This is a memory allocation issue that I\'ve never really understood.
void unleashMonkeyFish() { MonkeyFish * monkey_fish = new MonkeyFish(); std::string
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 *.