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I'm reading Accelerated C++ and in Chapter 4 they bring up the concept of lvalues. There's an example of something that shouldn't work, but after trying it myself I found that it does indeed work.

Specifically they state that, given these functions:

// return an empty vector
vector<double> emptyvec()
  vector<double> v;
  return v;

// read things from an input stream into a vector<double>
// (I'm leaving out the function body here because it's irrelevant)
istream &read_things(istream& in, vector<double>& hw);

This should not be allowed:

read_stuff( cin, emptyvec() );

Because emptyvec() is an expression and returns a temporary object (a non-lvalue as they called it in the book). However, this not only compiles but actually runs (Windows 7/VisualStudio 2010).

So, what's going on? Was this just a bad example on the authors' part, or is there something else happening that I don't understand.


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Can you double check the return type of emptyvec()? As given (vector<double> vector) it doesn't make much sense. Was this vector<double> & in the book (in which case the code is indeed an example of the problem the book mentions)? – jimwise Jan 24 '13 at 15:47
@jimwise That was a typo when I entered it here. The original program was actually written correctly :) – Alex Jan 24 '13 at 16:07
Which way? Returning vector<double> or vector<double> &? – jimwise Jan 24 '13 at 16:09
@jimwise (The original typo was vector<double> vector emptyvec() which is not valid syntax) I don't have the book here at the moment, but when I tried changing it from vector<double> to vector<double>& the compiler threw an error: warning C4172: returning address of local variable or temporary So that seems to be more in line with what the author was saying. – Alex Jan 24 '13 at 16:12
up vote 1 down vote accepted

The concern the book mentions would come into play if the function was declared as:

vector<double> &emptyvec()


vector<double> *emptyvec()

as it would then be returning a temporary value by reference, and by the time you used the reference, the value it referred to would be gone. Your compiler should issue a warning or an error for this case, and you might (but might not) see a variety of possible failures at run time (sadly, the type of error you would see could vary widely).

On the other hand, if the function is indeed declared as

vector<double> emptyvec()

the vector is being returned by value (i.e. a copy of the actual vector is being made into wherever you store the return value), and the program should work correctly.

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Ah yes, I just realized this after reading your earlier comment. So this was probably all down to me misreading the example in the book. The compiler throws an error now and I even get a runtime exception (access violation reading a memory location) and everything makes total sense now. Thanks for the help! – Alex Jan 24 '13 at 16:17
Glad I could help! :-) – jimwise Jan 24 '13 at 16:31
@Alex: Was it a typo in the book? Maybe it's worth contacting the author. – FrustratedWithFormsDesigner Jan 24 '13 at 16:56

It's a compiler extension on the part of Visual Studio. The code should indeed not compile, and it won't on the other major compilers like GCC and Clang, as a temporary cannot bind to a mutable reference.

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The vector is passed by value, which is perfectly fine. – tdammers Jan 24 '13 at 17:35
Why the downvotes? He's right. This shouldn't compile according to the C++ standard. There's no problem in emptyvec(), it's in read_things() binding a temporary (the return value of emptyvec()) to a non-const reference parameter. – Bwmat Jan 24 '13 at 22:25
It's certainly not perfectly fine when it tries to give a value vector to a mutable reference parameter. – DeadMG Jan 27 '13 at 22:44

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