Could someone explain this for me?

[CalebConnolly]

Hi, ive been reading these forums for a while and just now after reading something that makes no sense to me I thought I might as well ask on here for some help.

Basically I was reading through a textbook example showing how to determine if numbers are odd or even and it basically dosent explain how it works, just tells you that it does.

What is really bugging me, and I cant get my head round it is why does it state int (i) and then when it does modulus uses int a. Also the line before the while statement has even (i) - why is that and what does it do?
And is this an example of what is known as passing the integers to function even? - I was trying to find an example of a program doing that and I think this is.

I typed it up and its like this...


#include <iostream>
using namespace std;


void even (int a);

int main ()
{
int i;
do {
cout << "Type a number (0 to exit): ";
cin >> i;
even (i);
} while (i!=0);
return 0;
}

void even (int a)
{
if ((a%2)==0) cout << "Number is even.\n";
else cout << "odd" << endl;
}

[MarshMallow]

What is really bugging me, and I cant get my head round it is why does it state int (i) and then when it does modulus uses int a. Also the line before the while statement has even (i) - why is that and what does it do?

The reason is: variable name in the function prototype must not necessarily match the variable name of the argument;please read your book carefully and don't jump immediately to topics that you are not enough trained for.

[CalebConnolly]

Well I know that as that's why the text book uses it.

Could someone help me and explain how this program works in a few words?

Also am I right in thinking this a case of "passing integers to function even" - or is that something else?

[Airslash]

Beeing a student myself, i think i can help you without breaking forum rules. I will report some of the code and explain with steps what each part does. And why it does that.

int main ()
{
     int i;
     do {
          cout << "Type a number (0 to exit): ";
          cin >> i;
          even (i);
     } while (i!=0);
     return 0;
}
 

Ok this is the main method of your 'program'.
The first line int i; declares a new varialbe of the type integer. It has no value assigned yet, but thats not important right now.

The second line begins the while-loop. This is a reversed loop because you start with the do part, and perform the check at the end. Why this is done should be obvious from my previous point, if not you really should grab you c++ book again.

Inside the loop you ask the user for a number. You store that number inside the variable you declared. After storing the variable, you call the function even and pass it the variable you stored. So far still basic things.

After processing the even function, you check the conditition of your while loop, and repeat the process if the condition is met.

void even (int a)
{
if ((a%2)==0) cout << "Number is even.\n";
else cout << "odd" << endl;
}
 

This is a basic function declaration in C++. A function declaration has nothing to do with the implementation. What we do in this snippet, is we declare a new function called even which accepts a parameter of the type integer and returns nothing because the returntype is set to void. The syntax in C++ for declaring functions is the following :

returntype funcname (parameters) { code }
 

The code that is done inside the function body should be easy to understand.

My typing this piece of text, makes me wonder if you are already up to function definitions and passing arguments to functions.


anyways I hope it helps you.

[CalebConnolly]

Thats mint, thanks for the explanation, I really understand now. They should have had that in the textbook! Im up on my function definitions and have gone through quite a few programs its just he slightest new thing that i havent come across throws me, when it shouldn't.

Thanks again

[Ganon11]

Well, the modulus division operator (%) divides the two numbers, but returns the remainder. So, for example, 32 % 5 results in 2, because 5 divides into 30 6 times with a remainder of 2.

It is important to note that when you perform the general calculation a % b, there are b possible answers. Returning to our above example, these 5 answers are 0, 1, 2, 3, and 4. In general, the possible answers for a % b are all integers between 0 and b - 1, inclusive. (Test this out by manually seeing what 30 % 5 is, as well as 31 % 5, 32 % 5, 33 % 5, 34 % 5, 35 % 5, and 36 % 5).

Now we return to the original problem - how does a % 2 help us figure out if it is even or odd? Well, in this case, b == 2, so there are 2 possible answers (what are they?). There are also two possibilities for the number a - it is even, or it is odd. So that makes sense.

Now, what do all even numbers share in common? 24, 1056, 900000000002, and 6 (all even numbers) are all evenly divisible by 2. This means that, when they are divided by 2, an integer results (i.e. there is no remainder). In general, any even number e, when divided by 2, leaves no remainder. We can check this using modulus division! If a % 2 is 0 (there is no remainder), then 2 divides evenly into a; thus, a is even.

It is easy to see that, if a % 2 is not 0, then the number must be odd.

[MarshMallow]

in main you create a variable whose name i,and then you pass this variable to even ,which creates its own copy of "i" and calls it "a"(don't get stuck to this too much,just think that a procedure copy and rename the value that are passed into it),then it makes the computatione needed to find out wether the number is prime or not.

[CalebConnolly]

Many thanks again guys, thats really helpful.

Its falling into place with the things ive learn't, nice one

[weaknessforcats]

Code: ( text )
void even (int a)
{
if ((a%2)==0) cout << "Number is even.\n";
else cout << "odd" << endl;
}


This is a basic function declaration in C++. A function declaration has nothing to do with the implementation. What we do in this snippet, is we declare a new function called even which accepts a parameter of the type integer and returns nothing because the returntype is set to void. The syntax in C++ for declaring functions is the following :

Code: ( text )
returntype funcname (parameters) { code }

Not a bad explanation but I want to add that there is a difference between a declaration and a definition. A declaration says a thing exists. A definition allocates memory for it.

On that basis what you call a declaration is, in fact, the function definition. That is, the function definition and the function implementation are the same thing. The function declaration is the function prototype.