What is this way of assignment called? myVar(value); - c++

I was digging in somebody else's code where I noticed he assigns the public/private members of a class in the following way:
myMemberVar(Value);
instead of
myMemberVal=Value;
I'm wondering if this way of assigning is normal and can be really used interchangebally.

It's likely that you're looking at initialisations in the member initialisation list.
struct A
{
int x, y;
A()
: x(42) // <-- here
, y(12) // <-- and here
{}
{};
These are not "assignments", any more than the following are assignments:
void foo()
{
int x = 42;
int y = 42;
}
But recall that the C++03 initialisation syntax (as opposed to the above confusing, legacy, looks-like-assignment syntax†) is:
void foo()
{
int x(42);
int y(42);
}
And, going further, since C++11:
void foo()
{
int x{42};
int y{42};
}
And that's the syntax that's required in the member initialisation list.
Note that this means they're not generally interchangeable: only in an initialisation! Usually when you write = you're performing assignment and this initialisation syntax would be invalid there.
† Don't worry, I still prefer it too!

That is the syntax for construction, not assignment.

The first snippet of code doesn't assign anything; it attempts to call myMemberVar like a function.
Perhaps the real code was in the initialiser list of a constructor:
MyClass() : myMemberVar(Value) {}
while the second was in the constructor body:
MyClass() {
myMemberVar = Value;
}
The difference here is that the first performs direct-initialisation, not assignment; the second performs default-initialisation followed by assignment.
For simple types, both have the same effect, and can be regarded as interchangable. For class types, they might call different user-defined functions: a conversion/copy/move constructor in the first case, and a default constructor followed by an assignment operator in the second. Depending on how those functions are defined, the two options may have different effects or performance characteristics, and may not be possible if the required functions aren't accessible.

Related

How is the implementation of initializer_list changes the way the compiler reads?

I've just learned that we can use the standard library type initializer_list<T> to achieve initializing a class instance with a {}-delimited list of elements of type T. E.g.
class X
{
public:
X(initializer_list<int> lst)
{
cout << "initializer-list constructor\n";
}
X(int i, int j, int k)
{
cout << "constructor taking 3 ints\n";
}
};
int main()
{
X a{ 1,2,3 }; // initializer-list constructor
}
To my knowledge, when the compiler sees { 1,2,3 } in the above code, it will first seek for a constructor that takes a initializer_list<int>, and since there is one, it will construct a initializer_list<int> object out of { 1,2,3 }. Had there been no constructor which takes a initializer_list<int>, the compiler will generate code the same as for X a(1,2,3).
If the process I described above is correct, to me there is some "magic" going on here: I don't understand why defining a type (initializer_list) and let a function (the initializer-list constructor) accept an object of that type (say, lst as above) can "change" the way the function is called. By that I mean: the compiler can now understand {} notation so that we need not to have things like X a({1,2,3}), and in such cases prefers the initializer-list constructor to other constructors while treating {} notation as () notation when the initializer-list constructor is absent.
More to the point, can I define a type (say, my own initializer_list) such that when the compiler sees a special form of instantiation of a class (say, a {} initializer list), it will search for a constructor in that class that takes an object of my own initializer_list type, and will treat that special form of instantiation in some other way if such constructor does not exist?
In other words:
class X
{
public:
X(initializer_list<int> lst)
{
cout << "std\n";
}
X(my_initializer_list<int> lst)
{
cout << "user-defined\n";
}
};
int main()
{
X b{ 1,2,3 };
}
Can I implement a my_initializer_list in the global scope so that the X b{ 1,2,3 }; calls X::X(my_initializer_list<int> lst)?
My current guess is, it is a feature of the language itself that, when the compiler sees a {} initializer list, the compiler will always first seek for the constructor that takes a std::initializer_list, rather than a initializer_list defined in any other namespace (by that I mean, such behavior of the compiler is not implemented in the std::initializer_list, but is a feature designed into the compiler in the first place). May I ask if my guess is on the right track?
There is no magic. This is a type of list of initialization, the rules for which are available here: https://en.cppreference.com/w/cpp/language/list_initialization.
Can I implement a my_initializer_list in the global scope so that the X b{ 1,2,3 }; calls X::X(my_initializer_list lst)?
No, this won't do it. Read the above rules - there's a special handling for initializer-list. If there isn't a constructor taking in an intializer-list, the constructor which matches the argument list is picked.
To call X::X(my_initializer_list<int> lst) you'll have to do X b{my_initializer_list{1,2,3 }};
How the braces {} are interpreted when creating an instance of a class is a language feature. That is, we cannot change it so that the next time they are encountered, our custom my_initializer_list contstructor should be selected. What we can control is what happens once it is selected(assuming that it does gets selected by the language rules).
Can I implement a my_initializer_list in the global scope so that the X b{ 1,2,3 }; calls X::X(my_initializer_list<int> lst)?
So as far as i know, i don't think what you proposed is possible/doable as it requires the {} to be interpreted differently.
Imagine if you provided more than one my_initializer_list say like my_initializer_list1, my_initializer_list2, my_initializer_list3 and suppose your class has a constructor corrresponding to each of them, then when you use {} for creating an instance of the class, how can it be decided which of those constructor to select.

