## Pre and Post Operators

In this lesson you are going to have the opportunity to interact with Pre and Post Operators. A Pre-Operator is an operator that is stated in the code BEFORE the Variable. A Post-Operator is an Operator that is stated in the code AFTER the operator. these work very different than each other in the final results of how the solution is acquired.

A Pre-Operator will interact with the Variable before any other operations are accomplished.

Example:

int v;

v = 10;

v = 2 * ++v

From the example above the integer v which is equal to 10, is going to be multiplied by 2, BUT this will only happen AFTER v is increased by 1. The reason that this will happen is because the Pre-Operator is a higher priority than the multiplication operator.

So.... the way that this is figured is like this, (v + 1) * 10 = v

Result v = 22

A Post-Operator will interact with the Variable eventually some where along the course of the program.... But it will not really have any interaction with this equation

Example:

int v;

v = 10;

v = 2 * v++

From the example above the integer v which is equal to 10, is going to be multiplied by 2, eventually v will be increased by 1 to equal 11.

So..... the way that this is figured is like this, v * 10 = v, later on v will be increased by 1

Result v = 20

**Pre-Operator**A Pre-Operator will interact with the Variable before any other operations are accomplished.

Example:

int v;

v = 10;

v = 2 * ++v

From the example above the integer v which is equal to 10, is going to be multiplied by 2, BUT this will only happen AFTER v is increased by 1. The reason that this will happen is because the Pre-Operator is a higher priority than the multiplication operator.

So.... the way that this is figured is like this, (v + 1) * 10 = v

Result v = 22

**Post-Operator**A Post-Operator will interact with the Variable eventually some where along the course of the program.... But it will not really have any interaction with this equation

Example:

int v;

v = 10;

v = 2 * v++

From the example above the integer v which is equal to 10, is going to be multiplied by 2, eventually v will be increased by 1 to equal 11.

So..... the way that this is figured is like this, v * 10 = v, later on v will be increased by 1

Result v = 20

## Pre-Assignment video

## The Assignment

Using the code below as a starting point, build the new program so that it does the correct computations. i have provided the Psuedo-Code as a convenience for you. Enjoy!!!

## The Code

#include using namespace std;

int main(void)

{

int i, j, k;

cout << "Enter a value for i: ";

cin >> i;

cout << "Enter a value for j: ";

cin >> j;

// add i and j and assign it to k

// print k to the screen

// Pre-Operate 1 to i and Multiply it by j assign the result to k

// print k to the screen

// Post-Operate 1 to j and divide it by i assign the result to k

// print k to the screen

// Pre-Operate 1 to j and i, multiply i by j, add it k, assign the result to k

cout << k;

return 0;

}

int main(void)

{

int i, j, k;

cout << "Enter a value for i: ";

cin >> i;

cout << "Enter a value for j: ";

cin >> j;

// add i and j and assign it to k

// print k to the screen

// Pre-Operate 1 to i and Multiply it by j assign the result to k

// print k to the screen

// Post-Operate 1 to j and divide it by i assign the result to k

// print k to the screen

// Pre-Operate 1 to j and i, multiply i by j, add it k, assign the result to k

cout << k;

return 0;

}