is a 6 elements char array as the last one is '\0'.

The array has 6 elements, but the length of the string is 5, as reported by strlen(), is the number of characters up to, but not including the terminating null.

I can change the characters in myChar, it is not defined as constant. just has to be one by one.

char myChar[] = "Test";
myChar[0] = 'F'; myChar[1] = 'G'; etc.

If I do cout << myChar; I will get "FGst" as I changed the first two characters. I actually verified with the program.

myChar IS a constant, but myChar[0], myChar[1], etc. are not constants. You can change the individual characters in the string, provided that you include the index of the character (in brackets).
In short, you can't do this: myChar = "FGst";, but you can change individual characters as you did in your code example.
The constant that myChar represents is the address of the first byte in the array. Because myChar is a constant, you are not allowed to assign an new value (i.e., an address) to it.

 

[yungman]

From post #239:
The array has 6 elements, but the length of the string is 5, as reported by strlen(), is the number of characters up to, but not including the terminating null.
myChar IS a constant, but myChar[0], myChar[1], etc. are not constants. You can change the individual characters in the string, provided that you include the index of the character (in brackets).
In short, you can't do this: myChar = "FGst";, but you can change individual characters as you did in your code example.
The constant that myChar represents is the address of the first byte in the array. Because myChar is a constant, you are not allowed to assign an new value (i.e., an address) to it.

Thanks

So myChar is actually an address pointing to the first character of the array? I thought I asked and you said it's not somewhere. Is that true for int array, strings also?

Also, if I declare myChar[] = "Test", I forever locked myself that strlen() is going to be 4 for the rest of the program?

Thanks

 

[jtbell]

Is that true for int array, strings also?

With C-style arrays, whether of chars, ints or anything else, the name of an array by itself with no index (subscript) always "decays" to a pointer to the first item in the array.

This is not true for C++ std::string or std::array. The name of a string variable always represents the entire string, which is an object that contains more than just the chars. You can assign literal strings to them, and they automatically resize themselves accordingly.

#include <iostream>
#include <string>

using std::cout;
using std::endl;
using std::string;

int main ()
{
    string name = "Huey";
    cout << name << " has " << name.length() << " characters." << endl;
    name = "Dewey";
    cout << name << " has " << name.length() << " characters." << endl;
    return 0;
}

Output:

Huey has 4 characters.
Dewey has 5 characters.

(the first C++ program on my new iMac, after installing the compiler today! )

 

[yungman]

With C-style arrays, whether of chars, ints or anything else, the name of an array by itself with no index (subscript) always "decays" to a pointer to the first item in the array.

.......

What is the meaning of "decays"? Do you mean the array name is actually a POINTER ( address) to the first member of the array ( char, int and others)?

Thanks

I don't even look at strings and std::array. I just concentrate on array and C-String and resolve that, hopefully tomorrow I can get into strings. That's another can of worms. I just get the feeling that this is very very important to just go through the program and move on.

 

[Mark44]

What is the meaning of "decays"? Do you mean the array name is actually a POINTER ( address) to the first member of the array ( char, int and others)?

In a declaration like int numArray[5];, this is an array of five int values. However, if the name appears by itself, with no brackets, the name is a constant - the address of the first int value in the array -- a pointer. In this case, the address is of the first byte of the four bytes that hold an int value.
This business of array names being addresses will become more significant when you study functions and their parameters.

 

[yungman]

In a declaration like int numArray[5];, this is an array of five int values. However, if the name appears by itself, with no brackets, the name is a constant - the address of the first int value in the array -- a pointer. In this case, the address is of the first byte of the four bytes that hold an int value.
This business of array names being addresses will become more significant when you study functions and their parameters.

Sorry, I need to absolutely confirm:
numArray[5] is an array with 5 elements.
numArray is the address that point to the first element of array numArray[5].

Are std::strings like this also? Without brackets is the address pointing to the first element?

 

[yungman]

I just received this book today by Tony Gaddis
https://www.amazon.com/gp/product/0136022537/?tag=pfamazon01-20

Seems to be a really good book, I read the part on std::string section, it's very good. I should have waited for this book instead of wasting time on the first book. Now I am thinking whether I should stop and start over again following this new book.

I bought this used for cheap, it's in really bad shape. First time I bought a used book that is like this, pages wrinkled, pen marked all over the place, all worn out. I am going to read more, if it is that good, I might buy a better condition one.

 

[Mark44]

Sorry, I need to absolutely confirm:
numArray[5] is an array with 5 elements.
numArray is the address that point to the first element of array numArray[5].

