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    <p>Look into the ISO/IEC 9899:1999, chapter 6.10.3.1. It states that:</p> <blockquote> <p><strong>After</strong> the arguments for the invocation of a function-like macro have been <em>identified</em>, argument substitution takes place. A parameter in the replacement list, unless preceded by a # or ## preprocessing token or followed by a ## preprocessing token (see below), is replaced by the corresponding argument <strong>after</strong> all macros contained therein have been expanded. <strong>Before</strong> being substituted, <em>each argument’s preprocessing tokens are completely macro replaced</em> as if they formed the rest of the preprocessing file; no other preprocessing tokens are available.</p> </blockquote> <p>So in <em>va</em> the first argument <em>c</em> has one preprocessing token being <code>__VA_ARGS__</code>, which, according to this paragraph, must be <em>macro replaced</em> before being substituted to <em>c</em> (which still does not give the answer as to which compiler is right)</p> <p>But later:</p> <blockquote> <p>An identifier <code>__VA_ARGS__</code> that occurs in the replacement list shall be treated as if it were a <strong>parameter</strong>, and the variable arguments shall form the preprocessing tokens used to replace it.</p> </blockquote> <p>According to the first fragment, first the arguments to <em>var</em> are identified. They are: <code>2</code>, <code>3</code>, <code>printf</code>, <code>“%d %d %d\n”</code> and <code>1</code>. Now argument substitution takes place. This means that parameters from <em>var</em>'s replacement list are taken and replaced. However, the second fragment states, that identifier <code>__VA_ARGS__</code> is to be treated as a parameter, so it must be substituted to <code>printf, “%d %d %d\n”, 1</code>. Now there are no macros inside parameters, so no further substitution takes place, resulting in the expansion of <code>var(2, 3, printf, “%d %d %d\n”, 1);</code> into <code>va(printf, “%d %d %d\n”, 1, 2, 3);</code>. Because va is a macro, it is also expanded, giving the result of <code>printf(“%d %d %d\n”, 1, 2, 3);</code>. Now if you take VS 2008's reasoning, how can it <em>identify</em> the arguments to <em>va</em>, if one of them is <code>__VA_ARGS__</code> from <em>var</em>, and can contain many arguments? Well, it treats <code>__VA_ARGS__</code> as an argument to <em>va</em>, which in my opition is wrong, as, according to fragment one, argument substitution takes place only <em>after</em> the arguments for invocation have been identified.</p> <p>So it seems to me, that in <em>var</em> the preprocessor has to first identify the arguments to the invocation of the macro <em>va</em>, and then start expanding <em>va</em>. And this means that probably gcc is right here.</p> <p>It is also shown in <a href="https://stackoverflow.com/questions/1489932/c-preprocessor-and-concatenation">C preprocessor and concatenation</a> how processing tokens are macro-replaced in order, until there are no identifiers which can be further expanded as macros, or the preprocessor spots a recursion and terminates expansion.</p>
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    1. POthe problem about different treatment to __VA_ARGS__ when using VS 2008 and GCC
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    1. USMichał Trybus
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    1. COThank you a lot. Your answer is clear and concise. It seems to me too that gcc's answer makes more sense.
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    2. COWell, the fact that gcc's output was more sensible for a normal human being was what I started with, but I hope the answer was not biased. In my opinion the standard should be more precise on the exact order of expanding macros, especially when `__VA_ARGS__` come into play.
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