The test situation was to find if the last character in the string ‘abcdefghijklmnopqrstuvwxyz’ is ‘z’ and iterated 100000000 times.

First, the most intuitive which most would use. TickCount of 49546.

Right("abcdefghijklmnopqrstuvwxyz", 1) = "z"

Function Right takes in a Variant by default, by succeeding it with a dollar sign it accepts Strings by default. TickCount of 22828, a significant saving.

Right$("abcdefghijklmnopqrstuvwxyz", 1) = "z"

Using the equals sign would be seem the norm, but what if the String Compare function was used. TickCount of 18047.

StrComp(Right$("abcdefghijklmnopqrstuvwxyz", 1), "z", vbBinaryCompare) = 0

What if Mid was used to extract the last character instead of Right. TickCount of 28391.

StrComp(Mid$("abcdefghijklmnopqrstuvwxyz", Len("abcdefghijklmnopqrstuvwxyz"), 1), "z") = 0

Now what if we use the In String function. String length binary is faster than standard String length. TickCount of 14516.

InStrRev("abcdefghijklmnopqrstuvwxyz", "z", -1, vbBinaryCompare) = LenB("abcdefghijklmnopqrstuvwxyz") / 2

Finally, with out String length binary. TickCount of 11312.

InStrRev("abcdefghijklmnopqrstuvwxyz", "z", -1, vbBinaryCompare) = Len("abcdefghijklmnopqrstuvwxyz")

So using In String Reverse is 77% faster.

UPDATE next day: Three that I totally forgot about

In String starting at last character, TickCount 11266.

InStr(Len("abcdefghijklmnopqrstuvwxyz"), "abcdefghijklmnopqrstuvwxyz", "z", vbBinaryCompare) = Len("abcdefghijklmnopqrstuvwxyz")

And In String Binary which is naturally fast. TickCount 6672.

InStrB(LenB("abcdefghijklmnopqrstuvwxyz") - 1, "abcdefghijklmnopqrstuvwxyz", "z", vbBinaryCompare) = LenB("abcdefghijklmnopqrstuvwxyz") - 1

So using In String Binary is 70% faster.

Lets say we want to perform the following concatenation:

Dim i As Long
Dim s As String
For i = 1 To 100000000
	s = "A" & i & "B"
Next i

This takes approximately 6000 ticks. The faster approach, yet more complex functionality would be as follows:

Dim sourceLength As Long
sourceLength = 1
Dim source As String
s = "A B"

Dim i As Long
Dim s As String
For i = 1 To 100000000
	source = CStr(i)
	If Len(source) > sourceLength Then
		sourceLength = Len(source)
		s = "A" & Space$(sourceLength) & "B"
	End If
	Mid$(s, 2, sourceLength) = source
Next i

Which takes approximately 3700 ticks, a saving of nearly 40%.