The following code, as far as I can tell, correctly initializes the variables of the derived class B:
#include <utility>
struct A {
int i;
};
struct B : A {
int j;
explicit B(A&& a) : A(std::move(a)), j{i} { }
};
int main()
{
A a{3};
B b(std::move(a));
return 0;
}
Running cppcheck with --enable=all gives the warning:
[test.cpp:9]: (warning) Member variable 'A::i' is not initialized in
the constructor. Maybe it should be initialized directly in the class
A?
Is there a reason for this (false, I think) warning?
Yes, this looks like a false positive. Base class subobjects are initialized before direct member subobjects and A(std::move(a)) will use the implicit move constructor which initializes this->i with a.i, so this->i will be initialized before the initialization of this->j is performed (which reads this->i).
The argument given to the constructor in main is also completely initialized via aggregate initialization, so a.i's value will not be indeterminate either.
In C++17 code, I sometimes see something like:
void myfunction(const std::string& arg1, int arg2) {
//...
}
void otherfunction() {
myfunction("a",1);
myfunction({},2);
myfunction("b",{});
myfunction({},{});
}
What do the empty curly brace arguments means?
I found that I get an empty string and 0 integer, so I suppose it's some kind of default value.
This was introduced in C++17 and is called list-initialization.
When the braces are empty, it will perform value initialization:
For scalar types such as int, it will be as if you define it in global scope, i.e it will be zero-initialized (also true for class instances with default constructor that is not user-provided and not deleted, unless it has a non-trivial default constructor in which case it will be default initialized).
If class type with no default constructor or if the default constructor has been marked deleted, it will perform default initialization (ex. T t;), this is also true if there is a user defined constructor.
Examples:
struct A {
A() : x{1}, y{2} {}
int x,y;
};
/* A has a default user-provided constructor,
* calls default constructor */
A a{};
//////////////////////////
struct B {
B() {}
int x, y;
string z;
};
/* B has a user defined default constructor,
* since x and y are not mentioned in the initializer list,
* B::x and B::y contain intermediate values,
* while, B::z will be default initialized to "" */
B b{};
//////////////////////////
struct C {
int x, y;
};
/* C has an implicit default constructor,
* C::x and C::y will be zero-initialized */
C c{};
//////////////////////////
Someone gave me (part of) the following code:
struct MyStruct
{
int x = {};
int y = {};
};
I never saw this syntax before, what does initialization with {} mean?
This is default member initializer (since C++11),
Through a default member initializer, which is a brace or equals initializer included in the member declaration and is used if the member is omitted from the member initializer list of a constructor.
The initialization itself is copy-list-initialization (since C++11), as the effect, the data member x and y would be value-initialized (and zero-initialized as built-in type) to 0.
Since the C++11 standard there are two ways to initialize member variables:
Using the constructor initialization list as "usual":
struct Foo
{
int x;
Foo()
: x(0)
{
}
};
Use the new inline initialization where members are getting their "default" values using normal initialization syntax:
struct Foo
{
int x = 0;
};
Both these ways are for many values and types equivalent.
Are there any differences between the following three structure definitions according to the C++ standard?
struct Foo
{
int a;
};
struct Foo
{
int a{};
};
struct Foo
{
int a{0};
};
The last two are C++11.
Given the first definition, if you create an instance of Foo with automatic storage duration, a will be uninitialized. You can perform aggregate initialization to initialize it.
Foo f{0}; // a is initialized to 0
The second and third definitions of Foo will both initialize the data member a to 0.
In C++11, neither 2 nor 3 are aggregates, but C++14 changes that rule so that they both remain aggregates despite adding the brace-or-equal-initializer.
struct Foo
{
int a;
}bar;
bar.a is uninitialized if not in global scope or not-static.
struct Foo
{
int a{};
}bar;
bar.a is initialized to 0
struct Foo
{
int a{0};
}bar;
bar.a is initialized to 0
So constructs 2 and 3 are same. 1 is different.
For more details, you may want to read Initialization and Class Member Initialization
First one is POD type. Member a is initialized by 0.
Internally and about the generated code, is there a really difference between :
MyClass::MyClass(): _capacity(15), _data(NULL), _len(0)
{
}
and
MyClass::MyClass()
{
_capacity=15;
_data=NULL;
_len=0
}
thanks...
You need to use initialization list to initialize constant members,references and base class
When you need to initialize constant member, references and pass parameters to base class constructors, as mentioned in comments, you need to use initialization list.
struct aa
{
int i;
const int ci; // constant member
aa() : i(0) {} // will fail, constant member not initialized
};
struct aa
{
int i;
const int ci;
aa() : i(0) { ci = 3;} // will fail, ci is constant
};
struct aa
{
int i;
const int ci;
aa() : i(0), ci(3) {} // works
};
Example (non exhaustive) class/struct contains reference:
struct bb {};
struct aa
{
bb& rb;
aa(bb& b ) : rb(b) {}
};
// usage:
bb b;
aa a(b);
And example of initializing base class that requires a parameter (e.g. no default constructor):
struct bb {};
struct dd
{
char c;
dd(char x) : c(x) {}
};
struct aa : dd
{
bb& rb;
aa(bb& b ) : dd('a'), rb(b) {}
};
Assuming that those values are primitive types, then no, there's no difference. Initialization lists only make a difference when you have objects as members, since instead of using default initialization followed by assignment, the initialization list lets you initialize the object to its final value. This can actually be noticeably faster.
