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I was trying to understand what a->b referred to and I was able to understand it refers to 

(*a).b

where a is a pointer. What does a->b->c mean in those terms - i.e. is it (*((*a).b)).c?

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  • thanks! so I could re-write it as i did Commented Apr 14, 2015 at 1:24
  • If you check an operator precedence chart, you'll notice the associativity (which isn't precedence, but is often on the same chart) of -> is left-to-right. This is enough to arrive at Matt's comment of a->b->c being parsed as (a->b)->c. Commented Apr 14, 2015 at 1:28
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    And it can't be a->(b->c) as (b->c) is not a member of a. Commented Apr 14, 2015 at 1:35

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Yes, a->b->c is equivalent to (*(*a).b).c which is equivalent to (*((*a).b)).c.

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