Why a const reference doesn't extend the life of temporary object passed via function?

General Tech Bugs & Fixes 2 years ago

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Posted on 16 Aug 2022, this text provides information on Bugs & Fixes related to General Tech. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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manpreet Tuteehub forum best answer Best Answer 2 years ago

 

In the following simple example, why ref2 can't be bound to the result of min(x,y+1)?

#include 
template< typename T > const T& min(const T& a, const T& b){ return a < b ? a : b ; }

int main(){
      int x = 10, y = 2;
      const int& ref = min(x,y); //OK
      const int& ref2 = min(x,y+1); //NOT OK, WHY ?
      return ref2; // Compiles to return 0
}

live example - produces

main:
  xor eax, eax
  ret
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manpreet 2 years ago

 

It's by design. In a nutshell, only the named reference to which the temporary is bound directly will extend its lifetime.

[class.temporary]

5 There are three contexts in which temporaries are destroyed at a different point than the end of the full-expression. [...]

6 The third context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference except:

  • A temporary object bound to a reference parameter in a function call persists until the completion of the full-expression containing the call.
  • The lifetime of a temporary bound to the returned value in a function return statement is not extended; the temporary is destroyed at the end of the full-expression in the return statement.
  • [...]

You didn't bind directly to ref2, and you even pass it via a return statement. The standard explicitly says it won't extend the lifetime. In part to make certain optimizations possible. But ultimately, because keeping track of which temporary should be extended when a reference is passed in and out of functions is intractable in general.

Since compilers may optimize aggressively on the assumption that your program exhibits no undefined behavior, you see a possible manifestation of that. Accessing a value outside its lifetime is undefined, this is what return ref2; does, and since the behavior is undefined, simply returning zero is a valid behavior to exhibit. No contract is broken by the compiler.


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