Use literals where appropriate
TypeName |
UseLiteralsWhereAppropriate |
CheckId |
CA1802 |
Category |
Microsoft.Performance |
Breaking Change |
NonBreaking |
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";
}
}