How can a compiler generated default constructor be more efficient than a self-written one that does nothing but initialize members?

Triggered by this answer I was reading in the core guidelines:
C.45: Don’t define a default constructor that only initializes data
members; use in-class member initializers instead
The reasoning given is
Reason
Using in-class member initializers lets the compiler generate the
function for you. The compiler-generated function can be more
efficient.
Note that this is specifically about a default constructor that does nothing but initialize the members and the guideline suggests that one should not write such a constructor.
The "bad" example is:
Example, bad
class X1 { // BAD: doesn't use member initializers
string s;
int i;
public:
X1() :s{"default"}, i{1} { }
// ...
};
The "good" example is using in-class member initializers and no user declared constructor:
Example
class X2 {
string s = "default";
int i = 1;
public:
// use compiler-generated default constructor
// ...
};
What can the compiler generated constructor do more efficient than the user-provided one in that particular example (or in any other example)?
Is the initializer list not giving the same opportunities for optimization as in-class initializers?
Short Answer
A defaulted constructor should have the same generated assembly as the equivalent initializer constructor provided that the author includes the correct constexpr and noexcept statuses.
I suspect the "can be more efficient" is referring to the fact that, in general, it will generate more optimal code than the equivalent developer-authored one that misses opportunities such as inline, constexpr, and noexcept.
Long Answer
An important feature that defaulted constructors perform is that they interpret and deduce the correct status for both constexpr and noexcept
This is something that many C++ developers do not specify, or may not specify correctly. Since Core Guidelines targets both new and old C++ developers, this is likely why the "optimization" is being mentioned.
The constexpr and noexcept statuses may affect code generation in different ways:
constexpr constructors ensure that invocations of a constructor from values yielded from constant expressions will also yield a constant expression. This can allow things like static values that are not constant to not actually require a constructor invocation (e.g. no static initialize overhead or locking required). Note: this works for types that are not, themselves, able to exist in a constexpr context -- as long as the constexprness of the constructor is well-formed.
noexcept may generate better assembly of consuming code since the compiler may assume that no exceptions may occur (and thus no stack-unwinding code is necessary). Additionally, utilities such as templates that check for std::is_nothrow_constructible... may generate more optimal code paths.
Outside of that, defaulted constructors defined in the class-body also make their definitions visible to the caller -- which allows for better inlining (which, again, may otherwise be a missed-opportunity for an optimization).
The examples in the Core Guidelines don't demonstrate these optimizations very well. However, consider the following example, which illustrates a realistic example that can benefit from defaulting:
class Foo {
int a;
std::unique_ptr<int> b;
public:
Foo() : a{42}, b{nullptr}{}
};
In this example, the following are true:
A construction of Foo{} is not a constant expression
Construction of Foo{} is not noexcept
Contrast this to:
class Foo {
int a = 42;
std::unique_ptr<int> b = nullptr;
public:
Foo() = default;
};
On the surface, this appears to be the same. But suddenly, the following now changes:
Foo{} is constexpr, because std::unique_ptr's std::nullptr_t constructor is constexpr (even though std::unique_ptr cannot be used in a full constant expression)
