blocks|key|139576|text|您正在观察提升运算符的效果。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|139577|从C#+5规范的7.3.7节中：|139578|139579|提升运算符允许对不可为空值类型进行操作的预定义和用户定义运算符也可与这些类型的可为空形式一起使用。提升运算符由满足特定要求的预定义运算符和用户定义运算符构造而成，如下所述：|blockquote|139580|139581|139582|如果操作数和结果类型都是不可为空值类型，则存在|139583|+%2B++%2B%2B++-++--++!++~+unary运算符和提升形式的运算符。提升的形式是通过向操作数和结果类型添加一个?修饰符来构造的。如果操作数为空，则提升的运算符将生成空值。否则，提升的运算符将展开操作数，应用基础运算符，然后包装结果。|unordered-list-item|offset|length|style|CODE|139584|139585|因此，基本上，本例中的a%2B%2B是一个结果为null+(作为int?)的表达式，而变量保持不变。|139586|当你调用|139587|Console.WriteLine(a);|code-block|syntax|javascript|139588|它被装箱到object中，后者将其转换为空引用，并将其打印为空行。|139589|entityMap^0|0|0|0|0|0|0|0|1|I|1N|1|0|0|B|3|K|4|S|4|0|0|0|5|6|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|17|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|18|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|19|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|J|3|K|5|6|7|1C|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|1D|8|@$O|1E|P|1F|Q|R]|$O|1G|P|1H|Q|R]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|T|3|U|5|6|7|1J|8|@$O|1K|P|1L|Q|R]|$O|1M|P|1N|Q|R]|$O|1O|P|1P|Q|R]]|9|@]|A|$]]|$1|V|3|W|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|X|3|Y|5|Z|7|1R|8|@]|9|@]|A|$10|11]]|$1|12|3|13|5|6|7|1S|8|@$O|1T|P|1U|Q|R]]|9|@]|A|$]]|$1|14|3|-4|5|6|7|1V|8|@]|9|@]|A|$]]]|15|$]]

You're observing the effects of a lifted operator.
From section 7.3.7 of the C# 5 specification:
<blockquote>
Lifted operators permit predefined and user-defined operators that operate on non-nullable value types to also be used with nullable forms of those types. Lifted operators are constructed from predefined and user-defined operators that meet certain requirements, as described in the following:
<ul>
<li>For the unary operators <code>+ ++ - -- ! ~</code>
a lifted form of an operator exists if the operand and result types are both non-nullable value types. The lifted form is constructed by adding a single <code>?</code> modifier to the operand and result types. The lifted operator produces a null value if the operand is null. Otherwise, the lifted operator unwraps the operand, applies the underlying operator, and wraps the result.</li>
</ul>
</blockquote>
So basically, <code>a++</code> in this case is an expression with a result of <code>null</code> (as an <code>int?</code>) and the variable is left untouched.
When you call
<pre><code>Console.WriteLine(a);
</code></pre>
that's being boxed into <code>object</code>, which converts it to a null reference, which is printed as an empty line.

