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copied!<p>As far as C++ is concerned:</p> <p>C++11, [lex.icon] ¶2</p> <blockquote> <p>The type of an integer literal is the first of the corresponding list in Table 6 in which its value can be represented.</p> </blockquote> <p>And Table 6, for literals without suffixes and decimal constants, gives:</p> <pre><code>int long int long long int </code></pre> <p>(interestingly, for hexadecimal or octal constants also <code>unsigned</code> types are allowed - but each one come <em>after</em> the corresponding signed one in the list)</p> <p>So, it's clear that in that case the constant has been interpreted as a <code>long int</code> (or <code>long long int</code> if <code>long int</code> was too 32 bit).</p> <p>Notice that "too big literals" should result in a compilation error:</p> <blockquote> <p>A program is ill-formed if one of its translation units contains an integer literal that cannot be represented by any of the allowed types.</p> </blockquote> <p>(ibidem, ¶3)</p> <p>which is promptly seen <a href="http://ideone.com/9hujM">in this sample</a>, that reminds us that ideone.com uses 32 bit compilers.</p> <hr /> <p>I saw now that the question was about C... well, it's more or less the same:</p> <p>C99, §6.4.4.1</p> <blockquote> <p>The type of an integer constant is the first of the corresponding list in which its value can be represented.</p> </blockquote> <p>list that is the same as in the C++ standard.</p> <hr /> <p>Addendum: both C99 and C++11 allow also the literals to be of "extended integer types" (i.e. other implementation-specific integer types) if everything else fails. (C++11, [lex.icon] ¶3; C99, §6.4.4.1 ¶5 after the table)</p>

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