A tuple of references as a function parameter

Question

I do not understand well enough why the second call of func in the code below does not work:

#include <string> #include <iostream> #include <tuple> using Tuple = std::tuple<const int&, const std::string&>; void func(const Tuple& t) { std::cout << std::get<1u>(t) << std::endl; } int main() { const int n = 3; const std::string text = "xyz"; func(Tuple(n, text)); //Does not work: //func(Tuple(5, "abc")); return 0; }

When is the temporary string "abc" destroyed?

EDIT1

So, finally, according to 康桓瑋 this works:

 func(Tuple(5, std::string("abc")));

but this does not:

 Tuple t(5, std::string("abc")); func(t);

right? If yes what is the difference?

I'm not sure but I believe its because the lifetime of the temporary is extended to the lifetime of the reference it is bound to (the one in Tuple's constructor), so by the time you call func, the temporary string has been destroyed already and you're basically in UB land by that point. this question may help — Borgleader
– Borgleader, Commented Jul 13, 2022 at 3:48

康桓瑋 · Accepted Answer · 2022-07-13 04:10:51Z

Your second case is exactly what the standard wants to detect and forbid.

Since "abc" is not a std::string type, quote from P2255:

std::tuple<const std::string&> t("meow"); 
This construction always creates a dangling reference, because the std::string temporary is created inside the selected constructor of tuple (template<class... UTypes> tuple(UTypes&&...)), and not outside it. Thus, unlike string_view’s implicit conversion from rvalue strings, under no circumstances can this construction be correct.

And Tuple(5, "abc") is no longer well-formed in C++23 thanks to P2255.

To prevent the destruction of temporary strings, you can explicitly specify the std::string type such as func(Tuple(5, std::string("abc"))).

But that example is a named object that has a lifetime that extends past the statement. The OP's example is an unnamed Tuple object which has the same lifetime as the unnamed std::string object, so there will be no dangling.
@ChrisDodd This tuple is dangling as soon as it is constructed. Because we leave the constructor.
I did not understand the difference between std::string("abc") and "abc". My understanding was that "abc" converted std::string("abc"), right?
@Alexey Starinsky For tuple<const string&>(string("abc")), const string& is bound to the temporary string("abc") you pass in; But for "abc", const string& is bound to a temporary string variable created inside the tuple constructor, which is destroyed at the end of the constructor.

Adrian Mole · Accepted Answer · 2022-07-13 04:06:36Z

As hinted at in the comments, the problem is not with the parameter to the call to func, per se, but with the second parameter to the Tuple constructor.

You are attempting to assign a reference to a string constructed as a temporary object (from the given const char* literal) to the second member of the tuple; when the constructor has completed/returned, that reference will invalid.

Some compilers may accept the code (MSVC does) and it may seem to work. However, clang-cl (in Visual Studio 2022) gives the following error when uncommenting the 'offending' line in your code:

error : reference member '_Val' binds to a temporary object whose lifetime would be shorter than the lifetime of the constructed object

However, note the the exact same error is shown if we replace your one-line call to func with the following two lines:

 Tuple tup = Tuple(5, "abc"); // Error on this line func(tup);

Seems like clang's error message is patently wrong -- the lifetime of the temporary object bound is the same as the lifetime of the Tuple it is bound in. If you explicitly bind the Tuple to a name (which the OP does not do), only then is the lifetime of the Tuple extended and dangling problem occurs.
@ChrisDodd Not 100% sure, but I think the lifetime of the temporary string ends when the tuple's constructor ends, irrespective of whether that constructs a named/unnamed object.

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