我刚刚发现这个：
ushort i = 4;
i = i + 4;
给出编译器错误：
无法将类型“int”隐式转换为“ushort”。存在显式转换（您是否缺少演员表？）
我必须这样修复它：
ushort i = 4;
i = (ushort)(i + 4);
这背后的原因是什么？使用所有数据类型不应该是显而易见且易于使用的吗？
问问题
131 次
6 回答
6
文字4是 an int，加法i + 4也是，被提升为。此加法的结果也是，因此您不能在没有强制转换的情况下将其分配给 a ，因为 C# 不允许隐式转换为更小的数值类型。intiintintushort
于 2013-03-29T15:28:42.347 回答
3
原因是因为 ushort + ushort 实际上返回一个 int。查看此线程以获取有关为什么会这样的更多详细信息。
于 2013-03-29T15:29:54.797 回答
0
这是因为编译器将其4视为integer. 从较高数据类型到较低数据类型的转换需要显式转换。
于 2013-03-29T15:29:11.060 回答
0
根据 C# 语言标准，特别是关于整数文字的第 2.4.4.2 节：
整数文字的类型确定如下：
如果字面量没有后缀，则它具有可以表示其值的第一种类型：int、uint、long、ulong。如果文字以 U 或 u 为后缀，则它具有可以表示其值的第一种类型：uint、ulong。如果文字以 L 或 l 为后缀，则它具有可以表示其值的第一种类型：long、ulong。如果文字以 UL、Ul、uL、ul、LU、Lu、lU 或 lu 为后缀，则为 ulong 类型。
因此，您的号码被视为 int，这就是原因。
于 2013-03-29T15:34:09.140 回答
0
The +-operatior for integer types is defined only between pairs of int, uint, long, and ulong. To calculate i+4 first i has to be converted from ushort through an implicit numeric conversion into int, and the result is of the same type - also int So the type of i+4 is actually int. Since there is no implicit conversion that would allow assigning an int to a variable defined as ushort the compiler gives you an error for i = i + 4.
Note that you still can use the below because += involves an implicit cast:
i += 4;
于 2013-03-29T15:35:31.597 回答
0
I think the main thing to take away from the comments is that int32 was just the 'default' type to be used by the common architecture back in the day. I don't know about native support for smaller integer types by current processors, but I wouldn't be surprised if everything is still converted to int32 by the compiler under water even if it is a ushort.
So (u)short is more like a validation constraint for the programmer than a memory-saver. Even if it is a memory saver, I wouldn't be surprised if more CPU cycles are needed to convert back and to ushorts.
于 2013-03-29T15:37:22.967 回答