loop breakdown

[tobougg]

Hi there, i've been programming for 1 week now with C++. (my first ever programming language in fact).

I've gotten as far as loops, and up until now have been able to understand every concept in all lessons/chapters of the books i read. I'm having a little bit of a problem with loops though. Now i can write a loop, but i can't understand a part of it, and im just writing from memory.

Example code is at the bottom.
Now what im struggling to understand (for 2 days now) is the inner workings of that statement. those 3 lines. (well the top 2 actually). Can someone break every part between the ;'s and ()'s piece by piece for me. As im not grasping in loops and especially in the below example, how the output to console is showing me 7 *'s in the window if i type 7 in std::cin. How is that working, it looks like just 1 * in the code.

I guess what im asking is how the below works piece by piece, and when does the ++i come into it, and what happens after that to make the code print more than one *.

Thanks guys! counting on you to save my brain from meltdown

  for(int i = 1; i <= data1; ++i)
        cout << "*";
    cout << data1 << endl;

[Daleth]

for(/*...*/)
   cout << "*";
//Is equivalent to
for(/*...*/){
   cout << "*";
}


/*for loop*/
for(
   int i = 1; //Create a variable 1 and initialize it with 1. This will serve as a counter.
   i <= data1; //Keep running another iteration/loop while i is less than or equal to data1
   ++i //After each iteration/loop, increase i by 1
)
cout << "*"; //In each iteration/loop, print an asterisk.
/*end for loop*/

The last line is outside of the for loop.

[tobougg]

another quick example of the same thing that i need describing just as a "second case" so im sure of the same answer in multiple instances of a similar loop

int number, factorial;
cout << "Enter a number: ";
cin >> number;
factorial = 1;
for(int i = number; i >= 1; --i)
factorial *= i;
cout << number << "! equals " << factorial << endl;

[tobougg]

oh god thats perfect Daleth thank you so much! i really needed that info thanks alot!! wow. now i get it... damn i feel so stupid. can you do the same for the next (and last dont worry) one. cheers!

[Daleth]

The for loop in your second code snippet counts down instead of up in a range from number to 1. And in each iteration, the counter is multiplied to the factorial variable.

This is probably to make it easier to relate this to how a factorial works:
n! = n*(n-1)*(n-2)*(n-3)...*1

However, you can modify the loop to count up instead of down and still get the same result.

[tobougg]

you couldn't have made it easier to understand. thanks alot brother! i think im doing fine for only a week of coding for the first time ever. im just getting onto vectors once i clear this small loop issue up. then after that it's pointers.. maybe ill run into more trouble there

[tobougg]

oh one last thing. the book keep presuming im intelligent, so i have actually one last question. we typed

int square (int number)
{            int squared = number * number;  

now the book says "we could have done this in 2 statements, but as we know about declarations and initializations work, we'll only do it with 1.

what would that 2 statement one look like? just out of interest, as i dont get what it means.

and he has this line aswell in it
std::cout << number << "squared equals " << square(number) << endl;

how on earth would i ever know to add parenthesis around the number there.

[Daleth]

I think it is talking about the second line. You could, for example, write:

int squared;  //Declare a variable called squared
squared = number; //Assign it some number
squared *= number; //Now multiply it 

Vs.

   //Declare a variable squared and initialize it--immediately assign it a value--
   //   with number * number
int squared = number * number;

Extra:
And in the case of this function, you might have:

int square (int number)
{
   int squared = number * number;
   return squared;
}

But you can compress this even further to not create a variable:

int square (int number)
{
   return number * number;
}

[tobougg]

and the *= is multiply it and give the result in 1 statement? so times then ='s it together?

[Daleth]

Oh, sorry I assumed you already knew about those. Yes, *= is equivalent to * first then = second.
They are called compound assignment operators: +=, -=, *=, /=, %=, >>=, <<=, ^=, |=, &=.

They are used for shorthand notation, like:

i = i + j;
//Equivalent to
i += j;

For even shorter notations, there are the increment and decrement operators: ++ and --.

i = i - 1;
i -= 1;
--i;

There is also a difference between ++i (prefix) and i++ (postfix). ++i increments i by one and returns i. i++ increments i, but returns what i was before incrementing. The same is true for --i and i--.