Assignment, increment and decrement operators

Before we get into C++ loop studies, we need to know and learn how to use some operators that are important in studying repetition structures.

These operators will allow us to 'save' code, ie write less.
Obviously, we programmers are not lazy. We are just efficient, ok?

Assignment Operators

We have sometimes used expressions like:
x = x + 1;
This means "make variable x its old value plus 1".

However, this can be abbreviated to:
x += 1;
It means the same thing.

If we want to subtract y from x, we can do:
x = x - y;

Or more simply:
x -= y;

To make x receive the product value of x by y:
x = x * y;

Or in a lazier way:
x *= y;

The same goes for division and rest of division (module):
x /= y; is the same as x = x / y;
x %= y; is the same as x = x% y;

Very simple, no? Just one way to write less.

Increment and Decrement Operators

Do you think:
x + = 1;
x = 2;
Is it writing little?

There are two ways to do this for each expression:
x ++ or ++ x
x-- or --x

If we use these expressions alone:
x ++; and ++ x;
x--; and --x;

They will have the same effect: increment in the first case and decrement in the second case by one unit.

The variations: ++ x and x ++ are called preincrement and postincrement
The variations: --x and x-- are called pre-decrement and post-decrement.

They work as follows ... suppose the expression:
y = x++;

Two things happen in that order:

1. y gets the value of x
2. x is incremented by one unit

Already in the expression:
y = ++ x;

The following occurs:

1. x value is incremented by one unit
2. y gets new x value, incremented

See the following program. What prints on the screen?

#include <iostream>
using namespace std;

int main()
{
    int x=1;

    cout << x++;
    return 0;
}

The answer is 1. First print x, then increment x.
Put another cout for you to see the new x value later.

And the following program, which output will have?

#include <iostream>
using namespace std;

int main()
{
    int x=1;

    cout << ++x;
    return 0;
}

Now yes, directly prints 2.
First the increment (++) occurs, and only then the value of x is printed.

Now let's look at another sequence of operations:
x = 2;
y = 3;
z = x * ++y;

What are the values of x, y and z after these three procedures?
Come on.

First, y is incremented to 4.
So we have z = x * y = 2 * 4 = 8
The new values are 2, 4 and 8.

Now let's look at this sequence of operations:
x = 2;
y = 3;
z = x * y--;

The value of z is: z = x * y = 2 * 3 = 6

Since the decrement sign came after the variable y, only then is it decremented to become 2. The new values are 2, 2, and 6.