am Dienstag 23 Oktober 2007 11:05 schrieb Jim Langston:
"Thomas Lenz" <lenzth@gmx.de> wrote in message
news:471ce6c4$0$16664$9b4e6d93@newsspool3.arcor-online.net...
Please consider the following code snippet:
string myfunction()
{ ostringstream oss;
oss << "junk";
// do something more with oss; I can't make it const...
return oss.str();
}
Is the returned string object still valid after myfunction() has
returned? I
wonder because oss should be destroyed at the bottom '}', shouldn't it?
So what about its string? (The above code seems to work on my machine,
but is it guaranteed to work?)
Or is it safer to explicitly construct the string to return as a
temporary object, like
return string(oss.str());
?
As Victor says, you are returning by copy. You are returning a
std::string, so oss.str() is copied into a temporary, which is returned.
It would be a problem, however, if you returned a reference or a pointer.
std::string& myfunction()
{ ostringstream oss;
oss << "junk";
// do something more with oss; I can't make it const...
return oss.str();
}
Now you would have a problem, because you are returning a reference (a
type of pointer) to something that is destroyed when the function ends.
thanks everybody.
So there are two string objects involved: one being returned by oss.str(),
and will die when myfunction() returns, and a copied one that lives
outside
of myfunction(), right?
What happened if I change the return statement to
return string(oss.str());
? Would this yield 3 string objects? (one coming from oss.str(), one
produced by the string(...) Constructor, and one copied by the return
statement)? (Assuming the compiler doesn't optimize it away)
Yes, but be careful of "lives outside of myfunction". It only lives as long
as the stament. For instance:
The string that MyFunction returns is a temporary object. It will only live
as long as the statement that the MyFunction() call is on. So if you