: return语句只能用于从函数中返回一个值。但是，使用输出参数，可以从一个函数返回两个值。输出参数类似于引用参数，只是它们将数据传输出方法而不是传输到方法中。

下面的例子说明了这一点:

using System;

namespace CalculatorApplication
{
   class NumberManipulator
   {
      public void getValue(out int x )
      {
         int temp = 5;
         x = temp;
      }

      static void Main(string[] args)
      {
         NumberManipulator n = new NumberManipulator();
         /* local variable definition */
         int a = 100;

         Console.WriteLine("Before method call, value of a : {0}", a);

         /* calling a function to get the value */
         n.getValue(out a);

         Console.WriteLine("After method call, value of a : {0}", a);
         Console.ReadLine();

      }
   }
}

裁判: 引用形参是对变量内存位置的引用。与值参数不同，通过引用传递参数时，不会为这些参数创建新的存储位置。引用参数表示与提供给方法的实际参数相同的内存位置。

在c#中，使用ref关键字声明引用参数。下面的例子说明了这一点:

using System;
namespace CalculatorApplication
{
   class NumberManipulator
   {
      public void swap(ref int x, ref int y)
      {
         int temp;

         temp = x; /* save the value of x */
         x = y;   /* put y into x */
         y = temp; /* put temp into y */
       }

      static void Main(string[] args)
      {
         NumberManipulator n = new NumberManipulator();
         /* local variable definition */
         int a = 100;
         int b = 200;

         Console.WriteLine("Before swap, value of a : {0}", a);
         Console.WriteLine("Before swap, value of b : {0}", b);

         /* calling a function to swap the values */
         n.swap(ref a, ref b);

         Console.WriteLine("After swap, value of a : {0}", a);
         Console.WriteLine("After swap, value of b : {0}", b);

         Console.ReadLine();

      }
   }
}

由于您传递的是引用类型(类)，因此不需要使用ref，因为默认情况下只传递对实际对象的引用，因此您总是要更改引用后面的对象。

例子:

public void Foo()
{
    MyClass myObject = new MyClass();
    myObject.Name = "Dog";
    Bar(myObject);
    Console.WriteLine(myObject.Name); // Writes "Cat".
}

public void Bar(MyClass someObject)
{
    someObject.Name = "Cat";
}

只要传入一个类，如果你想在方法中改变对象，就不必使用ref。

裁判: ref关键字用于将参数作为引用传递。这意味着当该参数的值在方法中被更改时，它会反映在调用方法中。使用ref关键字传递的参数在传递给被调用方法之前必须在调用方法中初始化。

: out关键字也用于传递一个参数，如ref关键字，但参数可以在不给它赋值的情况下传递。使用out关键字传递的参数在返回调用方法之前必须在被调用方法中初始化。

public class Example
{
 public static void Main() 
 {
 int val1 = 0; //must be initialized 
 int val2; //optional

 Example1(ref val1);
 Console.WriteLine(val1); 

 Example2(out val2);
 Console.WriteLine(val2); 
 }

 static void Example1(ref int value) 
 {
 value = 1;
 }
 static void Example2(out int value) 
 {
 value = 2; 
 }
}

/* Output     1     2     

在方法重载中引用和out

ref和out不能同时用于方法重载。然而，ref和out在运行时的处理方式不同，但在编译时的处理方式相同(CLR在为ref和out创建IL时不区分两者)。

除了允许你将别人的变量重新分配给类的不同实例，返回多个值等，使用ref或out可以让别人知道你需要从他们那里得到什么，以及你打算用他们提供的变量做什么

You don't need ref or out if all you're going to do is modify things inside the MyClass instance that is passed in the argument someClass. The calling method will see changes like someClass.Message = "Hello World" whether you use ref, out or nothing Writing someClass = new MyClass() inside myFunction(someClass) swaps out the object seen by the someClass in the scope of the myFunction method only. The calling method still knows about the original MyClass instance it created and passed to your method You need ref or out if you plan on swapping the someClass out for a whole new object and want the calling method to see your change Writing someClass = new MyClass() inside myFunction(out someClass) changes the object seen by the method that called myFunction

