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CS253 Copy Elision

“elision”

• rhymes with “vision”
• “removal”
• “omission”
• “elimination”
• “suppression”

The Loud Class

The following examples use our class Loud, which displays a message for every method invoked.

int main() {
    Loud alpha;
}

Loud::Loud()
Loud::~Loud()

No surprise, here. alpha got created & destroyed.

More Construction

int main() {
    Loud beta, gamma;
}

Loud::Loud()
Loud::Loud()
Loud::~Loud()
Loud::~Loud()

Sure.

Copy ctor & assignment

int main() {
    Loud delta;
    Loud epsilon(delta);
    delta=epsilon;
}

Loud::Loud()
Loud::Loud(const Loud &)
Loud::operator=(const Loud &)
Loud::~Loud()
Loud::~Loud()

As expected.

More copying

Loud foo() {
    Loud zeta;
    return zeta;
}

int main() {
    Loud eta(foo());
}

Loud::Loud()
Loud::~Loud()

• Hey, wait. Where’s the copy ctor?
• Two distinct objects, zeta and eta, should have been created.
• They weren’t.

Ludicrous copying

int main() {
    Loud theta = Loud(Loud(Loud(Loud(Loud(Loud(Loud(Loud())))))));
}

Loud::Loud()
Loud::~Loud()

There should be more than that.

Elision

Loud foo() {
    Loud iota;
    return iota;
}

int main() {
    Loud kappa(foo());
}

Loud::Loud()
Loud::~Loud()

• The compiler is allowed to elide (rhymes with “ride”: suppress, omit) ctors, copy ctors, and dtors, even if the methods have side effects, like the Loud class.
• As of C++ 2017, copy elision is required in the zeta & theta examples.
• Another way to view it: the compiler can assume that ctors, copy ctors, and dtors only do creation⁠/⁠copying⁠/⁠destruction, even if they don’t (as in Loud).
• This is efficient. Copying takes time & space.
• There is no variable iota. Instead, the ctor in foo() constructs directly in kappa. No copying; it’s built in the right place.

Parameter Passing

• For this to make sense, we have to understand how parameters & results are passed to & from functions.
• How does something get passed as an argument?
  • On the stack, or in a register?
  • A double?
  • A double *?
  • An array of double?
  • A std::string?
  • A const std::string &?
• How does a a function return an int?
  • A std::string?

No Answers Yet

• The answer, of course, depends on the cpu architecture.
  • I first programmed on an IBM/360, which had no stack.
• However, in general, it can be assumed that registers are faster than memory.
• Therefore, we prefer passing arguments in registers.
• However, there are only so many registers.
• What if there are more arguments than available registers?
  • Usually, the extra arguments go onto the stack.

X86 Answers

The popular X86-64 CPU architecture has sixteen more-or-less general-purpose 64-bit registers, used for:

• Function arguments:
  • Scalars (int, double, pointers) are passed in registers
  • A reference is implemented as a pointer.
  • A C array is a special case, always passed by reference.
  • An object passed by value gets pushed onto the stack.
  • Methods get an additional this pointer argument.
  • If we run out of registers, push arguments onto the stack.
• Return value:
  • Scalars (int, double, pointers) are returned in a register.
  • An object returned by value is written through a pointer to a hidden parameter.

Parameter Passing

When a function returns an non-scalar object, compilers typically pass a hidden argument to the result.

// User code:
string tv() {
   string s="SNL";
   return s;
}

int main() {
   string fav=tv();


   cout << fav;
}

SNL

// Implementation:
void tv(string &out) {
   string s="SNL";
   out=s;
}

int main() {
   string result;
   tv(result);
   string fav=result;
   cout << fav;
}

SNL

// After optimization:
void tv(string &out) {
   out="SNL";

}

int main() {
   string fav;
   tv(fav);

   cout << fav;
}

SNL

All thanks to copy elision !