Hello! I just have a quick question: when is it appropriate/standard practice to include brackets to separate boolean expressions and why?
Below are a few examples of code given in the python code challenge for control flow.
NOTE: I was unable to include the image in this post because new users are only able to include 1 link per post. Please see the replies below for a full description of the problem.
For instance, here is an example from the exercise, where there are brackets surrounding the second line of code in the function.
In my code, I did not use brackets, yet I still got the correct outputs - why might this be?
Also, once seeing the brackets in use, I switched to using them; however for the second line of code this time I used two brackets:
if (num >= lower) and (num <= upper):
The answer only uses one larger bracket. But both my version and the one below give the same output. Why might this be and which one is more appropriate in standard practice?
Brackets are only needed for grouping to override precedence. There is no best practice other than not using what is not necessary. Know your precedence (order or operation) and determine if overriding is necessary to get the result you intend.
Ah! Very interesting. Thank you for this, @mtf.
So what might be an instance of overriding precedence with brackets?
Say I had the following code:
if (num1 >= num2 or num1 < num3) not num1 < num2:
return num1
Would the or operation be interpreted first - because it is in brackets?
Everything inside the brackets will be evaluated first, yes.
We should give your example a bit more analysis, just to be sure we’re clear what order things are actually happening in.
if (num1 >= num2 or num1 < num3) not num1 < num2:
Clearly () is the top of the food chain so we’ll chop that one out, first.
(num1 >= num2 or num1 < num3)
Membership operators have greater precedence than or so the first comparison is evaluated first. If that yields True it will short-circuit the OR that follows. If it yields False the yield will be determined in the second comparison.
So let’s call the yield of this portion, A.
Now we have,
A not num1 < num2
Oops, the wheels just came off. Can you envision why?
Okay, this is a fantastic, detailed explanation – thank you!
I am just trying to figure out the question you pose at the end. As I think I am clear now if num1 >= num2 is in fact determined False it would pose a contradiction with the following not statement!
Is this what you intended to hint toward?
Not exactly. We’ll reserve A for a moment.
Consider,
not num1 < num2
By precedence, not num1 has a lower precedence than num1 < num2 so if we call the latter, B, we have,
A not B
The wheels are still off but now we have the simplest form given the evaluation process, thus far.
For tickles and giggles let A equal 0 and B equal 1.
0 not 1
What is lacking is an operator that can see both sides of the not. NOT is unary, not binary. Its operand is on the right, not the left. The first operand is just an orphaned value with no meaning.
A and not B
A or not B
See how that puts the wheels back on?
Ah, I am beginning to see more clearly now ( I think). So what you are suggesting is that not by itself can not uphold a viable comparison (because it is unary), it would require another operator (such as and or or)?
Does that mean that both and and or are considered binary? Also, if we had:
A and B
or
A or B
would either of these work?
Are there other unary operators I should be aware of?
Yes, that is correct.
Yes, they both have two operands.
In this regard, NOT is the only unary.
Thank you so much for all your help! You’ve enriched my understanding of these concepts. So much to learn!