为什么我不能在范围内的switch语句中使用"常量"？

Question

使用此代码:

public partial class Form1 : Form
{
    private static readonly int TABCONTROL_BASICINFO = 0;
    private static readonly int TABCONTROL_CONFIDENTIALINFO = 1;
    private static readonly int TABCONTROL_ROLESANDSECURITY = 2;
    private static readonly int TABCONTROL_INACTIVEINFO = 3;
. . .
int ActiveTabPage = tabControlWorker.SelectedIndex;
switch (ActiveTabPage) {
    case TABCONTROL_BASICINFO:
        if (currentNode == "NodeBuckingham") {
        } else if (currentNode == "NodeNamath") {
        } else if (currentNode == "NodeParsons") {
        } else {
        }
    break;

...我必须用"0"替换"TABCONTROL_BASICINFO",或者我得到"预期值为常数"

天堂到Murgatroyd!不能查看并看到TABCONTROL_BASICINFO为0？

Answer 1

如果要在编译器方面使其成为常量表达式,请将其声明为const:

// Note that static readonly is implied here
private const int TABCONTROL_BASICINFO = 0;

或者遵循.NET命名约定......

private const int TabControlBasicInfo = 0;

或者使用枚举,因为你基本上得到了一组固定的值:

private enum TabControlType
{
    // Could ditch the explicit values here if you want
    BasicInfo = 0,
    ConfidentialInfo = 1,
    ...
}

顺便说一下,你也可以在C#中打开字符串,所以这个:

 if (currentNode == "NodeBuckingham") {
 } else if (currentNode == "NodeNamath") {
 } else if (currentNode == "NodeParsons") {
 } else {
 }

可以变成:

 switch (currentNode) {
     case "NodeBuckingham":
         ...
         break;
     case "NodeNamath":
         ...
         break;
     case "NodeParsons":
         ...
         break;
     default:
         ...
         break;
 }

Answer 2

甲readonly变量不是一个常数.该值在编译时是未知的,而是可以在声明中(如您所做)或在类构造函数(在本例中为您的类的静态构造函数)中初始化.

欲了解更多

http://msdn.microsoft.com/en-us/library/acdd6hb7(v=vs.71).aspx

您可以将其更改为:

private const int TABCONTROL_BASICINFO = 0; 

除非您需要计算某些内容来初始化变量,否则请将其声明为const.它会稍微高效一点.

Answer 3

这不是一个常数; 它是一个只读变量.如果您希望它是常量,则将其标记为常量.

private const int TabControlBasicInfo = 0;

甚至更好:

private enum TabControl
{
    BasicInfo,
    ...
}
...
switch(whatever)
{
    case TabControl.BasicInfo:

Answer 4

静态只读不是一个常量变量.它可以初始化.而" const "是不变的.