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

SG <>
Mon, 10 Nov 2008 12:26:54 CST
On 10 Nov., 03:01, Alan McKenney <> wrote:

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.

Old compilers still might accept this. But it's not exactly well-
formed code. You need to write

    std::string const& str_sum = str1 + str2;

But you're right. The life-time of this temporary is extended to that
of the reference. This rule only applies to "local" temporary objects.
operator+(string,string) returns a string object by value which will
be a local temporary object. The compiler will know exactly what it is
and where it is located (automatic memory) -- and even call the right
destructor for you in case you do something like this:

    Derived f();

    void g() {
       BaseClass const& obj = f();

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.

Right. It's not a "local" temporary. The function returns a reference
which could be anything (including a reference to a heap-allocated
object). Defining such a function is not ill-formed but a good
compiler should warn you about it. G++ says: "warning: reference to
local variable is returned".


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