Why do you need to confirm this ? it's exactly what I said, and which you quoted.

Are std::strings like this also?

No, they are very different.

Without brackets is the address pointing to the first element?

No. In @jtbell's post, name is a string instance, an object of type std::string. It contains data members such as size and capacity, as well as the data making up the characters in the string, and provides access to all of the function members that belong to the object. A C-type character array has only the characters in it, and none of the other capabilities of the Standard Template Library string class.

 

[jtbell]

Are std::strings like this also? Without brackets is the address pointing to the first element?

No, as I wrote in post #266.

 

[Jarvis323]

numArray is the address that point to the first element of array numArray[5].

Sort of.

There are two types of raw arrays, static and dynamic. This is a static array. The memory is allocated on the stack, which is limited in size. A dynamic array is allocated manually and stored on the heap, which has much more space.

int x[5]; // static array
int * y= new int[5]; // dynamic allocation
...
delete [] y; // must manually delete y

So you can see x and y are both "arrays", but y's type is a pointer/memory address. x and y are used the same way, by accessing their elements by index with the bracket operator.

So what is the bracket operator in this case? It's actually syntactic sugar (simplified syntax) for two operatotions put together: pointer arithmetic and dereferencing.

To understand pointer arithmetic, first understand that pointers store memory addresses to typed values each with specific sizes in bytes (e.g. an int is 4 bytes), but memory itself is addressed byte by byte. That is, say the first byte (8 bits) has address 1, then the next byte has address 2. An int starting at address 1 covers 4 bytes. So address 5 is the address 1 int beyond address 1. So pointer arithmetic means, say y is a pointer to an int starting at address 1, then y+2 is actually y's value + sizeof(int)*2, which is the address 9, 8 bytes (2 ints) passed the first one.

Now the bracket operator used to access elements in the array is actually doing this: y[2] is equivalent to *(y+2). y+2 gives you the address of the third element using pointer arithmetic. Then the * dereferences the result (another overload), which means it grabs those 4 bytes starting at the address of the first byte where the int value is stored and returns it as an int.

So why is x almost a pointer but not truly a pointer?

When you use the bracket operator to access elements, it's treated like a pointer. You can say it decays in that it is implicitly cast to a pointer first. So for x, the bracket operator is more like this: x[2] is *(((int *)x)+2). This is casting x to a pointer to an int, then doing pointer arithmetic, then dereferencing the pointer. So what is the difference between x and a pointer? Mainly it is that sizeof(x) is the size in bytes of the array, which is 5*4, whereas sizeof(y) is the size of a pointer, which is 8 on 64 bit executables or 4 on 32 bit executables.

std::string can't be explained until you've learned about classes. But you can think of it as an object that internally stores and manages a dynamically allocated c-string/char array.

 

[jtbell]

I just received this book today by Tony Gaddis
https://www.amazon.com/gp/product/0136022537/?tag=pfamazon01-20

This is an actual textbook, written for use in college/university programming courses. The link is to the 6th edition (2009). By the time it came out, I had already stopped teaching C++, so I never used it myself. That it uses std::string is a good sign. Does it also use std::vector?

It's now up to the 10th edition (2019), so it must be a popular textbook used by many schools. It's expensive because of the way the textbook "racket" works. The 8th edition (2013) probably includes some of the significant additions to C++ in the C++11 standard (like std::array and range-based for-loops). It might be worth watching for a cheap copy of that edition or a later one.

First time I bought a used book that is like this, pages wrinkled, pen marked all over the place, all worn out.

That means it was probably actually used by a student for two semesters in a real C++ course. Maybe two or three students in successive years, depending on when the next edition appeared.

Because it's a textbook, it probably has lots of programming exercises. Hint, hint...

 

[yungman]

This is an actual textbook, written for use in college/university programming courses. The link is to the 6th edition (2009). By the time it came out, I had already stopped teaching C++, so I never used it myself. That it uses std::string is a good sign. Does it also use std::vector?

It's now up to the 10th edition (2019), so it must be a popular textbook used by many schools. It's expensive because of the way the textbook "racket" works. The 8th edition (2013) probably includes some of the significant additions to C++ in the C++11 standard (like std::array and range-based for-loops). It might be worth watching for a cheap copy of that edition or a later one.
That means it was probably actually used by a student for two semesters in a real C++ course. Maybe two or three students in successive years, depending on when the next edition appeared.

Because it's a textbook, it probably has lots of programming exercises. Hint, hint...

