[英]Arithmetic operations with 8 and 16 bit integers
I do understand why this produce compile time-error: 我确实理解为什么这会产生编译时错误:
short x = 1;
short y = 2;
short z = x + y; // compile-time error
I've grasped why this runs without any issues: 我已经理解了为什么运行时没有任何问题:
short x = 1;
short y = 2;
x += y; // all right, because of c# specs section 7.17.2
But I've got no clue why this also works: 但是我不知道为什么这也行得通:
short x = (short)1 + (short)2;
I expected to get the same compile-time error as in the first example, but it runs successfully... Why? 我期望得到与第一个示例相同的编译时错误,但是它运行成功...为什么?
Since you're using constant values, the compiler can detect that it's allowable, evaluate it at compile time , and let it execute. 由于您使用的是常量值,因此编译器可以检测到它是允许的, 在编译时对其进行评估,然后让其执行。 The generated IL evaluates to the same as typing
short x = 3;
生成的IL与输入
short x = 3;
的结果相同short x = 3;
. 。
Note the following also works (for the same reason): 请注意,以下内容也可以工作(出于相同的原因):
const short x = 1;
const short y = 2;
short z = x + y;
But this fails: 但这失败了:
const short x = 32000;
const short y = 32001;
short z = x + y;
Note that this is covered in the C# Language Spec, 6.1.9 Implicit constant expression conversions: 请注意,C#语言规范6.1.9隐式常量表达式转换对此进行了介绍:
Your last snippet just compiles to constant 3
. 您的最后一个片段只是编译为常数
3
。 Compiler doesn't need to call any operators in int
it just computes and stores the value at compile time. 编译器不需要在
int
调用任何运算符,它只是在编译时计算并存储值。
it is same as short x = 3;
它与
short x = 3;
相同short x = 3;
Here is the generated IL 这是生成的IL
IL_0001: ldc.i4.3 //Load the constant 3 into evaluation stack
IL_0002: stloc.0 // stores the value in stack to x
I've got no clue why this also works:
我不知道为什么这也行得通:
short x = (short)1 + (short)2;
The compiler evaluates the rhs expression at compile time and can prove that the result is within bounds. 编译器在编译时评估rhs表达式,并可以证明结果在范围之内。
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