Initializing class using { * this } - c++

Question

It was suggested by a team member that using an intializer like this:
return Demo{ *this };
was better than:
return Demo(*this);
Assuming a simple class like this:
class Demo {
public:
int value1;
Demo(){}
Demo(Demo& demo) {
this->value1 = demo.value1;
}
Demo Clone() {
return Demo{ *this };
}
};
I admit to having not seen the { *this } syntax before, and couldn't find a reference that explained it well enough that I understood how the two options differed. Is there a performance benefit, a syntax choice, or something more?

Answer 1

Your colleague is missing a trick with "uniform initialization", there is no need for the type-name when it is known. E.g. when creating a return value. Clone could be defined as:
Demo Clone() {
return {*this};
}
This will call the Demo copy constructor as needed. Whether you think this is better or not, is up to you.
In GOTW 1 Sutter states as a guideline:
Guideline: Prefer to use initialization with { }, such as vector v = { 1, 2, 3, 4 }; or auto v = vector{ 1, 2, 3, 4 };, because it’s more consistent, more correct, and avoids having to know about old-style pitfalls at all. In single-argument cases where you prefer to see only the = sign, such as int i = 42; and auto x = anything; omitting the braces is fine. …
In particular, using braces can avoid confusion with:
Demo d(); //function declaration, but looks like it might construct a Demo
Demo d{}; //constructs a Demo, as you'd expect
The brace syntax will use a constructor that takes an initializer list first, if one exists. Otherwise it will use a normal constructor. It also prevents the chance of the vexing parse listed above.
There is also different behaviour when using copy initialization. With the standard way
Demo d = x;
The compiler has the option to convert x to a Demo if necessary and then move/copy the converted r-value into w. Something similar to Demo d(Demo(x)); meaning that more than one constructor is called.
Demo d = {x};
This is equivalent to Demo d{x} and guarantees that only one constructor will be called. With both assignments above explicit constructors are cannot be used.
As mentioned in the comments, there are some pitfalls. With classes that take an initializer_list and have "normal" constructors can cause confusion.
vector<int> v{5}; // vector containing one element of '5'
vector<int> v(5); // vector containing five elements.

Answer 2

This is just another syntax for calling your copy constructor (actually, for calling a constructor taking what is in the braces as parameters, in this case, your copy constructor).
Personally, I would say it's worse than before simply because it does the same... it just relies on C++ 11. So it adds dependencies without benefits. But your mileage may vary. You will have to ask your colleague.

Answer 3

I must admit that I'd never seen that before.
WikiPedia says this about C++11 initializer lists (search for "Uniform initialization"):
C++03 has a number of problems with initializing types. There are several ways to initialize types, and they do not all produce the same results when interchanged. The traditional constructor syntax, for example, can look like a function declaration, and steps must be taken to ensure that the compiler's most vexing parse rule will not mistake it for such. Only aggregates and POD types can be initialized with aggregate initializers (using SomeType var = {/stuff/};).
Then, later, they have this example,
BasicStruct var1{5, 3.2}; // C type struct, containing only POD
AltStruct var2{2, 4.3}; // C++ class, with constructors, not
// necessarily POD members
with the following explanation:
The initialization of var1 behaves exactly as though it were aggregate-initialization. That is, each data member of an object, in turn, will be copy-initialized with the corresponding value from the initializer-list. Implicit type conversion will be used where necessary. If no conversion exists, or only a narrowing conversion exists, the program is ill-formed. The initialization of var2 invokes the constructor.
They also have further examples for the case where initialiser list constructors were provided.
So based on the above alone: For the plain-old-data struct case, I don't know if there is any advantage. For a C++11 class, using the {} syntax may help avoid those pesky scenarios where the compiler thinks you're declaring a function. Maybe that is the advantage your colleague was referring to?

Answer 4

Sorry for comming late to this discussion but I want to add some points about the different types of initialization not mentioned by others.
Consider:
struct foo {
foo(int) {}
};
foo f() {
// Suppose we have either:
//return 1; // #1
//return {1}; // #2
//return foo(1); // #3
//return foo{1}; // #4
}
Then,
#1, #3 and #4 might call the copy/move constructor (if RVO isn't performed) whereas #2 won't call the copy/move constructor.
Notice that the most popular compilers do perform RVO and thus, in practice, all return statements above are equivalent. However, even when RVO is performed a copy/move constructor must be available (must be accessible to f and defined but not as deleted) for #1, #3 and #4 otherwise the compiler/linker will raise an error.
Suppose now that the constructor is explicit:
struct foo {
explicit foo(int) {}
};
Then,
#1 and #2 don't compile whereas #3 and #4 do compile.
Finally, if the constructor is explicit and no copy/move constructor is available:
struct foo {
explicit foo(int) {}
foo(const foo&) = delete;
};
none of the return statements compile/link.

Answer 5

This is known as list-initialization. The idea is that in C++11, you will have uniform initialization across the board, and avoid ambiguity where the compiler might think you may be making a function declaration (also known as a vexing parse). A small example:
vec3 GetValue()
{
return {x, y, z}; // normally vec(x, y, z)
}
One reason you would want to avoid list-initialization is where your class takes an initializer list constructor that does something different than you would expect.