Use literals where appropriate

Cause

A field is declared static and readonly (Shared and ReadOnly in Visual Basic), and is initialized with a value that is computable at compile time.

Rule Description

The value of a static readonly field is computed at runtime when the static constructor for the declaring type is called. If the static readonly field is initialized when it is declared and a static constructor is not declared explicitly, the compiler emits a static constructor to initialize the field.

The value of a const field is computed at compile time and stored in the metadata, which increases runtime performance when compared to a static readonly field.

Because the value assigned to the targeted field is computable at compile time, change the declaration to a const field so that the value is computed at compile time instead of at runtime.

How to Fix Violations

To fix a violation of this rule, replace the static and readonly modifiers with the const modifier.

When to Exclude Warnings

It is safe to exclude a warning from this rule, or disable the rule entirely, if performance is not of concern.

Example

The following example shows a type, UseReadOnly, that violates the rule and a type, UseConstant, that satisfies the rule.

Imports System

Namespace PerformanceLibrary

   ' This class violates the rule.
   Public Class UseReadOnly

      Shared ReadOnly x As Integer = 3
      Shared ReadOnly y As Double = x + 2.1
      Shared ReadOnly s As String = "readonly"

   End Class

   ' This class satisfies the rule.
   Public Class UseConstant

      Const x As Integer = 3
      Const y As Double = x + 2.1
      Const s As String = "const"

   End Class

End Namespace

using System;

namespace PerformanceLibrary
{
   // This class violates the rule.
   public class UseReadOnly
   {
      static readonly int x = 3;
      static readonly double y = x + 2.1;
      static readonly string s = "readonly";
   }

   // This class satisfies the rule.
   public class UseConstant
   {
      const int x = 3;
      const double y = x + 2.1;
      const string s = "const";
   }
}