你是正确的，在语义上，ref同时提供“in”和“out”功能，而out只提供“out”功能。有一些事情需要考虑:

out requires that the method accepting the parameter MUST, at some point before returning, assign a value to the variable. You find this pattern in some of the key/value data storage classes like Dictionary<K,V>, where you have functions like TryGetValue. This function takes an out parameter that holds what the value will be if retrieved. It wouldn't make sense for the caller to pass a value into this function, so out is used to guarantee that some value will be in the variable after the call, even if it isn't "real" data (in the case of TryGetValue where the key isn't present). out and ref parameters are marshaled differently when dealing with interop code

Also, as an aside, it's important to note that while reference types and value types differ in the nature of their value, every variable in your application points to a location of memory that holds a value, even for reference types. It just happens that, with reference types, the value contained in that location of memory is another memory location. When you pass values to a function (or do any other variable assignment), the value of that variable is copied into the other variable. For value types, that means that the entire content of the type is copied. For reference types, that means that the memory location is copied. Either way, it does create a copy of the data contained in the variable. The only real relevance that this holds deals with assignment semantics; when assigning a variable or passing by value (the default), when a new assignment is made to the original (or new) variable, it does not affect the other variable. In the case of reference types, yes, changes made to the instance are available on both sides, but that's because the actual variable is just a pointer to another memory location; the content of the variable--the memory location--didn't actually change.

与ref关键字一起传递表示原始变量和函数参数实际上都指向相同的内存位置。同样，这只影响赋值语义。如果将一个新值赋给其中一个变量，那么由于其他变量指向相同的内存位置，新值将反映在另一侧。

关于c# 7的额外注意事项: 在c# 7中，不需要使用out预先声明变量。代码是这样的:

public void PrintCoordinates(Point p)
{
  int x, y; // have to "predeclare"
  p.GetCoordinates(out x, out y);
  WriteLine($"({x}, {y})");
}

可以这样写:

public void PrintCoordinates(Point p)
{
  p.GetCoordinates(out int x, out int y);
  WriteLine($"({x}, {y})");
}

来源:c# 7的新特性。

如果计划读取和写入参数，则需要使用ref。如果你只打算写，你需要使用out。实际上，当您需要多个返回值时，或者当您不想使用正常的返回机制进行输出时(但这种情况应该很少发生)，就会使用out。

有一些语言机制可以帮助这些用例。Ref形参在传递给方法之前必须已经初始化(强调它们是可读可写的)，out形参在被赋值之前不能被读取，并且保证在方法结束时已经被写入(强调它们只能被写入)。违反这些原则将导致编译时错误。

int x;
Foo(ref x); // error: x is uninitialized

void Bar(out int x) {}  // error: x was not written to

例如，int。TryParse返回bool类型，并接受out int形参:

int value;
if (int.TryParse(numericString, out value))
{
    /* numericString was parsed into value, now do stuff */
}
else
{
    /* numericString couldn't be parsed */
}

这是需要输出两个值的一个明确示例:数值结果和转换是否成功。CLR的作者决定在这里选择out，因为他们不关心int之前可能是什么。

作为参考，你可以看联锁。增量:

int x = 4;
Interlocked.Increment(ref x);

联锁。Increment自动地增加x的值。因为需要读取x来增加它，所以在这种情况下ref更合适。你完全关心x在传递给Increment之前是什么。

在c#的下一个版本中，甚至可以声明变量In out形参，更加强调它们只输出的性质:

if (int.TryParse(numericString, out int value))
{
    // 'value' exists and was declared in the `if` statement
}
else
{
    // conversion didn't work, 'value' doesn't exist here
}

它取决于编译上下文(参见下面的示例)。

out和ref都表示通过引用传递变量，但ref要求变量在传递之前进行初始化，这在封送处理上下文中是一个重要的区别(Interop: UmanagedToManagedTransition或反之亦然)。

MSDN警告说:

不要将按引用传递的概念与引用类型的概念混淆。这两个概念是不一样的。不管方法形参是值类型还是引用类型，都可以通过ref来修改。通过引用传递值类型时，没有对值类型进行装箱。

来自MSDN官方文档:

