Enumerable.Min Method (IEnumerable<Int32>)

May 11, 2014

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
)

Parameters

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.

Usage Note

In Visual Basic and C#, you can call this method as an instance method on any object of type IEnumerable<Int32>. When you use instance method syntax to call this method, omit the first parameter.

ExceptionCondition
ArgumentNullException

source is null.

InvalidOperationException

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 Enumerable.Min.

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

NoteNote:

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();

      outputBlock.Text += String.Format("The smallest number is {0}.", min) + "\n";

      /*
       This code produces the following output:

       The smallest number is -2E+103.
      */



Windows Phone OS

Supported in: 8.1, 8.0, 7.1, 7.0

Windows Phone

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