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    <p>It isn't that an address can't be taken (the compiler could always order it shoved on the stack, which it does with ref-to-const), it's a question of programmers intent. With an interface that takes a A&amp;, it is saying "I will modify what is in this parameter so you can read after the function call". If you pass it a temporary, then the thing it "modified" doesn't exist after the function. This is (probably) a programming error, so it is disallowed. For instance, consider:</p> <pre><code>void plus_one(int &amp; x) { ++x; } int main() { int x = 2; float f = 10.0; plus_one(x); plus_one(f); cout &lt;&lt; x &lt;&lt; endl &lt;&lt; f &lt;&lt; endl; } </code></pre> <p>This doesn't compile, but if temporaries could bind to a ref-to-non-const, it would compile but have surprising results. In plus_one(f), f would be implicitly converted to an temporary int, plus_one would take the temp and increment it, leaving the underlying float f untouched. When plus_one returned, it would have had no effect. This is almost certainly not what the programmer intended.</p> <hr> <p>The rule does occasionally mess up. A common example (described <a href="http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2004/n1690.html" rel="noreferrer">here</a>), is trying to open a file, print something, and close it. You'd want to be able to do:</p> <pre><code>ofstream("bar.t") &lt;&lt; "flah"; </code></pre> <p>But you can't because operator&lt;&lt; takes a ref-to-non-const. Your options are break it into two lines, or call a method returning a ref-to-non-const:</p> <pre><code>ofstream("bar.t").flush() &lt;&lt; "flah"; </code></pre>
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    1. COI agree that this sort of thing is usually a programmer error. In case you're curious, the class I'm trying to pass as a reference is a sort of smart pointer, which has an underlying heap-allocated pointer member. I actually do want the receiving function to do things with the underlying pointer. I just want to be sure the temporary smart pointer is not destroyed (decrementing the reference count) until after the receiving function returns.
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    2. COIf that's your intent, either declare the smart pointer on the stack explicitly and pass it in or pass it in by value. If you pass it in by value you are guaranteed that it will not be destroyed until the function exits.
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    3. CO@Ben - added some to my post about it. Can't say if it is a good idea for you without seeing your code, but there are some use cases (and it won't be destroyed until it returns)
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