Foo{} is a noexcept expression
You can compare the generated assembly with this Live Example. Note that the default case does not require any instructions to initialize foo; instead it simply assigns the values as constants through compiler directive (even though the value is not constant).
Of course, this could also be written:
class Foo {
int a;
std::unique_ptr<int> b;
public:
constexpr Foo() noexcept :a{42}, b{nullptr};
};
However, this requires prior knowledge that Foo is able to be both constexpr and noexcept. Getting this wrong can lead to problems. Worse yet, as code evolves over time, the constexpr/noexcept state may become incorrect -- and this is something that defaulting the constructor would have caught.
Using default also has the added benefit that, as code evolves, it may add constexpr/noexcept where it becomes possible -- such as when the standard library adds more constexpr support. This last point is something that would otherwise be a manual process every time code changes for the author.
Triviality
If you take away the use of in-class member initializers, then there is one last worthwhile point mentioning: there is no way in code to achieve triviality unless it gets compiler-generated (such as through defaulted constructors).
class Bar {
int a;
public:
Bar() = default; // Bar{} is trivial!
};
Triviality offers a whole different direction on potential optimizations, since a trivial default-constructor requires no action on the compiler. This allows the compiler to omit any Bar{} entirely if it sees that the object is later overwritten.
I think that it's important to assume that C.45 refers to constants (example and enforcement):
Example, bad
class X1 { // BAD: doesn't use member initializers
string s;
int i; public:
X1() :s{"default"}, i{1} { }
// ... };
Example
class X2 {
string s = "default";
int i = 1; public:
// use compiler-generated default constructor
// ... };
Enforcement
(Simple) A default constructor should do more than just initialize
member variables with constants.
With that in mind, it's easier to justify (via C.48) why we should prefer in-class initializers to member initializers in constructors for constants:
C.48: Prefer in-class initializers to member initializers in
constructors for constant initializers
Reason
Makes it explicit that the same value is expected to be used in all
constructors. Avoids repetition. Avoids maintenance problems. It leads
to the shortest and most efficient code.
Example, bad
class X { // BAD
int i; string s;
int j; public:
X() :i{666}, s{"qqq"} { } // j is uninitialized
X(int ii) :i{ii} {} // s is "" and j is uninitialized
// ... };
How would a maintainer know whether j was deliberately uninitialized
(probably a poor idea anyway) and whether it was intentional to give s
the default value "" in one case and qqq in another (almost certainly
a bug)? The problem with j (forgetting to initialize a member) often
happens when a new member is added to an existing class.
Example
class X2 {
int i {666};
string s {"qqq"};
int j {0}; public:
X2() = default; // all members are initialized to their defaults
X2(int ii) :i{ii} {} // s and j initialized to their defaults
// ... };
Alternative: We can get part of the benefits from default arguments to
constructors, and that is not uncommon in older code. However, that is
less explicit, causes more arguments to be passed, and is repetitive
when there is more than one constructor:
class X3 { // BAD: inexplicit, argument passing overhead
int i;
string s;
int j; public:
X3(int ii = 666, const string& ss = "qqq", int jj = 0)
:i{ii}, s{ss}, j{jj} { } // all members are initialized to their defaults
// ... };
Enforcement
(Simple) Every constructor should initialize every member variable (either explicitly, via a delegating ctor call or via default
construction).
(Simple) Default arguments to constructors suggest an in-class initializer may be more appropriate.

