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    <p>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 <em>that</em> far.</p> <p>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 <code>++x</code> and <code>x+1</code> may be evaluated in this order, in the opposite order, <em>or in some interleaved way</em>. When you modify a variable and try to access its value in parallel, the behavior is undefined.</p> <p>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.</p> <p>However, a compiler could generate code like this:</p> <ol> <li>Load x into register <code>r1</code>. </li> <li>Calculate <code>x+1</code> by adding 1 to <code>r1</code>.</li> <li>Calculate <code>++x</code> by adding 1 to <code>r1</code>. That's ok because <code>x</code> has been loaded into <code>r1</code>. Given how the compiler was designed, step 2 cannot have modified <code>r1</code>, because that could only happen if <code>x</code> was read as well as written between two sequence points. Which is forbidden by the C standard.</li> <li>Store <code>r1</code> into <code>x</code>.</li> </ol> <p>And on this (hypothetical, but correct) compiler, the program would print 3.</p> <p>(<strong>EDIT:</strong> passing an extra argument to <code>printf</code> is correct (§7.19.6.1-2 in <a href="http://www.open-std.org/jtc1/sc22/WG14/www/docs/n1256.pdf" rel="nofollow noreferrer">N1256</a>; thanks to <a href="https://stackoverflow.com/users/165520/prasoon-saurav">Prasoon Saurav</a>) for pointing this out. Also: added an example.)</p>
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    1. POIs this undefined C behaviour?
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    1. COFirst, printf with a format of "%d" expects one integer parameter, but you've passed two. The behavior is undefined . No it is not. The second parameter is merely evaluated but the behavior is not undefined because of this reason. Note : printf("%d %d",x++);is Undefined because number of format specifiers is greater than the number of arguments.
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    2. CO@Prasoon: thanks, you're right. I've corrected my answer.
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    3. COWhat would make it a correct (though theoretical, and so unexistant) compiler? Such a compiler would fail for printf("%d %d", x+1, x+1) too: 1) load x into r1; 2) add 1 to r1 (x+1) 3) add 1 to r1 (x+1)... ops, it's a bugged compiler... once r1 is an "alias" for x into memory, the result of eval of x+1 put in r1 again will be like ++x, that is wrong; or r1 is temporary (meaning that when compiler has to eval ++x it loads x from memory again), or the result of x+1 must not be put in r1 again
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