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POunion for uint32_t and uint8_t[4] undefined behavior?
 

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  1. POunion for uint32_t and uint8_t[4] undefined behavior?
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    <p>In the comments of <a href="https://stackoverflow.com/a/8568531/1025391">this answer</a> it is said that it would be undefined behavior to split up an integer into their bytes using a union like follows. The code given at that place is similar though not identical to this, please give a note if have I changed undefined-behavior-relevant aspects of the code.</p> <pre><code>union addr { uint8_t addr8[4]; uint32_t addr32; }; </code></pre> <p>Up to now I thought this would be a fine approach to do things like <code>addr = {127, 0, 0, 1};</code> and get the corresponding <code>uint32_t</code> in return. (I acknowledge that this may yield different results depending on the endianness of my system. The question however remains.)</p> <p>Is this undefined behavior? If so, why? (I don't know what means <em>What's UB in C++ is to access inactive union members.</em>)</p> <hr> <p><strong>C99</strong></p> <ul> <li>C99 is apparantly pretty close to C++03 in this point.</li> </ul> <p><strong>C++03</strong></p> <ul> <li><em>In a union, at most one of the data members can be active at any time, that is, the value of at most one of the data members can be stored in a union at any time.</em> C++03, Section 9.5 (1), page 162</li> </ul> <p>However</p> <ul> <li><em>If a POD-union contains several POD-structs that share a common initial sequence [...] it is permitted to inspect the common initial sequence of any of POD-struct members</em> ibid.</li> <li><em>Two POD-struct [...] types are layout-compatible if they have the same number of nonstatic data members, and corresponding nonstatic data members (in order) have layout-compatible types</em> C++03, Section 9.2 (14), page 157</li> <li><em>If two types T1 and T2 are the same type, then T1 and T2 are layout-compatible types.</em> C++03, Section 3.9 (11), page 53</li> </ul> <p>Conclusion</p> <ul> <li>as <code>uint8_t[4]</code> and <code>uint32_t</code> are not the same type (I guess, a <a href="https://stackoverflow.com/a/99010/1025391">strict aliasing thing</a>) (plus both not being POD-structs/union) the above is indeed UB?</li> </ul> <p><strong>C++11</strong></p> <ul> <li><em>Note that aggregate type does not include union type because an object with union type can only contain one member at a time.</em> C++11, Footnote 46, page 42</li> </ul>
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    1. COFWIW, I have litb on record [saying this is not UB](http://chat.stackoverflow.com/transcript/10?m=3138021#3138021).
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    2. COTechnically, it is UB; if you read out of union member other than the one that was last written to, you technically get UB. However, unless compiler writers are deliberately trying to make life impossible, it will work OK, with a major caveat about endian-ness in the context of 127.0.0.1 (which looks suspiciously like a big-endian localhost loopback address in IPv4).
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      1. USJonathan Leffler
    3. CO@JonathanLeffler Acknowledged. Please assume these variables as being anonymized!
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