[RESOLVED] String Concatenation Question...

[Logophobic]

Hey there
Perhaps a visual aide will help?

Repeat(16, "abc")

dilettante's method, using the Mid() statement:

"................................................"
"abc............................................."
"abcabc.........................................."
"abcabcabc......................................."
"abcabcabcabc...................................."
"abcabcabcabcabc................................."
"abcabcabcabcabcabc.............................."
"abcabcabcabcabcabcabc..........................."
"abcabcabcabcabcabcabcabc........................"
"abcabcabcabcabcabcabcabcabc....................."
"abcabcabcabcabcabcabcabcabcabc.................."
"abcabcabcabcabcabcabcabcabcabcabc..............."
"abcabcabcabcabcabcabcabcabcabcabcabc............"
"abcabcabcabcabcabcabcabcabcabcabcabcabc........."
"abcabcabcabcabcabcabcabcabcabcabcabcabcabc......"
"abcabcabcabcabcabcabcabcabcabcabcabcabcabcabc..."
"abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"

anhn's method, using concatenation:

"abcabc"
"abcabcabcabc"
"abcabcabcabcabcabcabcabc"
"abcabcabcabcabcabcabcabcabcabcabcabcabcabcabcabc"

anhn uses Log to determine how many concatenations are needed and a recursive call to get the exact number of reps.

anhn's method clearly has fewer steps, but the slowness of concatenation makes the end result only marginally faster. (Or so it seems, I didn't compare them significantly)

A blend of the two should prove to be faster still:

Function Repeat(ByVal Number As Long, ByVal Text As String) As String
  Dim lngLength As Long
  Dim n As Long
  
  If Number < 1 Then Exit Function
  n = Len(Text)
  lngLength = Number * n
  Repeat = Space(lngLength)
  Mid(Repeat, 1, n) = Text
  Do While n < lngLength
    Mid(Repeat, n + 1, n) = Mid$(Repeat, 1, n)
    n = n + n
  Loop
End Function