blocks|key|139644|text|Jon的回答是正确的，但我想补充一些补充说明。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|139645|139646|为什么Console.WriteLine(null)会出现编译错误？|blockquote|offset|length|style|CODE|139647|139648|139649|Console.WriteLine有19个重载，其中三个重载适用于null：接受string的重载、接受char[]的重载和接受object的重载。C#无法确定您所指的是这三个中的哪一个，因此它会给出一个错误。Console.WriteLine((object)null)将是合法的，因为现在它是明确的。|139650|139651|为什么Console.WriteLine(a)会写一个空行？|139652|139653|139654|a为空int?。重载解析选择方法的object版本，因此int?被装箱为空引用。所以这基本上和Console.WriteLine((object)null)是一样的，它写一个空行。|139655|139656|为什么增量上没有NullReferenceException？|139657|139658|139659|您担心的空引用在哪里？a是一个空int?，从一开始就不是引用类型！请记住，可为空的值类型是值类型，而不是引用类型，因此不要期望它们具有引用语义，除非它们被装箱为引用类型。加法中没有装箱。|BOLD|139660|entityMap^0|0|0|3|N|0|0|0|0|H|X|4|14|6|1G|6|1S|6|2X|V|0|0|3|K|0|0|0|0|1|3|4|H|6|S|4|1B|V|0|0|8|M|0|0|0|5|2|19|3|1G|4|B|1|G|4|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|15|8|@]|9|@]|A|$]]|$1|C|3|D|5|E|7|16|8|@$F|17|G|18|H|I]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|19|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1B|8|@$F|1C|G|1D|H|I]|$F|1E|G|1F|H|I]|$F|1G|G|1H|H|I]|$F|1I|G|1J|H|I]|$F|1K|G|1L|H|I]|$F|1M|G|1N|H|I]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]|$1|O|3|P|5|E|7|1P|8|@$F|1Q|G|1R|H|I]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1U|8|@$F|1V|G|1W|H|I]|$F|1X|G|1Y|H|I]|$F|1Z|G|20|H|I]|$F|21|G|22|H|I]|$F|23|G|24|H|I]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|25|8|@]|9|@]|A|$]]|$1|V|3|W|5|E|7|26|8|@$F|27|G|28|H|I]]|9|@]|A|$]]|$1|X|3|-4|5|6|7|29|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|2A|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|2B|8|@$F|2C|G|2D|H|11]|$F|2E|G|2F|H|11]|$F|2G|G|2H|H|11]|$F|2I|G|2J|H|I]|$F|2K|G|2L|H|I]]|9|@]|A|$]]|$1|12|3|-4|5|6|7|2M|8|@]|9|@]|A|$]]]|13|$]]

Jon's answer is correct but I would add some additional notes.

<blockquote>
 Why does <code>Console.WriteLine(null)</code> give a compilation error?
</blockquote>

There are 19 overloads of <code>Console.WriteLine</code> and three of them are applicable to a <code>null</code>: the one that takes a <code>string</code>, the one that takes a <code>char[]</code> and the one that takes an <code>object</code>. C# cannot determine which of these three you mean, so it gives an error. <code>Console.WriteLine((object)null)</code> would be legal because now it is clear.

<blockquote>
 why does <code>Console.WriteLine(a)</code> write an empty line?
</blockquote>

<code>a</code> is a null <code>int?</code>. Overload resolution chooses the <code>object</code> version of the method, so the <code>int?</code> is boxed to a null reference. So this is basically the same as <code>Console.WriteLine((object)null)</code>, which writes an empty line.

<blockquote>
 Why there is not <code>NullReferenceException</code> on the increment?
</blockquote>

Where's the null reference that you are worried about? <code>a</code> is a null <code>int?</code> which is not a reference type to begin with! Remember, nullable value types are value types, not reference types, so don't expect them to have reference semantics unless they are boxed to a reference type. There is no boxing in the addition.

