Introduction
For a technical overview - skip to this answer.
For common cases where copy elision occurs - skip to this answer.
Copy elision is an optimization implemented by most compilers to prevent extra (potentially expensive) copies in certain situations. It makes returning by value or pass-by-value feasible in practice (restrictions apply).
It's the only form of optimization that elides (ha!) the as-if rule - copy elision can be applied even if copying/moving the object has side-effects.
The following example taken from Wikipedia:
struct C {
C() {}
C(const C&) { std::cout << "A copy was made.\n"; }
};
C f() {
return C();
}
int main() {
std::cout << "Hello World!\n";
C obj = f();
}
Depending on the compiler & settings, the following outputs are all valid:
Hello World!
A copy was made.
A copy was made.
Hello World!
A copy was made.
Hello World!
This also means fewer objects can be created, so you also can't rely on a specific number of destructors being called. You shouldn't have critical logic inside copy/move-constructors or destructors, as you can't rely on them being called.
If a call to a copy or move constructor is elided, that constructor must still exist and must be accessible. This ensures that copy elision does not allow copying objects which are not normally copyable, e.g. because they have a private or deleted copy/move constructor.
C++17: As of C++17, Copy Elision is guaranteed when an object is returned directly, and in this case, the copy or move constructor need not be accessible or present:
struct C {
C() {}
C(const C&) { std::cout << "A copy was made.\n"; }
};
C f() {
return C(); //Definitely performs copy elision
}
C g() {
C c;
return c; //Maybe performs copy elision
}
int main() {
std::cout << "Hello World!\n";
C obj = f(); //Copy constructor isn't called
}
Unless that value is 0 (in which case you can omit some part of the initializer
and the corresponding elements will be initialized to 0), there's no easy way.
Don't overlook the obvious solution, though:
int myArray[10] = { 5, 5, 5, 5, 5, 5, 5, 5, 5, 5 };
Elements with missing values will be initialized to 0:
int myArray[10] = { 1, 2 }; // initialize to 1,2,0,0,0...
So this will initialize all elements to 0:
int myArray[10] = { 0 }; // all elements 0
In C++, an empty initialization list will also initialize every element to 0.
This is not allowed with C until C23:
int myArray[10] = {}; // all elements 0 in C++ and C23
Remember that objects with static storage duration will initialize to 0 if no
initializer is specified:
static int myArray[10]; // all elements 0
And that "0" doesn't necessarily mean "all-bits-zero", so using the above is
better and more portable than memset(). (Floating point values will be
initialized to +0, pointers to null value, etc.)
Best Answer
It is guaranteed to exit with 0.
t
ands
are different objects, pointers to them can't compare equal.About the code that you removed:
From https://port70.net/~nsz/c/c11/n1570.html#6.2.4p2 :
It may return 1 or 0 or perform a trap.