Yes. In the first case you can declare _capacity, _data and _len as constants:
class MyClass
{
private:
const int _capacity;
const void *_data;
const int _len;
// ...
};
This would be important if you want to ensure const-ness of these instance variables while computing their values at runtime, for example:
MyClass::MyClass() :
_capacity(someMethod()),
_data(someOtherMethod()),
_len(yetAnotherMethod())
{
}
const instances must be initialized in the initializer list or the underlying types must provide public parameterless constructors (which primitive types do).
I think this link http://www.cplusplus.com/forum/articles/17820/ gives an excellent explanation - especially for those new to C++.
The reason why intialiser lists are more efficient is that within the constructor body, only assignments take place, not initialisation. So if you are dealing with a non-built-in type, the default constructor for that object has already been called before the body of the constructor has been entered. Inside the constructor body, you are assigning a value to that object.
In effect, this is a call to the default constructor followed by a call to the copy-assignment operator. The initialiser list allows you to call the copy constructor directly, and this can sometimes be significantly faster (recall that the initialiser list is before the body of the constructor)
I'll add that if you have members of class type with no default constructor available, initialization is the only way to construct your class.
A big difference is that the assignment can initialize members of a parent class; the initializer only works on members declared at the current class scope.
Depends on the types involved. The difference is similar between
std::string a;
a = "hai";
and
std::string a("hai");
where the second form is initialization list- that is, it makes a difference if the type requires constructor arguments or is more efficient with constructor arguments.
The real difference boils down to how the gcc compiler generate machine code and lay down the memory. Explain:
(phase1) Before the init body (including the init list): the compiler allocate required memory for the class. The class is alive already!
(phase2) In the init body: since the memory is allocated, every assignment now indicates an operation on the already exiting/'initialized' variable.
There are certainly other ways to handle const type members. But to ease their life, the gcc compiler writers decide to set up some rules
const type members must be initialized before the init body.
After phase1, any write operation only valid for non-constant members.
There is only one way to initialize base class instances and non-static member variables and that is using the initializer list.
If you don't specify a base or non-static member variable in your constructor's initializer list then that member or base will either be default-initialized (if the member/base is a non-POD class type or array of non-POD class types) or left uninitialized otherwise.
Once the constructor body is entered, all bases or members will have been initialized or left uninitialized (i.e. they will have an indeterminate value). There is no opportunity in the constructor body to influence how they should be initialized.
You may be able to assign new values to members in the constructor body but it is not possible to assign to const members or members of class type which have been made non-assignable and it is not possible to rebind references.
For built in types and some user-defined types, assigning in the constructor body may have exactly the same effect as initializing with the same value in the initializer list.
If you fail to name a member or base in an initializer list and that entity is a reference, has class type with no accessible user-declared default constructor, is const qualified and has POD type or is a POD class type or array of POD class type containing a const qualified member (directly or indirectly) then the program is ill-formed.
If you write an initializer list, you do all in one step; if you don't write an initilizer list, you'll do 2 steps: one for declaration and one for asign the value.
There is a difference between initialization list and initialization statement in a constructor.
Let's consider below code:
#include <initializer_list>
#include <iostream>
#include <algorithm>
#include <numeric>
class MyBase {
public:
MyBase() {
std::cout << __FUNCTION__ << std::endl;
}
};
class MyClass : public MyBase {
public:
MyClass::MyClass() : _capacity( 15 ), _data( NULL ), _len( 0 ) {
std::cout << __FUNCTION__ << std::endl;
}
private:
int _capacity;
int* _data;
int _len;
};
class MyClass2 : public MyBase {
public:
MyClass2::MyClass2() {
std::cout << __FUNCTION__ << std::endl;
_capacity = 15;
_data = NULL;
_len = 0;
}
private:
int _capacity;
int* _data;
int _len;
};
int main() {
MyClass c;
MyClass2 d;
return 0;
}
When MyClass is used, all the members will be initialized before the first statement in a constructor executed.
But, when MyClass2 is used, all the members are not initialized when the first statement in a constructor executed.
In later case, there may be regression problem when someone added some code in a constructor before a certain member is initialized.
Here is a point that I did not see others refer to it:
class temp{
public:
temp(int var);
};
The temp class does not have a default ctor. When we use it in another class as follow:
class mainClass{
public:
mainClass(){}
private:
int a;
temp obj;
};
the code will not compile, cause the compiler does not know how to initialize obj, cause it has just an explicit ctor which receives an int value, so we have to change the ctor as follow:
mainClass(int sth):obj(sth){}
So, it is not just about const and references!