blocks|key|2709319|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2709320|您是否正在递增null？|blockquote|2709321|2709322|2709323|int?+a+=+null;
a%2B%2B;|code-block|syntax|javascript|2709324|这句话的意思就是null%2B%2B，即null%2B1。|offset|length|style|CODE|2709325|根据本文档，可空类型可以表示其基础值类型的值的正确范围，外加附加的空值。可以为可空类型分配从-2147483648到2147483647的任何值，也可以为其分配空值，发音为“Nullable+of+Int32”|2709326|在这里你递增null，然后它也将变成null+value+not+0或任何其他整数。|2709327|2709328|为什么打印空白而不是错误？？|2709329|2709330|2709331|当您使用null值打印一个可空类型时，它会打印空白而不是错误，因为您打印的是一个变量，即内存位置的值。它可以是null，也可以是任意整数。|2709332|但是当您尝试使用Console.WriteLine(null)打印null时，因为null不是一个变量，所以它不会引用任何内存位置。因此它会给出错误"NullReferenceException"。|2709333|那么，如何使用Console.WriteLine(2);打印任意整数呢?？|2709334|2709335|2709336|2709337|在这种情况下，2将在内存中的临时位置获取，并且指针指向要打印的内存位置。|2709338|entityMap^0|0|0|0|0|0|8|6|0|0|0|0|0|0|0|0|8|N|22|O|0|7|L|0|0|0|0|0^^$0|@$1|2|3|-4|4|5|6|1C|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|1D|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|1E|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|1F|7|@]|8|@]|9|$]]|$1|F|3|G|4|H|6|1G|7|@]|8|@]|9|$I|J]]|$1|K|3|L|4|5|6|1H|7|@$M|1I|N|1J|O|P]]|8|@]|9|$]]|$1|Q|3|R|4|5|6|1K|7|@]|8|@]|9|$]]|$1|S|3|T|4|5|6|1L|7|@]|8|@]|9|$]]|$1|U|3|-4|4|5|6|1M|7|@]|8|@]|9|$]]|$1|V|3|W|4|C|6|1N|7|@]|8|@]|9|$]]|$1|X|3|-4|4|5|6|1O|7|@]|8|@]|9|$]]|$1|Y|3|-4|4|5|6|1P|7|@]|8|@]|9|$]]|$1|Z|3|10|4|5|6|1Q|7|@]|8|@]|9|$]]|$1|11|3|12|4|5|6|1R|7|@$M|1S|N|1T|O|P]|$M|1U|N|1V|O|P]]|8|@]|9|$]]|$1|13|3|14|4|5|6|1W|7|@$M|1X|N|1Y|O|P]]|8|@]|9|$]]|$1|15|3|-4|4|C|6|1Z|7|@]|8|@]|9|$]]|$1|16|3|-4|4|5|6|20|7|@]|8|@]|9|$]]|$1|17|3|-4|4|5|6|21|7|@]|8|@]|9|$]]|$1|18|3|19|4|5|6|22|7|@]|8|@]|9|$]]|$1|1A|3|-4|4|5|6|23|7|@]|8|@]|9|$]]]|1B|$]]

<blockquote>
 Are You incrementing null???
</blockquote>

<pre><code>int? a = null;
a++;
</code></pre>

This statement simply means <code>null++</code> i.e. null+1.

As per this document, A nullable type can represent the correct range of values for its underlying value type, plus an additional null value.A Nullable, pronounced "Nullable of Int32," can be assigned any value from -2147483648 to 2147483647, or it can be assigned the null value

Here you are incrementing null, then also it will become null value not 0 or any other integer.

<blockquote>
 Why it prints blank instead of error??
</blockquote>

when you print a nullable type with null value it prints blank instead of error because you are printing a variable i.e. value of a memory location. which maybe null or any integer.

But when you try to print null using <code>Console.WriteLine(null)</code>, as null is not a variable, so it doesn't refer to any memory location. And hence it gives error <code>"NullReferenceException"</code>. 

<blockquote>
 Then how can you print any integer using <code>Console.WriteLine(2);</code> ??
</blockquote>

In this case, 2 will gets in memory at temporary location, and the pointer points to that memory location to print.

Could you please explain, why does Console.WriteLine write empty line (<code>Console.WriteLine(null)</code> give me compilation error) and why there isn't NullReferenceException (even <code>a+=1</code> shouldn't raise it)?

<pre><code>int? a = null;
a++; // Why there is not NullReferenceException? 
Console.WriteLine(a); // Empty line
</code></pre>

Why doesn't incrementing Nullable<int> throw an exception?

你能解释一下，为什么Console.WriteLine写空行(Console.WriteLine(null)给了我编译错误)，为什么没有NullReferenceException (即使是a+=1也不应该抛出它)？int? a = null;a++; // Why there is not NullReference...

问为什么递增Nullable<int>不抛出异常？
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问为什么递增Nullable<int>不抛出异常？
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