# Manipulating Values in Your Programs

This chapter is from the book

## Working with Mathematical/Arithmetic Operators

Among the most commonly used operators are the mathematical operators. All the basic math functions are available within C#, including addition, subtraction, multiplication, division, and modulus (remaindering). Additionally, compound operators make doing some of these operations more concise.

For addition and subtraction, you use the additive operators. As you should expect, for addition, the plus operator (+) is used. For subtraction, the minus (-) operator is used. The general format of using these variables is as follows:

```NewVal = Value1 + Value2;
NewVal2 = Value1 – Value2;```

In the first statement, Value2 is being added to Value1 and the result is placed in NewVal. When this command is done, Value1 and Value2 remain unchanged. Any pre-existing values in NewVal are overwritten with the result.

For the subtraction statement, Value2 is subtracted from Value1 and the result is placed in NewVal2. Again, Value1 and Value2 remain unchanged, and the value in NewVal2 is overwritten with the result.

Value1 and Value2 can be any of the value data types, constants, or literals. You should note that NewVal must be a variable; however, it can be the same variable as Value1 or Value2. For example, the following is legal as long as Variable1 is a variable:

`Variable1 = Variable1 – Variable2;`

In this example, the value in Variable2 is subtracted from the value in Variable1. The result is placed into Variable1, thus overwriting the previous value that Variable1 held. The following example is also valid:

`Variable1 = Variable1 – Variable1;`

In this example, the value of Variable1 is subtracted from the value of Variable1. Because these values are the same, the result is 0. This 0 value is then placed into Variable1, overwriting any previous value.

If you want to double a value, you enter the following:

`Variable1 = Variable1 + Variable1;`

Variable1 is added to itself, and the result is placed back into Variable1. The end result is that you double the value in Variable1.

### Doing Multiplicative Operations

An easier way to double the value of a variable is to multiply it by two. Three multiplicative operators commonly are used in C#:

• For multiplication, the multiplier (or times) operator, which is an asterisk (*)

• For division, the divisor operator, which is a forward slash (/)

• For obtaining remainders, the remaindering (also called modulus) operator, which is the percentage sign (%)

Multiplication and division are done in the same manner as addition and subtraction. To multiply two values, you use the following format:

`NewVal = Value1 * Value2;`

For example, to double the value in Val1 and place it back into itself (as seen with the last addition example), you can enter the following:

`Val1 = Val1 * 2;`

This is the same as this line:

`Val1 = 2 * Val1;`

Again, division is done the same way:

`NewVal = Value1 / Value2;`

This example divides Value1 by Value2 and places the result in NewVal. To divide 2 by 3, you write the following:

`answer = 2 / 3;`

Sometimes when doing division, you want only the remainder. For example, I know that 3 will go into 4 one time; however, I also would like to know that I have 1 remaining. You can get this remainder using the remaindering (also called modulus) operator, which is the percentage sign (%). For example, to get the remainder of 4 divided by 3, you enter this:

`Val = 4 % 3;`

The result is that Val is 1.

Consider another example that is near and dear to my heart. You have three pies that can be cut into six pieces. If 13 people each want a piece of pie, how many pieces of pie are left over for you?

To solve this, take a look at Listing 3.2.

#### Listing 3.2 Pie.cs—Number of Pieces of Pie for Me

``` 1: // Pie.cs - Using the modulus operators
2: //------------------------------------------------
3: class Pie
4: {
5:   public static void Main()
6:   {
7:    int PiecesForMe = 0;
8:    int PiecesOfPie = 0;
9:
10:    PiecesOfPie = 3 * 6;
11:
12:    PiecesForMe = PiecesOfPie % 13;
13:
14:    System.Console.WriteLine("Pieces Of Pie = {0}", PiecesOfPie);
15:    System.Console.WriteLine("Pieces For Me = {0}", PiecesForMe);
16:   }
17: }
Pieces Of Pie = 18
Pieces For Me = 5```

Listing 3.2 presents the use of the multiplication and modulus operators. Line 10 illustrates the multiplication operator, which is used to determine how many pieces of pie there are. In this case, there are six pieces in three pies (so, 6 x 3). Line 12 then uses the modulus operator to determine how many pieces are left for you. As you can see from the information printed in Lines 14–15, there are 18 pieces of pie, and 5 will be left for you.