c++ variable assignment, is this a normal way..?

This may be a silly question, but still I'm a bit curious...
Recently I was working on one of my former colleague projects, and I've noticed that he really loved to use something like this:
int foo(7);
instead of:
int foo = 7;
Is this a normal/good way to do in C++ language?
Is there some kind of benefits to it? (Or is this just some silly programming style that he was into..?)
This really reminds me a bit of a good way how class member variables can be assigned in the class constructor... something like this:
class MyClass
{
public:
MyClass(int foo) : mFoo(foo)
{ }
private:
int mFoo;
};
instead of this:
class MyClass
{
public:
MyClass(int foo)
{
mFoo = foo;
}
private:
int mFoo;
};
For basic types there's no difference. Use whichever is consistent with the existing code and looks more natural to you.
Otherwise,
A a(x);
performs direct initialization, and
A a = x;
performs copy initialization.
The second part is a member initializer list, there's a bunch of Q&As about it on StackOverflow.
Both are valid. For builtin types they do the same thing; for class types there is a subtle difference.
MyClass m(7); // uses MyClass(int)
MyClass n = 3; // uses MyClass(int) to create a temporary object,
// then uses MyClass(const MyClass&) to copy the
// temporary object into n
The obvious implication is that if MyClass has no copy constructor, or it has one but it isn't accessible, the attempted construction fails. If the construction would succeed, the compiler is allowed to skip the copy constructor and use MyClass(int) directly.
All the answers above are correct. Just add that to it that C++11 supports another way, a generic one as they say to initialize variables.
int a = {2} ;
or
int a {2} ;
Several other good answers point out the difference between constructing "in place" (ClassType v(<constructor args>)) and creating a temporary object and using the copy constructor to copy it (ClassType v = <constructor arg>). Two additional points need to be made, I think. First, the second form obviously has only a single argument, so if your constructor takes more than one argument, you should prefer the first form (yes, there are ways around that, but I think the direct construction is more concise and readable - but, as has been pointed out, that's a personal preferance).
Secondly, the form you use matters if your copy constructor does something significantly different than your standard constructor. This won't be the case most of the time, and some will argue that it's a bad idea to do so, but the language does allow for this to be the case (all surprises you end up dealing with because of it, though, are your own fault).
It's a C++ style of initializing variables - C++ added it for fundamental types so the same form could be used for fundamental and user-defined types. this can be very important for template code that's intended to be instantiated for either kind of type.
Whether you like to use it for normal initialization of fundamental types is a style preference.
Note that C++11 also adds the uniform initialization syntax which allows the same style of initialization to be used for all types - even aggregates like POD structs and arrays (though user defined types may need to have a new type of constructor that takes an initialization list to allow the uniform syntax to be used with them).
Yours is not a silly question at all as things are not as simple as they may seem. Suppose you have:
class A {
public:
A() {}
};
and
class B {
public:
class B(A const &) {}
};
Writing
B b = B(A());
Requires that B's copy constructor be accessible. Writing
B b = A();
Requires also that B's converting constructor B(A const &) be not declared explicit. On the other hand if you write
A a;
B b(a);
all is well, but if you write
B b(A());
This is interpreted by the compiler as the declaration of a function b that takes a nameless argument which is a parameterless function returning A, resulting in mysterious bugs. This is known as C++'s most vexing parse.
I prefer using the parenthetical style...though I always use a space to distinguish from function or method calls, on which I don't use a space:
int foo (7); // initialization
myVector.push_back(7); // method call
One of my reasons for preferring using this across the board for initialization is because it helps remind people that it is not an assignment. Hence overloads to the assignment operator will not apply:
#include <iostream>
class Bar {
private:
int value;
public:
Bar (int value) : value (value) {
std::cout << "code path A" << "\n";
}
Bar& operator=(int right) {
value = right;
std::cout << "code path B" << "\n";
return *this;
}
};
int main() {
Bar b = 7;
b = 7;
return 0;
}
The output is:
code path A
code path B
It feels like the presence of the equals sign obscures the difference. Even if it's "common knowledge" I like to make initialization look notably different than assignment, since we are able to do so.
It's just the syntax for initialization of something :-
SomeClass data(12, 134);
That looks reasonable, but
int data(123);
Looks strange but they are the same syntax.

How to prevent default initialization of a const variable with a class type

I have a custom class that I want to behave like a built-in type.
However I have noticed that you can initialise a const variable of that class without providing an initial value. My class currently has an empty default constructor.
Here is a comparison of int and my class foo:
int a; // Valid
int a = 1; // Valid
const int a = 1; // Valid
const int a; // Error
foo a; // Valid
foo a = 1; // Valid
const foo a = 1; // Valid
const foo a; // Should cause an error, but it compiles
As you can see I need to prevent
const foo a;
from compiling.
Any ideas from C++ gurus?
It compiles only if it has a default constructor, and it compiles because it has it, which means that it is initialized. If you don't want that line to compile, just disable the default constructor (will also make foo a; an error as an unwanted side effect). Without a definition of foo or what you want to do, this is as far as I can get.
I don't think there is any way of achieving what you want (i.e. allow the non-const variable to be default initialized, while having the const version fail compilation and allowing the other use cases --that require providing constructors)
The rules of C++ simply say that default-initialization (e.g. new T;) and value-initialization (e.g. new T();) are the same for objects of class type, but not for objects of fundamental type.
There's nothing you can do to "override" this distinction. It's a fundamental part of the grammar. If your class is value-initializable, then it is also default-initializable.
There is a sort-of exception for classes without any user-defined constructors: In that case, initialization of members is done recursively (so if you default-init the object, it tries to default-init all members), and this will fail if any of the class members are themselves fundamental, or again of this nature.
For example, consider the following two classes:
struct Foo { int a; int b; };
struct Goo { int a; int b; Goo(){} };
//const Foo x; // error
const Goo y; // OK
The implicit constructor for Foo is rejected because it doesn't initialize the fundamental members. However, y is happily default-initialized, and y.a and y.b are now "intentionally left blank".
But unless your class doesn't have any user-defined constructors, this information won't help you. You cannot "forward" the initialization type to a member (like Foo() : INIT_SAME_AS_SELF(a), b() { }).

