Bitwise shift right

Projects Programming
1

When I use digitalWrite(selectPin[2], 1 >> 2 & 1)
The result is: digitalWrite(selectPin[2], 0 ) Pin is set LOW. Remark: Shift occurs.
Serial.Print(1 >> 2 & 1 ) shows 0.

However
Calculating 2 & 1 bitwise gives a result 0. No Shift should uccur. Pin must be set to HIGH
What is the difference?
ToBr

2

Precedence my dear Tobr.

Serial.Print(1 >> 2 & 1 ) shows 0

Serial.Print(1 >> (2 & 1) ) shows ....

Serial.Print((1 >> 2) & 1 ) shows ....

please fill in the dots...

3

( 1 >> 2 & 1 ) shows 0 a)
( 1 >> ( 2 & 1) ) shows 1 b)
( (1 >> 2) & 1 ) shows 0 c)

I copied the a) code from an article, and it works, also code c) works
In my calculation I calculated first 2 & 1, it gives 0 as a result.

My conclusion: without ( ): first Bitwise shift, then Bitwise AND.
Indeed better to use ( ) to prevent miscalculations.

Where can I find an overview of the order in calculation?
I know the rule MVDWOA.
But what about the comparison operators, boolean operators, bitwise ..., compound.
What about spaces? Are they important?
Thanks for your help.
ToBr

4

ToBr:
Where can I find an overview of the order in calculation?

http://en.cppreference.com/w/cpp/language/operator_precedence

5

I know the rule MVDWOA.

From the Netherlands/Belgium?