Hat ^ operator vs Math.Pow()

Question

仔细阅读了^(hat)运算符和Math.Pow()函数的MSDN文档后,我发现没有明显的区别.有吗？

显然有一个区别是一个是一个函数而另一个被认为是一个运算符,例如这不起作用:

Public Const x As Double = 3
Public Const y As Double = Math.Pow(2, x) ' Fails because of const-ness

但这会:

Public Const x As Double = 3
Public Const y As Double = 2^x

但他们如何产生最终结果有所不同吗？是否Math.Pow()做更多的安全检查例子吗？或者只是另一种别名的某种别名？

Answer 1

找出答案的一种方法是检查IL.对于:

Dim x As Double = 3
Dim y As Double = Math.Pow(2, x)

IL是:

IL_0000:  nop         
IL_0001:  ldc.r8      00 00 00 00 00 00 08 40 
IL_000A:  stloc.0     // x
IL_000B:  ldc.r8      00 00 00 00 00 00 00 40 
IL_0014:  ldloc.0     // x
IL_0015:  call        System.Math.Pow
IL_001A:  stloc.1     // y

并为:

Dim x As Double = 3
Dim y As Double = 2 ^ x

IL 也是:

IL_0000:  nop         
IL_0001:  ldc.r8      00 00 00 00 00 00 08 40 
IL_000A:  stloc.0     // x
IL_000B:  ldc.r8      00 00 00 00 00 00 00 40 
IL_0014:  ldloc.0     // x
IL_0015:  call        System.Math.Pow
IL_001A:  stloc.1     // y

IE编译器已将其^转换为调用Math.Pow- 它们在运行时是相同的.