Is this undefined C behaviour?

Question

Our class was asked this question by the C programming prof:

You are given the code:

int x=1;
printf(\"%d\",++x,x+1);

What output w

Answer 1

Any time the behavior of a program is undefined, anything can happen — the classical phrase is that "demons may fly out of your nose" — although most implementations don't go that far.

The arguments of a function are conceptually evaluated in parallel (the technical term is that there is no sequence point between their evaluation). That means the expressions ++x and x+1 may be evaluated in this order, in the opposite order, or in some interleaved way. When you modify a variable and try to access its value in parallel, the behavior is undefined.

With many implementations, the arguments are evaluated in sequence (though not always from left to right). So you're unlikely to see anything but 2 in the real world.

However, a compiler could generate code like this:

1. Load x into register r1.
2. Calculate x+1 by adding 1 to r1.
3. Calculate ++x by adding 1 to r1. That's ok because x has been loaded into r1. Given how the compiler was designed, step 2 cannot have modified r1, because that could only happen if x was read as well as written between two sequence points. Which is forbidden by the C standard.
4. Store r1 into x.

And on this (hypothetical, but correct) compiler, the program would print 3.

(EDIT: passing an extra argument to printf is correct (§7.19.6.1-2 in N1256; thanks to Prasoon Saurav) for pointing this out. Also: added an example.)