:

out关键字使参数通过引用传递。这类似于ref关键字，不同的是ref要求在传递变量之前对变量进行初始化

裁判:

ref关键字使参数按引用传递，而不是按值传递。通过引用传递的效果是，对方法中参数的任何更改都反映在调用方法中的底层参数变量中。引用形参的值始终与基础实参变量的值相同。

当参数被赋值时，我们可以验证out和ref确实是相同的:

CIL的例子:

考虑下面的例子

static class outRefTest{
    public static int myfunc(int x){x=0; return x; }
    public static void myfuncOut(out int x){x=0;}
    public static void myfuncRef(ref int x){x=0;}
    public static void myfuncRefEmpty(ref int x){}
    // Define other methods and classes here
}

在CIL中，myfuncOut和myfuncRef的指令与预期相同。

outRefTest.myfunc:
IL_0000:  nop         
IL_0001:  ldc.i4.0    
IL_0002:  starg.s     00 
IL_0004:  ldarg.0     
IL_0005:  stloc.0     
IL_0006:  br.s        IL_0008
IL_0008:  ldloc.0     
IL_0009:  ret         

outRefTest.myfuncOut:
IL_0000:  nop         
IL_0001:  ldarg.0     
IL_0002:  ldc.i4.0    
IL_0003:  stind.i4    
IL_0004:  ret         

outRefTest.myfuncRef:
IL_0000:  nop         
IL_0001:  ldarg.0     
IL_0002:  ldc.i4.0    
IL_0003:  stind.i4    
IL_0004:  ret         

outRefTest.myfuncRefEmpty:
IL_0000:  nop         
IL_0001:  ret         

Nop:不操作，ldloc: load local, stloc: stack local, ldarg: load参数，bs。S:分支到目标....

(见:CIL指令列表)

你是正确的，在语义上，ref同时提供“in”和“out”功能，而out只提供“out”功能。有一些事情需要考虑:

out requires that the method accepting the parameter MUST, at some point before returning, assign a value to the variable. You find this pattern in some of the key/value data storage classes like Dictionary<K,V>, where you have functions like TryGetValue. This function takes an out parameter that holds what the value will be if retrieved. It wouldn't make sense for the caller to pass a value into this function, so out is used to guarantee that some value will be in the variable after the call, even if it isn't "real" data (in the case of TryGetValue where the key isn't present). out and ref parameters are marshaled differently when dealing with interop code

Also, as an aside, it's important to note that while reference types and value types differ in the nature of their value, every variable in your application points to a location of memory that holds a value, even for reference types. It just happens that, with reference types, the value contained in that location of memory is another memory location. When you pass values to a function (or do any other variable assignment), the value of that variable is copied into the other variable. For value types, that means that the entire content of the type is copied. For reference types, that means that the memory location is copied. Either way, it does create a copy of the data contained in the variable. The only real relevance that this holds deals with assignment semantics; when assigning a variable or passing by value (the default), when a new assignment is made to the original (or new) variable, it does not affect the other variable. In the case of reference types, yes, changes made to the instance are available on both sides, but that's because the actual variable is just a pointer to another memory location; the content of the variable--the memory location--didn't actually change.

与ref关键字一起传递表示原始变量和函数参数实际上都指向相同的内存位置。同样，这只影响赋值语义。如果将一个新值赋给其中一个变量，那么由于其他变量指向相同的内存位置，新值将反映在另一侧。

基本上ref和out都用于在方法之间传递对象/值

out关键字使参数通过引用传递。这类似于ref关键字，只不过ref要求在传递变量之前对其进行初始化。

out:参数没有初始化，必须在方法中初始化

ref:参数已经初始化，可以在方法中读取和更新。

引用类型的“ref”有什么用?

您可以将给定的引用更改为不同的实例。

你知道吗?

Although the ref and out keywords cause different run-time behavior, they are not considered part of the method signature at compile time. Therefore, methods cannot be overloaded if the only difference is that one method takes a ref argument and the other takes an out argument. You can't use the ref and out keywords for the following kinds of methods: Async methods, which you define by using the async modifier. Iterator methods, which include a yield return or yield break statement. Properties are not variables and therefore cannot be passed as out parameters.