还有其他程序员

他们想知道你将如何处理他们的数据。假设您正在编写一个将被数百万开发人员使用的库。你想让他们知道当他们调用你的方法时你要对他们的变量做什么

Using ref makes a statement of "Pass a variable assigned to some value when you call my method. Be aware that I might change it out for something else entirely during the course of my method. Do not expect your variable to be pointing to the old object when I'm done" Using out makes a statement of "Pass a placeholder variable to my method. It doesn't matter whether it has a value or not; the compiler will force me to assign it to a new value. I absolutely guarantee that the object pointed to by your variable before you called my method, will be different by the time I'm done

顺便说一下，在c# 7.2中也有一个in修饰符

And that prevents the method from swapping out the passed in instance for a different instance. Think of it like saying to those millions of developers "pass me your original variable reference, and I promise not to swap your carefully crafted data out for something else". in has some peculiarities, and in some cases such as where an implicit conversion might be required to make your short compatible with an in int the compiler will temporarily make an int, widen your short to it, pass it by reference and finish up. It can do this because you've declared you're not going to mess with it.

微软对数值类型的.TryParse方法做到了这一点:

int i = 98234957;
bool success = int.TryParse("123", out i);

通过将参数标记为out他们在这里积极地声明"我们肯定会将你苦心制作的98234957值更改为其他值"

当然，对于像解析值类型这样的事情，它们有点不得不这样做，因为如果parse方法不允许将值类型替换为其他类型，那么它就不能很好地工作。但是想象一下在你创建的库中有一些虚构的方法:

public void PoorlyNamedMethod(out SomeClass x)

你可以看到它是一个out，因此你可以知道，如果你花了几个小时处理数字，创建一个完美的SomeClass:

SomeClass x = SpendHoursMakingMeAPerfectSomeClass();
//now give it to the library
PoorlyNamedMethod(out x);

那真是浪费时间，花那么多时间来做一节完美的课。它肯定会被丢弃，并被PoorlyNamedMethod取代

: return语句只能用于从函数中返回一个值。但是，使用输出参数，可以从一个函数返回两个值。输出参数类似于引用参数，只是它们将数据传输出方法而不是传输到方法中。

下面的例子说明了这一点:

using System;

namespace CalculatorApplication
{
   class NumberManipulator
   {
      public void getValue(out int x )
      {
         int temp = 5;
         x = temp;
      }

      static void Main(string[] args)
      {
         NumberManipulator n = new NumberManipulator();
         /* local variable definition */
         int a = 100;

         Console.WriteLine("Before method call, value of a : {0}", a);

         /* calling a function to get the value */
         n.getValue(out a);

         Console.WriteLine("After method call, value of a : {0}", a);
         Console.ReadLine();

      }
   }
}

裁判: 引用形参是对变量内存位置的引用。与值参数不同，通过引用传递参数时，不会为这些参数创建新的存储位置。引用参数表示与提供给方法的实际参数相同的内存位置。

在c#中，使用ref关键字声明引用参数。下面的例子说明了这一点:

using System;
namespace CalculatorApplication
{
   class NumberManipulator
   {
      public void swap(ref int x, ref int y)
      {
         int temp;

         temp = x; /* save the value of x */
         x = y;   /* put y into x */
         y = temp; /* put temp into y */
       }

      static void Main(string[] args)
      {
         NumberManipulator n = new NumberManipulator();
         /* local variable definition */
         int a = 100;
         int b = 200;

         Console.WriteLine("Before swap, value of a : {0}", a);
         Console.WriteLine("Before swap, value of b : {0}", b);

         /* calling a function to swap the values */
         n.swap(ref a, ref b);

         Console.WriteLine("After swap, value of a : {0}", a);
         Console.WriteLine("After swap, value of b : {0}", b);

         Console.ReadLine();

      }
   }
}

ref和out就像c++中传递引用和传递指针一样。

对于ref，参数必须声明并初始化。

对于out，实参必须声明，但可以初始化，也可以不初始化

        double nbr = 6; // if not initialized we get error
        double dd = doit.square(ref nbr);

        double Half_nbr ; // fine as passed by out, but inside the calling  method you initialize it
        doit.math_routines(nbr, out Half_nbr);