When is it not a good idea to pass by reference?

Question

This is a memory allocation issue that I\'ve never really understood.

void unleashMonkeyFish()  
{  
    MonkeyFish * monkey_fish = new MonkeyFish();
    std::string          


        

Answer 1

If you use a temporary variable to assign the name (as in your sample code) you will eventually have to copy the string to your MonkeyFish object in order to avoid the temporary string object going end-of-scope on you.

As Andrew Flanagan mentioned, you can avoid the string copy by using a local static variable or a constant.

Assuming that that isn't an option, you can at least minimize the number of string copies to exactly one. Pass the string as a reference pointer to setName(), and then perform the copy inside the setName() function itself. This way, you can be sure that the copy is being performed only once.

Answer 2

When the compiler sees ...

std::string localname = "Wanda";  

... it will (barring optimization magic) emit 0x57 0x61 0x6E 0x64 0x61 0x00 [Wanda with the null terminator] and store it somewhere in the the static section of your code. Then it will invoke std::string(const char *) and pass it that address. Since the author of the constructor has no way of knowing the lifetime of the supplied const char *, s/he must make a copy. In MonkeyFish::setName(const std::string &), the compiler will see std::string::operator=(const std::string &), and, if your std::string is implemented with copy-on-write semantics, the compiler will emit code to increment the reference count but make no copy.

You will thus pay for one copy. Do you need even one? Do you know at compile time what the names of the MonkeyFish shall be? Do the MonkeyFish ever change their names to something that is not known at compile time? If all the possible names of MonkeyFish are known at compile time, you can avoid all the copying by using a static table of string literals, and implementing MonkeyFish's data member as a const char *.

Answer 3

Just to clarify the terminology, you've created MonkeyFish from the heap (using new) and localname on the stack.

Ok, so storing a reference to an object is perfectly legit, but obviously you must be aware of the scope of that object. Much easier to pass the string by reference, then copy to the class member variable. Unless the string is very large, or your performing this operation a lot (and I mean a lot, a lot) then there's really no need to worry.

Can you clarify exactly why you don't want to copy the string?

Edit

An alternative approach is to create a pool of MonkeyName objects. Each MonkeyName stores a pointer to a string. Then get a new MonkeyName by requesting one from the pool (sets the name on the internal string *). Now pass that into the class by reference and perform a straight pointer swap. Of course, the MonkayName object passed in is changed, but if it goes straight back into the pool, that won't make a difference. The only overhead is then the actual setting of the name when you get the MonkeyName from the pool.

... hope that made some sense :)