Enumerable.Min Method (IEnumerable<Int32>)


Returns the minimum value in a sequence of Int32 values.

Namespace:   System.Linq
Assembly:  System.Core (in System.Core.dll)

public static int Min(
	this IEnumerable<int> source


Type: System.Collections.Generic.IEnumerable<Int32>

A sequence of Int32 values to determine the minimum value of.

Return Value

Type: System.Int32

The minimum value in the sequence.

Exception Condition

source is null.


source contains no elements.

The Min(IEnumerable<Int32>) method uses the Int32 implementation of IComparable<T> to compare values.

In Visual Basic query expression syntax, an Aggregate Into Min() clause translates to an invocation of Min.

The following code example demonstrates how to use Min(IEnumerable<Double>) to determine the minimum value in a sequence.


This code example uses an overload of this overloaded method that is different from the specific overload that this topic describes. To extend the example to this topic, substitute the elements of the source sequence with elements of the appropriate numerical type.

double[] doubles = { 1.5E+104, 9E+103, -2E+103 };

double min = doubles.Min();

Console.WriteLine("The smallest number is {0}.", min);

 This code produces the following output:

 The smallest number is -2E+103.

Universal Windows Platform
Available since 4.5
.NET Framework
Available since 3.5
Portable Class Library
Supported in: portable .NET platforms
Available since 2.0
Windows Phone Silverlight
Available since 7.0
Windows Phone
Available since 8.1
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