Is this undefined C behaviour?

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The output is likely to be 2 in every reasonable case. In reality, what you have is undefined behavior though.

Specifically, the standard says:

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.

There is a sequence point before evaluating the arguments to a function, and a sequence point after all the arguments have been evaluated (but the function not yet called). Between those two (i.e., while the arguments are being evaluated) there is not a sequence point (unless an argument is an expression includes one internally, such as using the && || or , operator).

That means the call to printf is reading the prior value both to determine the value being stored (i.e., the ++x) and to determine the value of the second argument (i.e., the x+1). This clearly violates the requirement quoted above, resulting in undefined behavior.

The fact that you’ve provided an extra argument for which no conversion specifier is given does not result in undefined behavior. If you supply fewer arguments that conversion specifiers, or if the (promoted) type of the argument disagrees with that of the conversion specifier you get undefined behavior — but passing an extra parameter does not.

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