Question

Very recently I’ve started to learn Ruby and I was experimenting with how Ruby calls methods on individual objects. However, the following code piece stuck me hard as I am not realising how it is actually working to

a = 4
b = -3
c = 2

puts a*b-c                      # operator precedence preserved
puts a . * b . - c              # operator precedence not preserved
puts a.send(:*, b).send(:-, c)  # operator precedence preserved
puts a-b*c                      # operator precedence preserved
puts a . - b . * c              # operator precedence preserved
puts a.send(:-, b).send(:*, c)  # operator precedence not preserved

Outputs:

-14
-20    
-14
10
10
14

Can anyone able to explain how the operator precedence working here? I assumed all the three syntax in each part should reflect the same meaning. I apologise first if this question has been asked or explained already.

Answer 1

Operator precedence only applies when using operators. All of these examples:

puts a . * b . - c              # operator precedence not preserved
puts a.send(:*, b).send(:-, c)  # operator precedence preserved
puts a . - b . * c              # operator precedence preserved
puts a.send(:-, b).send(:*, c)  # operator precedence not preserved

are direct method calls, and happen to be in either the wrong order or the right order as compared to their corresponding operators.

Maybe parentheses make it more clear?

puts a.*(b.-(c))                # .- called first, .* with the return value of .-
puts a.send(:*, b).send(:-, c)  # .* called first, .- with the return value of .*
puts a.-(b.*(c))                # .* called first, .- with the return value of .*
puts a.send(:-, b).send(:*, c)  # .- called first, .* with the return value of .-