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Hat ^ operator vs Math.Pow()

Having perused the MSDN documentation for both the ^ (hat) operator and the Math.Pow() function, I see no explicit difference. Is there one?

There obviously is the difference that one is a function while the other is considered an operator, e.g. this will not work:

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

But this will:

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

But is there a difference in how they produce the end result? Does Math.Pow() do more safety checking for example? Or is one just some kind of alias for the other?

like image 329
Toby Avatar asked Jul 03 '26 10:07

Toby


1 Answers

One way to find out is to inspect the IL. For:

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

The IL is:

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

And for:

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

The IL also is:

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 the compiler has turned the ^ into a call to Math.Pow - they're identical at runtime.

like image 123
James Thorpe Avatar answered Jul 05 '26 00:07

James Thorpe



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