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** A: The error is quite clear here: the const& line will not compile. But the const& line will be required for the get() function. So, you should modify your code like this: template T get(const auto &el) { return static_cast(*el.key); } The most important change here is that the template parameter T is now deduced from the type T::key, since this is the required line for the get() function. After reading the comment from Raphael, I’d like to add a bit more explanation. To be able to call el.key, you need to make sure that you have a proper value for the key variable. And to have a value for this key variable, you need to make sure that you have some kind of an object to call the key function of. If a non-const object is used here, like you did with el, the standard will throw the error in its current form. And what’s the alternative you saw in your question? You had a pointer and you want to take the value of it. That’s ok, as long as you’re sure, that there is something in that pointer. const *el; *el.key; *(el.key); el->key; Of course, the last three lines are wrong because you want to take the value from the key function, but I want to show you what you could do in case, el is a pointer. In your original code, you had a problem, because you didn’t do any casting here. If you call a function with a reference, you cannot assign a reference to this function’s return value. The compiler will always reject this construction. I think that’s what your compiler is telling you. I personally like the compiler error messages, but they’re designed to ensure that you get your code right. And here, there is no “you”. You were simply not using the right function and the compiler warned you about that. /* * Copyright 2010-2015 Amazon.com, Inc. or its affiliates. All Rights Reserved. * * Licensed under the Apache License, Version 2.0 (the “License”). * You may not use this file except in compliance with the License. * A copy of the License is located at * c6a93da74d
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