Re: Simple question: extending lifetime of temporaries with a reference

From:
olekk <sashakster@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Mon, 10 Nov 2008 12:26:26 CST
Message-ID:
<1c077c9c-c8e5-45c2-80bd-d9a236b906c3@z6g2000pre.googlegroups.com>

I've heard of "extending the lifetime of a temporary
by binding it to a reference", but am not entirely
sure of when that applies.

My impression is that if you do, say:

     // sample 1
     std::string str1( "String 1\n");
     std::string str2( "string 2\n");
     std::string &str_sum = str1 + str2;
     std::cout << str_sum << str::endl;

then the temporary object "str1 + str2" won't be destroyed
until str_sum goes out of scope.

On the other hand, I'm fairly convinced that

     // sample 2
     std::string &f() { std::string temp( "temp value"); return temp; }
     void g() {
         std::string &ret_val = f();
         std::cout << ret_val << std::endl;
     }

won't work.

1. Is this correct, or am I just terminally bewildered?


What you do here is returning a reference to local variable that goes
out of scope.
If you made f() returning std::string,
           std::string f() {return "temp object;}
then in
           std::string& ret_val = f();
ret_val actually the reference to temporary object.

Regards

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