Does anyone know why integer division in C# returns an integer and not a float?

What is the idea behind it? (Is it only a legacy of C/C++?)

In C#:

```
float x = 13 / 4;
//== operator is overridden here to use epsilon compare
if (x == 3.0)
print 'Hello world';
```

Result of this code would be:

```
'Hello world'
```

Strictly speaking, there is no such thing as integer division (division by definition is an operation which produces a rational number, integers are a very small subset of which.)

##
Solution #1:

While it is common for new programmer to make this mistake of performing integer division when they actually meant to use floating point division, in actual practice integer division is a very common operation. If you are assuming that people rarely use it, and that every time you do division you’ll always need to remember to cast to floating points, you are mistaken.

First off, integer division is quite a bit faster, so if you only need a whole number result, one would want to use the more efficient algorithm.

Secondly, there are a number of algorithms that use integer division, and if the result of division was always a floating point number you would be forced to round the result every time. One example off of the top of my head is changing the base of a number. Calculating each digit involves the integer division of a number along with the remainder, rather than the floating point division of the number.

Because of these (and other related) reasons, integer division results in an integer. If you want to get the floating point division of two integers you’ll just need to remember to cast one to a `double`

/`float`

/`decimal`

.

##
Solution #2:

See C# specification. There are three types of division operators

- Integer division
- Floating-point division
- Decimal division

In your case we have Integer division, with following rules applied:

The division rounds the result towards zero, and the absolute value of

the result is the largest possible integer that is less than the

absolute value of the quotient of the two operands. The result is zero

or positive when the two operands have the same sign and zero or

negative when the two operands have opposite signs.

I think the reason why C# use this type of division for integers (some languages return floating result) is hardware – integers division is faster and simpler.

##
Solution #3:

Each data type is capable of overloading each operator. If both the numerator and the denominator are integers, the integer type will perform the division operation and it will return an integer type. If you want floating point division, you must cast one or more of the number to floating point types before dividing them. For instance:

```
int x = 13;
int y = 4;
float x = (float)y / (float)z;
```

or, if you are using literals:

```
float x = 13f / 4f;
```

Keep in mind, floating points are not precise. If you care about precision, use something like the decimal type, instead.

##
Solution #4:

Since you don’t use any suffix, the literals `13`

and `4`

are interpreted as integer:

If the literal has no suffix, it has the first of these types in which its value can be represented:

`int`

,`uint`

,`long`

,`ulong`

.

Thus, since you declare `13`

as integer, integer division will be performed:

For an operation of the form x / y, binary operator overload resolution is applied to select a specific operator implementation. The operands are converted to the parameter types of the selected operator, and the type of the result is the return type of the operator.

The predefined division operators are listed below. The operators all compute the quotient of x and y.

Integer division:

`int operator /(int x, int y); uint operator /(uint x, uint y); long operator /(long x, long y); ulong operator /(ulong x, ulong y);`

And so rounding down occurs:

The division rounds the result towards zero, and the absolute value of the result is the largest possible integer that is less than the absolute value of the quotient of the two operands. The result is zero or positive when the two operands have the same sign and zero or negative when the two operands have opposite signs.

If you do the following:

```
int x = 13f / 4f;
```

You’ll receive a compiler error, since a floating-point division (the `/`

operator of `13f`

) results in a float, which cannot be cast to int implicitly.

If you want the division to be a floating-point division, you’ll have to make the result a float:

```
float x = 13 / 4;
```

Notice that you’ll still divide integers, which will implicitly be cast to float: the result will be `3.0`

. To explicitly declare the operands as float, using the `f`

suffix (`13f`

, `4f`

).

##
Solution #5:

Might be useful:

```
double a = 5.0/2.0;
Console.WriteLine (a); // 2.5
double b = 5/2;
Console.WriteLine (b); // 2
int c = 5/2;
Console.WriteLine (c); // 2
double d = 5f/2f;
Console.WriteLine (d); // 2.5
```

##
Solution #6:

It’s just a **basic operation**.

Remember when you learned to divide. In the beginning we solved `9/6 = 1 with remainder 3`

.

```
9 / 6 == 1 //true
9 % 6 == 3 // true
```

The /-operator in combination with the %-operator are used to retrieve those values.

##
Solution #7:

The result will always be of type that has the greater range of the numerator and the denominator. The exceptions are byte and short, which produce int (Int32).

```
var a = (byte)5 / (byte)2; // 2 (Int32)
var b = (short)5 / (byte)2; // 2 (Int32)
var c = 5 / 2; // 2 (Int32)
var d = 5 / 2U; // 2 (UInt32)
var e = 5L / 2U; // 2 (Int64)
var f = 5L / 2UL; // 2 (UInt64)
var g = 5F / 2UL; // 2.5 (Single/float)
var h = 5F / 2D; // 2.5 (Double)
var i = 5.0 / 2F; // 2.5 (Double)
var j = 5M / 2; // 2.5 (Decimal)
var k = 5M / 2F; // Not allowed
```

There is no implicit conversion between floating-point types and the decimal type, so division between them is not allowed. You have to explicitly cast and decide which one you want (Decimal has more precision and a smaller range compared to floating-point types).

##
Solution #8:

As a little trick to know what you are obtaining you can use *var*, so the compiler will tell you the type to expect:

```
int a = 1;
int b = 2;
var result = a/b;
```

your compiler will tell you that *result* would be of type *int* here.