I spent more time on the book reading std::string and even some pointers. It is a REALLY REALLY good book. This is the one my grandson used, he even gave me the pdf file. But due to my neck problem, I cannot read on the monitor for long time, besides I want to ruin my own book by writing notes. I bought the used one for cheap, I am printing the book out chapter by chapter to read now.

So far, it's very clear, I think I only have to print the function and one or two chapters more, those condition logical comparison, boolean stuffs and even the looping stuffs are quite easy, I think I got the idea from even the worst book...the first one. It's the structures, naming stuffs that's hard for me. All the instance, class, headers, objects stuffs. The programming is actually not hard as long as I get all the names straight.

I somehow have a mental block of names, so often I ask the person for phone number, I remember the phone number and didn't remember the name!

 

[yungman]

I am so stoked reading the GADDIS book, it is so simple. I read the std::string, but I decided to stop and go through what I learn in the last 2 weeks or so. Now I think I have a better feel of the names...

That object is just a self contained module with both data and programs, it just carry out request from the main program with given parameter and return results. It is like a module that the main program call upon to perform a task.

That cout and cin are object that the main program can call to write a stream onto the console or read as stream from the keyboard, that << and >> are streaming operator.

That header files contain a set of object files of particular type of function that the program to call on. Just like cout and cin are object in iostream for streaming. That we #include <iostream> in order to access to cout and cin.

These were all a blur to me before, the first book doesn't explain any of these. When you guys explain, it's really over my head. This book really starting from basic on up. I went through quite a few pages, I think it's worth my while to even have to stop for a few days and catch up with all the holes I missed.

Before, all the names and terms is just like people's names, why you call John? Why your last name is Jones? It's like I just blindly remember the names and it's hard.

I learn a lot on strings also, but I don't even want to talk about it until I back track all the stuffs and revisit again.

I am stoked.

 

[yungman]

I am working of cin.get(). I think I got what I want from the program, BUT, it's really not what I want.

I want the program to read in any character from the keyboard, display it and show the ASCII number. Then loop back to ask another character. I want it to exit when I hit ENTER ( ASCII = 10)

I understand for cin, I need to hit the character and then ENTER to get it going or else it will just sit there and wait. cin.get can recognize ENTER as character. BUT, I still have to hit ENTER again before it will exit the loop. this is because I have to put in cin.ignore to prevent the program from reading the ENTER that is in the stream and exit without looping. Is there any way to exit right when I hit ENTER if I choose to exit?

#include <iostream>
using namespace std;

int main()
{
    int Ascii = 0;
    char userSelection = 'A';
    do
    {
    cout << "hit any key:  ";
    cin.get(userSelection);
    cin.ignore();
    Ascii = userSelection;
    cout << "you hit: " << userSelection << ",   the ASCII is =  " << Ascii << endl;
    } while (userSelection != 10) ;
    return 0;

}

thanks

 

[pbuk]

Rather than try and fit an answer into what you wrote, I'm going to totally rewrite it for you. In doing this you will see that nearly every line is different. Apart from the difference between this

int main()
{
  // Code here.

and this

int main() {
  // Code here.

, and 2 spaces of indent vs. 4, which are just different style choices, all the other differences are important. Some of them (like

  int ascii = 0;

vs

    int Ascii = 0;

wont stop your code working, but they can make it more vulnerable to mistakes and you should get into the right habits now.

So here we go, you can also see this working at https://repl.it/repls/RealSubduedAtom. This code is missing comments of course, you might like to add them as you work through it.

#include <iostream>
using namespace std;

#define EOL '\n'

int main() {
  int ascii = 0;
  char userSelection = 0;

  while (userSelection != EOL) {
    cout << "type a key and [enter]: ";
    cin.get(userSelection);
    ascii = userSelection;

    if (userSelection == EOL) continue;

    cin.ignore(255, EOL);
    cout << "the first key was: " << userSelection
      << ", the ASCII is " << ascii << endl;
  }
  return 0;
}

 

[yungman]

Thanks, that's what I want, but I don't understand.

So ENTER = '\n' ? I did not know about this!

I don't remember whether I can use const char EOL = '\n' instead of #define EOL '\n'

What is continue? I don't see in while loop have anything to do with this.

 

[jtbell]

What is continue? I don't see in while loop have anything to do with this.

There are two ways to interrupt a cycle of a loop in the middle:

• break; exits the loop immediately. The next statement executed is the first statement after the loop's closing-brace.
• continue; jumps back to the beginning of the loop. In a while loop, the loop condition is tested again, then the loop continues or exits accordingly.

Are you sure it's not in your (Gaddis's) book? Have you tried the index at the end of the book? (I don't have it, so I can't check it for myself.)