CA1802: Use Literals Where Appropriate


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TypeName UseLiteralsWhereAppropriate
CheckId CA1802
Category Microsoft.Performance
Breaking Change Non-breaking


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 it is 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 Suppress Warnings

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


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

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";
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