Using this operator is almost the same as specifying result = result + expression,except that result is only evaluated once.

所以他们不是相同的，这就是为什么x = 1会更有效率。

更新：我只是注意到我的MSDN库链接到JScript页面而不是VB page，它不包含相同的报价。

因此，进一步的研究和测试，这个答案不适用于VB.NET。我错了。下面是一个示例控制台应用程序：

Module Module1

Sub Main()
    Dim x = 0
    Console.WriteLine(PlusEqual1(x))
    Console.WriteLine(Add1(x))
    Console.WriteLine(PlusEqual2(x))
    Console.WriteLine(Add2(x))
    Console.ReadLine()
End Sub

Public Function PlusEqual1(ByVal x As Integer) As Integer
    x += 1
    Return x
End Function

Public Function Add1(ByVal x As Integer) As Integer
    x = x + 1
    Return x
End Function

Public Function PlusEqual2(ByVal x As Integer) As Integer
    x += 2
    Return x
End Function

Public Function Add2(ByVal x As Integer) As Integer
    x = x + 2
    Return x
End Function

End Module

IL对于PlusEqual1和Add1确实是相同的：

.method public static int32 Add1(int32 x) cil managed
{
.maxstack 2
.locals init (
    [0] int32 Add1)
L_0000: nop 
L_0001: ldarg.0 
L_0002: ldc.i4.1 
L_0003: add.ovf 
L_0004: starg.s x
L_0006: ldarg.0 
L_0007: stloc.0 
L_0008: br.s L_000a
L_000a: ldloc.0 
L_000b: ret 
}

PlusEqual2和Add2的IL几乎相同：

.method public static int32 Add2(int32 x) cil managed
{ 
.maxstack 2
.locals init (
    [0] int32 Add2)
L_0000: nop 
L_0001: ldarg.0 
L_0002: ldc.i4.2 
L_0003: add.ovf 
L_0004: starg.s x
L_0006: ldarg.0 
L_0007: stloc.0 
L_0008: br.s L_000a
L_000a: ldloc.0 
L_000b: ret 
}