Initializing in constructors, best practice?

I've been programming in C++ a while and I've used both methods:
class Stuff {
public:
Stuff( int nr ) : n( nr ) { }
private:
int n;
}
Or
class Stuff {
public:
Stuff( int nr ) {
n = nr;
}
private:
int n;
}
Note: This is not the same as this, similar but not the same.
What is considered best practice?
Initializer lists are preferred. See FAQ 10.6
One big advantage to using initializers: If an exception is thrown anywhere within the initializer list, the destructors will be called for those members that had already been initialized -- and only for those members.
When you use the contructor body to initialize the object, it's up to you to handle exceptions properly and unwind the object as appropriate. This is usually much harder to get right.
Use the initializer list when possible. For an int, it doesn't matter much either way, but for a more complex member object, you'd end up with the default constructor of the object being called, followed by an assignment to that object, which is likely to end up being slower.
Plus, you have to do it that way anyway for const members or members which don't have a default constructor.
If possible, use the first version.
The first is initializing using intializer lists, and actually calls the constructors of the members.
The second is assignment. If n was of a type with a default constructor, it the would have already been called, and then you'd be assigning to it. If n didn't have a default constructor, you'd be forced to use the first type. Likewise if n was a reference: int &n.
If there are no constructors of you members that directly take one of the parameters to your constructor, it may be worthwhile to add private static functions that can do the conversion for you.
I generally try to do the initializer list when I can. For one thing, this makes it explicit that you are initializing code in the constructor. const memebers have to be initialized this way.
If you just put code in the constructor's body, it is quite possible someone may decide to come along and move a big chunk of it into a non-constructor "setup" routine later.
It can be taken overboard though. I have a coworker who likes to create classes that have 2 pages of initilizer code, no constructor code, and perhaps 2 pages for the entire rest of the class' code. I find that really tough to read.
I want to add that you don't need to declare the initializer list on the Header (.h). It can be done at the implementation of the constructor (which is very common).
So then:
//Stuff.h
class Stuff {
public:
Stuff( int nr );
private:
int n;
}
//Stuff.cpp
Stuff::Stuff(int nr)
: n(nr)
{
//initalize complex members
}
is legal and imo concentrates the initialization of fields where it matters. Sometimes we need to initialize complex members in the body, so you have your initializer list and the complex initialization all in the .cpp file.
The second option is not initialization but assignment. With types that have user defined default constructors, the second option will call the default constructor and later on call the assignment operator (whether user defined or not) to assign the value.
Some types cannot be default initialized: If you have an attribute without default constructor, hold references (constant or not) or have constant attributes they must be initialized in the initializer list.
Arrays can be value-initialized in the initialization list, but not in the constructor body:
class X {
public:
X() : array() {} // value-initializes the array
// equivalent to:
// X() { for ( int i = 0; i < 10; ++i ) array[i]=0; }
private:
int array[10];
};
For POD types, you can value-initialize them in the initialization list but not inside the brackets:
class X {
public:
X() : pod() {} // value-initializes
// equivalent to (but easier to read and subtly faster as it avoids the copy):
// X() { pod = {}; }
private:
PODType pod;
};
Finally, some classes offer functionality through the use of constructors that will be more complex (if achievable) after default construction.
class X
{
public:
X() : v(10) {} // construct a vector of 10 default initialized integers
// equivalent to:
// X() { for ( int i = 0; i < 10; ++i ) v.push_back(0); }
private:
std::vector<int> v;
};
Last, whenever they are in fact equivalent, initialization lists are more idiomatic in C++.