Shell Programming and Scripting

i am new to shell scripting. this is similar to the code which i was writing for some other work. i want the variable 'x' to have the value which it will finally have at the end of the loop ( the number of directories ). but the value of 'x' only changes inside the loop and it remains '0' out-side of the loop.
(please note that i dont want to count the number of directories. but i am doing something simillar to this.)

Code:

#!/usr/bin/env bash

echo Begin

x=0
find $PWD -type d | while read filename
do
     echo $x
      x=`echo "$x + 1" | bc`
done

echo x=$x

echo End

result I got was this

Code:

Begin
0
1
2
x=0
End

But this is the result i really needed

Code:

Begin
0
1
2
x=2
End

but i know that this works for a normal loop which is like this

Code:

#!/usr/bin/env bash

echo Begin

x=0
while [ $x -lt 3 ]
do
     echo $x
      x=`echo "$x + 1" | bc`
done

echo x=$x

echo End

sorry if i am being so stupid.. please help... i am learning shell scripting by my self.... i am a bit of a C programmer... so is this because the value is passed inside instead of the variable refference or memory location... is there any other way to do this...
(i need the while loop to perform some tasks for each path that 'find' command gives.)
thank you...

oh thanks.... thank you very much....

This is the case for bash but not for ksh93. OP's example works in ksh93.