### Working with the Compound Arithmetic Assignment Operators

You've learned about the basic assignment operator; however, there are also other assignment operators—the compound assignment operators (see Table 3.1).

#### Table 3.1 Compound Arithmetic Assignment Operators

 Operator Description Noncompound Equivalent += x += 4 x = x + 4 -= x -= 4 x = x – 4 *= x *= 4 x = x * 4 /= x /= 4 x = x / 4 %= x %= 4 x = x % 4

The compound operators provide a concise method for performing a math operation and assigning it to a value. For example, if you want to increase a value by 5, you use the following:

`x = x + 5;`

Or, you can use the compound operator:

`x += 5;`

As you can see, the compound operator is much more concise.

TIP

Although the compound operators are more concise, they are not always the easiest to understand in code. If you use the compound operators, make sure that what you are doing is clear, or remember to comment your code.

### Doing Unary Math

All the arithmetic operators that you have seen so far have been binary. Each has required two values to operate. A number of unary operators also work with just one value or variable. The unary arithmetic operators include the increment operator (++) and the decrement operator (--).

These operators add 1 to the value or subtract 1 from the value of a variable. The following example adds 1 to x:

`++x;`

It is the same as saying this:

`x = x + 1;`

Additionally, the following subtracts 1 from x:

`--x;`

It is the same as saying this:

`x = x – 1;`

TIP

The increment and decrement operators are handy when you need to step through a lot of values one by one.

The increment and decrement operators have a unique feature that causes problems for a lot of newer programmers. Assume that the value of myNbr is 10. Look at the following line of code:

`NewNbr = ++myNbr;`

After this statement executes, what will the values of myNbr and newNbr be? You should be able to guess that the value of myNbr will be 11 after it executes. The value of newNbr will also be 11. Now consider the following line of code; again consider the value of myNbr to start at 10.

`newNbr = myNbr++;`

After this statement executes, what will the values of myNbr and newNbr be? If you said that they would both be 11 again, you are wrong! After this line of code executes, myNbr will be 11; however, newNbr will be 10. Confused?

It is simple: The increment operator can operate as a pre-increment operator or a post-increment operator. If it operates as a pre-increment operator, the value is incremented before everything else. If it operates as a post-increment operator, it happens after everything else. How do you know whether it is pre- or post-? Easy. If it is before the variable, ++myNbr, it is pre-. If it is after the variable, myNbr++, it is post-. The same is true of the decrement operator. Listing 3.3 illustrates the pre- and post- operations of the increment and decrement operators.

#### Listing 3.3 Prepost.cs—Using the Increment and Decrement Unary Operators

``` 1: // Prepost.cs - Using pre- versus post-increment operators
2: //----------------------------------------------------
3:
4: class Prepost
5: {
6:   public static void Main()
7:   {
8:    int Val1 = 0;
9:    int Val2 = 0;
10:
11:    System.Console.WriteLine("Val1 = {0} Val2 = {1}", Val1, Val2);
12:
13:    System.Console.WriteLine("Val1 (Pre) = {0} Val2 = (Post) {1}",
14:       ++Val1, Val2++);
15:
16:    System.Console.WriteLine("Val1 (Pre) = {0} Val2 = (Post) {1}",
17:       ++Val1, Val2++);
18:
19:    System.Console.WriteLine("Val1 (Pre) = {0} Val2 = (Post) {1}",
20:       ++Val1, Val2++);
21:   }
22: }
Val1 = 0 Val2 = 0
Val1 (Pre) = 1 Val2 = (Post) 0
Val1 (Pre) = 2 Val2 = (Post) 1
Val1 (Pre) = 3 Val2 = (Post) 2```

It is important to understand what is happening in Listing 3.3. In Lines 8–9, two variables are again being initialized to 0. These values are printed in Line 11. As you can see from the output, the result is that Val1 and Val2 equal 0. Line 13, which continues to Line 14, prints the values of these two variables again. The values printed, though, are ++Val1 and Val2++. As you can see, the pre-increment operator is being used on Val1, and the post-increment operator is being used on Val2. The results can be seen in the output. Val1 is incremented by 1 and then printed. Val2 is printed and then incremented by 1. Lines 16 and 19 repeat these same operations two more times.

Do use the compound operators to make your math routines concise.

Don't confuse the post-increment and pre-increment operators. Remember that the pre-increment adds before the variable, and the post-increment adds after it.

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