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Ryan_001

This error has me perplexed.

32 posts in this topic

The thing is, you're not supposed to be able to take an l-value reference to an r-value. Yes, there are some ways that an r-value reference may be modified (through calling non-const member functions, or if you're using move semantics), but they're not supposed to be modified through any other means.
 
But it's possible Visual Studio allows it. I just know that in standard C++, it's illegal.

 
But you bind rvalues to lvalue references all the time.  For example:

string s("hello");
cout << s.find(string("l") + string("o")) << endl;

 

This may be a contrived example, but temporary values are constantly bound to lvalue references.

Edited by Ryan_001
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I found it, a silly oversight on my part (as usual).

 

I compiled the library with the /vmg compiler option, and I didn't do the same with the program, I had forgotten it.

 

Thank-you very much for your time and insight.  I know I checked my compiler and linker options repeatedly, I'm not sure how I had missed that.

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This may be a contrived example, but temporary values are constantly bound to lvalue references.

The rule is that temporaries can't (normally) be bound to non-const lvalue references. Binding a temporary to a const lvalue reference is perfectly fine. Binding a temporary to a non-const lvalue reference can be done, but requires trickery or a compiler extension like MSVC has.
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The thing is, you're not supposed to be able to take an l-value reference to an r-value. Yes, there are some ways that an r-value reference may be modified (through calling non-const member functions, or if you're using move semantics), but they're not supposed to be modified through any other means.
 
But it's possible Visual Studio allows it. I just know that in standard C++, it's illegal.

 
But you bind rvalues to lvalue references all the time.  For example:

string s("hello");
cout << s.find(string("l") + string("o")) << endl;

 

This may be a contrived example, but temporary values are constantly bound to lvalue references.

Sorry, poor wording on my part. I meant non-const lvalue reference (i.e. the kind of reference we've been talking about in the code).

Edited by Cornstalks
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Well even if it wasn't the problem its good to know Corn/SiCrane.  Its something I will not do, despite not making much sense IMHO.

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Well even if it wasn't the problem its good to know Corn/SiCrane.  Its something I will not do, despite not making much sense IMHO.

Yeah, it's kind of weird. I think it kind of makes sense (helping you not assign to temporaries that are about to get nuked can help avoid some bugs), but at the same time, it seems like an unnecessary restriction. I think that's the main reason this question has so many upvotes, but answers don't (that is, the question is a good, interesting question, but the answers (provided by the standard) aren't very satisfactory).

 

I think the best answer to this question is actually in one of the comments in that question (from sbi):

The reason Stroustrup gives (in [Design and Evolution of C++]) for disallowing the binding of rvalues to non-const references is that, if Alexey's g() would modify the object (which you'd expect from a function taking a non-const reference), it would modify an object that's going to die, so nobody could get at the modified value anyway. He says that this, most likely, is an error.

 

It seems kinda weird that C++ is trying to hold your hand here (seeing that the rest of the standard doesn't care if you shoot your leg off if you want), but oh well.

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The problem isn't throwing the object. The problem is modifying the temporary object, which (by standard C++) is undefined behavior. The problem is binding a non-constant reference to a temporary object and then modifying the object (not throwing the object). In the ThrowException() function, e is a reference to a temporary, and you are modifying it, which is undefined behavior. e should be a const reference, and you should not modify it (feel free to throw it).

Is not the reference valid for the lifetime of the temporary object?  I was certain pg.245 of N3337 implies that the temporary object the reference was bound to will exist for the duration of the function, in particular 12.2.5.  Am I reading this wrong?
Nope, it's not. A const-reference extends the lifetime of a temporary object, a non-const-reference does not.
See GotW #88:
http://herbsutter.com/2008/01/01/gotw-88-a-candidate-for-the-most-important-const/ Edited by iMalc
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