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    <p>"what is the exact behaviour?"</p> <p>The problem is that <code>push_back</code> takes a reference parameter. You can use the <em>value</em> of a <code>static const int</code> member variable without providing a separate definition of the object, but you can't use a reference to the object itself (since it doesn't exist). The meaning of "using" the member itself is defined in the section of the standard on the One Definition Rule, 3.2/2.</p> <p>One fix is to provide a definition in exactly one translation unit:</p> <pre><code>const int check::init; </code></pre> <p>If you do this, you can also choose to move the <code>= 1</code> initialization from the declaration (inside the class) to the definition (outside the class).</p> <p>Another fix is to create a temporary from the member variable (this only uses the value, it doesn't care where the object is located and hence doesn't care whether it exists), then pass a reference to the temporary:</p> <pre><code>dummy.push_back(int(init)); </code></pre> <p>Of course there's a potential maintenance issue there, that if the types of <code>init</code> and <code>dummy</code> both change to, say, <code>long long</code>[*], and the value changes from <code>1</code> to something bigger than <code>INT_MAX</code>, then you're in trouble. For that reason you could use <code>+init</code>, since the unary + operator also creates a temporary for its result. Readers and future maintainers might be a bit puzzled by it, though.</p> <p>[*] Supposing your implementation has <code>long long</code>.</p>
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    1. POusing compile time constant throws error
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    1. USSteve Jessop
    1. USSteve Jessop
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    1. COOf course the easiest fix is to define `init` like `enum { init = 1; }. That can't be used with long long, but otherwise it does not need to be defined separately, never gets a storage and compiler always creates appropriate temporary automatically if it needs one.
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      1. USJan Hudec
    2. COAnother alternative that I have seen (not that I am recommending it) is using the static constant in an expression like `check::init + 0`. The addition does not require an lvalue and thus the member is not considered *used*.
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      1. USDavid Rodríguez - dribeas
    3. CO@Jan: good point. That assumes there isn't anywhere in the program that genuinely needs to use the fact that `check::init` has an address. Very likely, since otherwise there would be more errors in the program than just this one, but I failed to do it :-)
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      1. USSteve Jessop
 

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