Operator precedence for And/&& in Ruby [duplicate]

Question

I have a question regarding the and/&&/= keywords in Ruby.

The ruby docs say that the precedence for the mentioned keywords is: (1)&&, (2)=, (3)and.

I have this snippet of code I wrote:

def f(n) 
 n
end

if a = f(2) and  b = f(4) then  
    puts "1) #{a} #{b}" 
 end

if a = f(2) &&  b = f(4) then   
    puts "2) #{a} #{b}"     
end

The output is:

1) 2 4 [Expected]

2) 4 4 [Why?]

For some reason using the && causes both a and b to evaluate to 4?

Answer 1

I don't quite understand the question you are asking. I mean, you have already given the answer yourself, before even asking the question: && binds tighter than = while and binds less tightly than =.

So, in the first case, the expression is evaluated as follows:

( a=f(2) )  and  ( b=f(4) )
( a=  2  )  and  ( b=f(4) )
      2     and  ( b=f(4) ) # a=2
      2     and  ( b=  4  ) # a=2
      2     and        4    # a=2; b=4
                       4    # a=2; b=4

In the second case, the evaluation is as follows:

a   =   (  f(2) && ( b=f(4) )  )
a   =   (    2  && ( b=f(4) )  )
a   =   (    2  && ( b=  4  )  )
a   =   (    2  &&       4     ) # b=4
a   =                    4       # b=4
                         4       # b=4; a=4

Answer 2

The reason is simple: precedence. As you say, the order is:

1. &&
2. =
3. and

Since && has precedence over =, the statement is evaluated like this:

if a = (f(2) && (b = f(4))) then 

Which results in:

if a = (2 && 4) then

When x and y are integers, x && y returns y. Thus 2 && 4 results in a = 4.

For comparison's sake, the first one is evaluated like this:

if (a = f(2)) and  (b = f(4)) then