Replace 3th letter in a string

[Afrow UK]

code:
Function Replace3rdLetter

Exch $R0

Exch

Exch $R1

Exch

Exch 2

Exch $R2

Exch 2

Push $R3

StrCpy $R3 $R0 $R2

IntOp $R2 $R2 + 1

StrCpy $R0 $R0 "" $R2

StrCpy $R0 $R3$R1$R0

Pop $R3

Pop $R2

Pop $R1

Exch $R0

FunctionEnd

!macro Replace3rdLetter RetVar ReplaceInStr ReplaceWithChar ReplaceAtIndex

 Push "${ReplaceAtIndex}"

 Push "${ReplaceWithChar}"

 Push "${ReplaceInStr}"

  Call Replace3rdLetter

 Pop "${RetVar}"

!macroend

!define Replace3rdLetter '!insertmacro Replace3rdLetter'

Usage:
${Replace3rdLetter} $R0 "123456" "5" "4"
$R0 = 123556

Made this in college without a compiler so it is untested.
Also, you said you want the 3th (3rd) character replaced, but your example of 123456 > 123556 replaces the 4th character. The last parameter of ${Replace3rdLetter} specifies the index to replace; replacing it with the value of the 3rd parameter.

-Stu