实际上有一种方法可以做到这一点，这是c#允许的。这又回到了c++，并且违背了c#的面向对象结构。

使用这个方法要小心!

下面是使用可选参数声明和编写函数的方法:

unsafe public void OptionalOutParameter(int* pOutParam = null)
{
    int lInteger = 5;
    // If the parameter is NULL, the caller doesn't care about this value.
    if (pOutParam != null) 
    { 
        // If it isn't null, the caller has provided the address of an integer.
        *pOutParam = lInteger; // Dereference the pointer and assign the return value.
    }
}

然后像这样调用函数:

unsafe { OptionalOutParameter(); } // does nothing
int MyInteger = 0;
unsafe { OptionalOutParameter(&MyInteger); } // pass in the address of MyInteger.

为了进行编译，您需要在项目选项中启用不安全代码。这是一个通常不应该使用的解决方案，但如果你为了一些奇怪的，神秘的，管理灵感的决定，确实需要一个可选的out参数在c#中，那么这将允许你这样做。

对于c# 6.0及以下版本，使用不带out形参的重载方法调用带out形参的方法。当被特别问到c# 4.0是否可以有一个可选的out参数时，我不确定为什么用于。net Core的c# 7.0甚至是这个线程的正确答案。答案是否定的!

正如已经提到的，这是不允许的，我认为这是非常有意义的。 然而，为了增加更多的细节，这里引用了c# 4.0规范21.1节:

Formal parameters of constructors, methods, indexers and delegate types can be declared optional: fixed-parameter:     attributesopt parameter-modifieropt type identifier default-argumentopt default-argument:     = expression A fixed-parameter with a default-argument is an optional parameter, whereas a fixed-parameter without a default-argument is a required parameter. A required parameter cannot appear after an optional parameter in a formal-parameter-list. A ref or out parameter cannot have a default-argument.

这个直接的问题已经在其他得到好评的答案中得到了回答，但有时根据你想要实现的目标考虑其他方法是值得的。

If you're wanting an optional parameter to allow the caller to possibly request extra data from your method on which to base some decision, an alternative design is to move that decision logic into your method and allow the caller to optionally pass a value for that decision criteria in. For example, here is a method which determines the compass point of a vector, in which we might want to pass back the magnitude of the vector so that the caller can potentially decide if some minimum threshold should be reached before the compass-point judgement is far enough away from the origin and therefore unequivocally valid:

public enum Quadrant {
    North,
    East,
    South,
    West
}

// INVALID CODE WITH MADE-UP USAGE PATTERN OF "OPTIONAL" OUT PARAMETER
public Quadrant GetJoystickQuadrant([optional] out magnitude)
{
    Vector2 pos = GetJoystickPositionXY();
    float azimuth = Mathf.Atan2(pos.y, pos.x) * 180.0f / Mathf.PI;
    Quadrant q;
    if (azimuth > -45.0f && azimuth <= 45.0f) q = Quadrant.East;
    else if (azimuth > 45.0f && azimuth <= 135.0f) q = Quadrant.North;
    else if (azimuth > -135.0f && azimuth <= -45.0f) q = Quadrant.South;
    else q = Quadrant.West;
    if ([optonal.isPresent(magnitude)]) magnitude = pos.Length();
    return q;
}

在这种情况下，我们可以将“最小大小”逻辑移动到方法中，并以一个更清晰的实现结束，特别是因为计算大小涉及到一个平方根，所以如果我们只想做大小的比较，那么计算效率就会很低，因为我们可以用平方值来比较:

public enum Quadrant {
    None, // Too close to origin to judge.
    North,
    East,
    South,
    West
}

public Quadrant GetJoystickQuadrant(float minimumMagnitude = 0.33f)
{
    Vector2 pos = GetJoystickPosition();
    if (minimumMagnitude > 0.0f && pos.LengthSquared() < minimumMagnitude * minimumMagnitude)
    {
        return Quadrant.None;
    }
    float azimuth = Mathf.Atan2(pos.y, pos.x) * 180.0f / Mathf.PI;
    if (azimuth > -45.0f && azimuth <= 45.0f) return Quadrant.East;
    else if (azimuth > 45.0f && azimuth <= 135.0f) return Quadrant.North;
    else if (azimuth > -135.0f && azimuth <= -45.0f) return Quadrant.South;
    return Quadrant.West;
}

当然，这可能并不总是可行的。由于其他答案提到了c# 7.0，如果你真正要做的是返回两个值，并允许调用者可选地忽略其中一个，那么惯用的c#将返回两个值的元组，并使用c# 7.0的元组与位置初始化器和_ "discard"参数:

public (Quadrant, float) GetJoystickQuadrantAndMagnitude()
{
    Vector2 pos = GetJoystickPositionXY();
    float azimuth = Mathf.Atan2(pos.y, pos.x) * 180.0f / Mathf.PI;
    Quadrant q;
    if (azimuth > -45.0f && azimuth <= 45.0f) q = Quadrant.East;
    else if (azimuth > 45.0f && azimuth <= 135.0f) q = Quadrant.North;
    else if (azimuth > -135.0f && azimuth <= -45.0f) q = Quadrant.South;
    else q = Quadrant.West;
    return (q, pos.Length());
}

(Quadrant q, _) = GetJoystickQuadrantAndMagnitude();
if (q == Quadrant.South)
{
    // Do something.
}

实际上有一种方法可以做到这一点，这是c#允许的。这又回到了c++，并且违背了c#的面向对象结构。

使用这个方法要小心!

下面是使用可选参数声明和编写函数的方法:

unsafe public void OptionalOutParameter(int* pOutParam = null)
{
    int lInteger = 5;
    // If the parameter is NULL, the caller doesn't care about this value.
    if (pOutParam != null) 
    { 
        // If it isn't null, the caller has provided the address of an integer.
        *pOutParam = lInteger; // Dereference the pointer and assign the return value.
    }
}

然后像这样调用函数:

unsafe { OptionalOutParameter(); } // does nothing
int MyInteger = 0;
unsafe { OptionalOutParameter(&MyInteger); } // pass in the address of MyInteger.

为了进行编译，您需要在项目选项中启用不安全代码。这是一个通常不应该使用的解决方案，但如果你为了一些奇怪的，神秘的，管理灵感的决定，确实需要一个可选的out参数在c#中，那么这将允许你这样做。

ICYMI:包括在这里列举的c# 7.0的新特性中，“discards”现在被允许以_的形式作为out参数，让你忽略你不关心的out参数:

p.GetCoordinates(out var x, out _);//我只关心x

附注:如果你也对“输出变量x”的部分感到困惑，请阅读链接上关于“输出变量”的新功能。