title: Is the prime number 532660159 somehow "connected" to the STL?
body: <p>If you do a <code>std::vector&lt;double&gt; v = std::vector&lt;double&gt;();</code> your vector is going to have a capacity equal to zero, fine.</p>
  
  <p>Now I have an API having a function having a <code>std::vector&lt;double&gt;</code> reference parameter to which I pass a vector definied as above. And at debug, I see a capacity equal to ... <code>532660159</code>.</p>
  
  <p>At the beginning I thought it was an <code>INT_MAX</code> of some sort, until I check that it is a prime number (between 2^28 and 2^29, and it's not even the biggest prime number smaller than 2^29.)</p>
  
  <p>Just to be really sure (I am already sure of it in fact, but before harassing the API guys) : is this number connected somehow to the STL ?</p>
  
  <p><em>Some precisions.</em> The real setting is that I am in Python, using an API binding c++ code, in fact binding <code>std::vector&lt;double&gt;</code> and a c++ function <code>F</code> having as parameter a reference to such a vector. I see a normal capacity (zero) after initializing the vector binder in the python code, and I see the capacity being this prime number immediately after entering the c++ function binding <code>F</code>.</p>
  
  <p><em>Remark.</em> Following a comment of <a href="https://stackoverflow.com/users/2193020/hnefatl">hnefatl</a> I did a little track down and :
  <code>d:\Program Files (x86)\Microsoft Visual Studio\2017\Community\VC\Tools\MSVC\14.12.25827\include\xhash</code> has <code>#define _HASH_SEED    (size_t)0xdeadbeef</code> and <code>size_t hash_value(const _Kty&amp; _Keyval)</code> uses it, and it appears in a comment in ... <code>c:\PYTHON\PYTHON_OFFICIAL\python-3.6.3\include\abstract.h</code> regarding the <code>long PyObject_Hash(PyObject *o)</code> which is implemented elsewhere, and used indeed I guess ... and I indeed rely on <code>python-3.6.3</code> ...</p>
  
  <p>Of course, I would like to give a minimal reproducing example, but it's not easy --> I will give it this week-end.</p>