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C# / .NET - Math.Min() method example

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Created by:
Mohammad-Oneal
207

The Math.Min() function returns the smallest number from given numbers.

using System;

public class Program
{
    public static void Main(string[] args)
    {
        int min = Math.Min(2, 5);
        Console.WriteLine( min); // 2

        Console.WriteLine( Math.Min(1, 2)); // 1
        Console.WriteLine( Math.Min(1.5, 2.0)); // 1.5

        Console.WriteLine( Math.Min(2, Double.NaN)); // NaN
    }
}

1. Documentation

Syntax
namespace System
{
    public static class Math
    {
        // ...
        public static byte Min(byte val1, byte val2) { ... }
        public static decimal Min(decimal val1, decimal val2) { ... }
        public static double Min(double val1, double val2) { ... }
        public static short Min(short val1, short val2) { ... }
        public static int Min(int val1, int val2) { ... }
        public static long Min(long val1, long val2) { ... }
        public static sbyte Min(sbyte val1, sbyte val2) { ... }
        public static float Min(float val1, float val2) { ... }
        public static ushort Min(ushort val1, ushort val2) { ... }
        public static uint Min(uint val1, uint val2) { ... }
        public static ulong Min(ulong val1, ulong val2) { ... }
        // ...
    }
}
Parametersval1, val2 - values to compare.
Result

Minimal number value (primitive value).

It returns NaN if at least one of the provided values is not a number.

DescriptionMin is a static method that takes number arguments and returns the smallest one value.

2. Getting min value from array examples

2.1. With Min() method from System.Linq.

using System;
using System.Linq;

public class Program
{
    public static void Main(string[] args)
    {
        int[] array = new int[] { 1, 5, 2, 16, -5 };

        int min = array.Min();
        Console.WriteLine(min); // -5
    }
}

2.2. By comparing all the elements in the array

using System;

public class Program
{
    static double Min(double[] array)
    {
        double result = Double.PositiveInfinity;
        foreach (double currentNumber in array)
        {
            if (result > currentNumber) result = currentNumber;
        }
        return result;
    }

    public static void Main(string[] args)
    {
        double[] numbersArray = new double[] { 1.4, 5.0, 2.2, 16.0, -5.0 };

        Console.WriteLine(Min(numbersArray)); // -5
    }
}

References

  1. Maxima and minima - Wikipedia

C# / .NET - Math object

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