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Why does using a temporary object in the range-based for initializer result in a crash?

Tags:

c++

c++11

The range initialization line of a for(:) loop does not extend lifetime of anything but the final temporary (if any). Any other temporaries are discarded prior to the for(:) loop executing.

Now, do not despair; there is an easy fix to this problem. But first a walk through of what is going wrong.

The code for(auto x:exp){ /* code */ } expands to, basically:

{
  auto&& __range=exp;
  auto __it=std::begin(__range);
  auto __end=std::end(__range);
  for(; __it!=__end;++__it){
    auto x=*__it;
    /* code */
  }
}

(With a modest lies on the __it and __end lines, and all variables starting with __ have no visible name. Also I am showing C++17 version, because I believe in a better world, and the differences do not matter here.)

Your exp creates a temporary object, then returns a reference to within it. The temporary dies after that line, so you have a dangling reference in the rest of the code.

Fixing it is relatively easy. To fix it:

std::string const& func() const& // notice &
{
    return m.find("key")->second;
}
std::string func() && // notice &&
{
    return std::move(m.find("key")->second);
}

do rvalue overloads and return moved-into values by value when consuming temporaries instead of returning references into them.

Then the

auto&& __range=exp;

line does reference lifetime extension on the by-value returned string, and no more dangling references.

As a general rule, never return a range by reference to a parameter that could be an rvalue.


Appendix: Wait, && and const& after methods? rvalue references to *this?

C++11 added rvalue references. But the this or self parameter to functions is special. To select which overload of a method based on the rvalue/lvalue-ness of the object being invoked, you can use & or && after the end of the method.

This works much like the type of a parameter to a function. && after the method states that the method should be called only on non-const rvalues; const& means it should be called for constant lvalues. Things that don't exactly match follow the usual precidence rules.

When you have a method that returns a reference into an object, make sure you catch temporaries with a && overload and either don't return a reference in those cases (return a value), or =delete the method.


S().func()

This constructs a temporary object, and invokes a method that returns a reference to a std::string that's owned (indirectly) by the temporary object (the std::string is in the container that's a part of the temporary object).

After obtaining the reference, the temporary object gets destroyed. This also destroys the std::string that was owned (indirectly) by the temporary object.

After that point, any further usage of the referenced object becomes undefined behavior. Such as iterating over its contents.

This is a very common pitfall, when it comes to using range iteration. Yours truly is also guilty of getting tripped over this.