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Integer overflow and arithmetic assignment operators in C#: short = short
In C#, when two short
integers are added, the result will be a int
type with a larger range of values than short
. Integer overflow occurs if the sum of two short
integers exceeds the maximum value of short
.
Consider the following code:
<code class="language-csharp">short a, b; a = 10; b = 10; a = a + b; // 错误:无法隐式转换类型“int”为“short”。</code>
Here, the addition operation result of a
and b
is of type int
, and the compiler cannot implicitly convert it to type short
. This is because assigning int
to short
without an explicit conversion results in data loss.
However, the following code does not produce an error:
<code class="language-csharp">a += b; // 但这段代码可以成功运行,为什么?</code>
This is because C# handles the arithmetic assignment operator =
differently. When using =
, the compiler internally does the following:
<code class="language-csharp">a = (short)(a + b);</code>
The compiler will automatically convert the result of the addition operation to the a
type before assigning the result to short
. This ensures that the value is correctly stored in the int
variable even if the intermediate result is of type short
.
This behavior is similar to other compound assignment operators, such as -=
, *=
, and /=
, which automatically convert the result to the type of the left operand.
Thus, the arithmetic assignment operator short
can be safely used on =
variables even if the sum exceeds the range of short
. The compiler handles type conversion internally to avoid integer overflow errors.
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