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Comment by vhf | Hacker News Reader

It is the opposite, like in many other languages. e.g.

  int i = 0;
  printf("%i %i", i++, ++i); // prints "0 2"

Same goes in C, C++, Java, PHP, ...

[EDIT] Turns out this is a bad example, as "the order in which function arguments are evaluated is undefined" (cf. below) Correct is :

  int i = 0;
  printf("%i", i++); // prints 0
  printf("%i", ++i); // prints 2

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Actually, the order in which function arguments are evaluated is undefined. http://c-faq.com/expr/comma.html

You can use a comma in other expressions to introduce a sequence point: http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=expr#seqpoints

Thanks for your clever comment and for those emitting similar concerns.

You surely are right. I wanted to give a quick example, turns out it was a bad one. Next time I'll write :

  int i = 0;
  printf("%i", i++); // prints 0
  printf("%i", ++i); // prints 2

turns out it was a bad one

Given that you were making a point on an article about the complexity of C, I'd say it was an unintentionally excellent example.

If you really want to make your brain hurt, there was an article on HN a while ago about the following statement:

    i = i++;

Evaluating that expression takes a real dive into the guts of the C spec.

I believe you trigger an undefined behaviour there. You modify i twice in the same statement.

The standard says, evaluation order of function arguments is undefined.. but in general, printf evaluates right to left. This was a common puzzle as I remember.

Is that defined behaviour?

Is there any rule saying that args to a function have to be evaluated in a particular order - ie is ',' a function point?

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