Does operator precedence explain the difference in these expressions involving multiplication of a numeric with a negated logical?

Question

I have three expressions, each involving multiplication with a logical or its negation. These logicals and their negation represent indicator variables, so that the expressions are conditionally evaluated:

-2*3*!T + 5*7*T
5*7*T + -2*3*!T 
(-2*3*!T) + 5*7*T

I expect the above to produce the same result. However:

> -2*3*!T + 5*7*T
[1] 0          # unexpected!
> 5*7*T + -2*3*!T 
[1] 35
> (-2*3*!T) + 5*7*T
[1] 35

I am sure this has something to do with operator precedence and type coercion, but I can't work out how it makes sense to even evaluate !T after the *.

Answer 1

You're exactly right that this is about operator precedence. As ?base::Syntax (which you link above) states, ! has lower precedence than all of the arithmetic operators, so the first expression is equivalent to

(-2*3)*!(T + 5*7*T)  

(because the expression containing ! has to be evaluated before the final multiplication can be done) or

-6*!(36)  # T coerced to 1 in numeric operations

or

-6*FALSE  # non-zero numbers coerced to TRUE in logical operations

or

-6*0      # FALSE coerced to 0 in numeric operations