StackOverflow2013

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copied!<p>It appears that as far as your first result goes, VC++ still follows the rule from C89/90, which said (§ 6.1.3.2):</p> <blockquote> <p>The type of an integer constant is the first of the corresponding list in which its value can be represented. Unsuffixed decimal: <code>int</code>, <code>long int</code>, <code>unsigned long int</code>; [...]</p> </blockquote> <p>So, since 4294967295 can be represented as an <code>unsigned long int</code>, that's what it's using.</p> <p>In C++98/03, this is still permitted, but no longer required -- you're using a value larger than can be represented in an <code>long int</code>, which gives undefined behavior (§ 2.13.1/2):</p> <blockquote> <p>If it is decimal and has no suffix, it has the first of these types in which its value can be represented: <code>int</code>, <code>long int</code>; <strong>if the value cannot be represented as a long int, the behavior is undefined.</strong></p> </blockquote> <p>[emphasis added]</p> <p>C++11 adds <code>long long int</code> to the list, so that's the type <code>4294967295</code> <em>should</em> become, but even in VC++ 2013 RC, it still follows the C89/90 standard in this respect and gives it type <code>unsigned long int</code>.</p> <p>Note that the the string produced by <code>typeid</code> is implementation defined, so it doesn't have to correspond directly to the proper name of the type. If you use overloading, we can see that <code>0x100000000</code> and <code>4294967296</code> have type <code>long long</code> though:</p> <pre><code>#include <iostream> void f(unsigned long){ std::cout << "unsigned long\n"; } void f(long long) { std::cout << "long long\n"; } void f(unsigned long long) { std::cout << "unsigned long long\n"; } void f(unsigned) { std::cout << "unsigned int\n"; } int main(){ f(4294967295); f(4294967296); f(0xffffffff); f(0x100000000); } </code></pre> <p>Result with VC++ 2008 and VC++ 2013 RC:</p> <pre><code>unsigned long long long unsigned int long long </code></pre> <p>I don't have all the intervening versions installed, but given that 2008 and 2013 match, I think it's fair to guess that versions in between them act the same way as well.</p>

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