问题
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- What is this weird colon-member (“ : ”) syntax in the constructor? 12 answers
What does the colon operator (\":\") do in this constructor? Is it equivalent to MyClass(m_classID = -1, m_userdata = 0);?
class MyClass {
public:
MyClass() : m_classID(-1), m_userdata(0) {
}
int m_classID;
void *m_userdata;
};
回答1:
This is an initialization list, and is part of the constructor's implementation.
The constructor's signature is:
MyClass();
This means that the constructor can be called with no parameters. This makes it a default constructor, i.e., one which will be called by default when you write MyClass someObject;.
The part : m_classID(-1), m_userdata(0) is called initialization list. It is a way to initialize some fields of your object (all of them, if you want) with values of your choice, instead of leaving them as undefined.
After executing the initialization list, the constructor body (which happens to be empty in your example) is executed. Inside it you could do more assignments, but once you have entered it all the fields have already been initialized - either to random, unspecified values, or to the ones you chose in your initialization list. This means the assignments you do in the constructor body will not be initializations, but changes of values.
回答2:
It is an initialization list.
By the time you get in the body of the constructor, all fields have already been constructed; if they have default constructors, those were already called. Now, if you assign a value to them in the body of the constructor, you are calling the copy assignment operator, which may mean releasing and reacquiring resources (e.g. memory) if the object has any.
So in the case of primitive types like int, there's no advantage compared to assigning them in the body of the constructor. In the case of objects that have a constructor, it is a performance optimization because it avoids going through two object initializations instead of one.
An initialization list is necessary if one of the fields is a reference because a reference can never be null, not even in the brief time between object construction and the body of the constructor. The following raises error C2758: 'MyClass::member_' : must be initialized in constructor base/member initializer list
class MyClass {
public :
MyClass(std::string& arg) {
member_ = arg;
}
std::string& member_;
};
The only correct way is:
class MyClass {
public :
MyClass(std::string& arg)
: member_(arg)
{
}
std::string& member_;
};
回答3:
It denotes the beginning of an initialiser list, which is for initialising member variables of your object.
As to: MyClass(m_classID = -1, m_userdata = 0);
That declares a constructor which can take arguments (so I could create a MyClass using MyClass m = MyClass(3, 4), which would result in m_classID being 3, and m_userdata being 4). If I were to pass no arguments to the MyClass constructor, it would result in an equivalent object being created to the version with the initialiser list.
回答4:
It signals the beginning of an initializer list.
Also it is not equivalent to MyClass(m_classId=-1,m_userData=0). This is attempting to define a constructor with 2 parameters that have default values. However the values lack types and it should not compile at all.
回答5:
It's a initialization list. In your example, it's rather something like this (something like this - doesn't mean it's equivalent in all cases):
class MyClass {
public:
MyClass(){
m_classID = -1;
m_userdata = 0;
}
int m_classID;
void *m_userdata;
};回答6:
That is called the member initialization list. It is used to call the superclass constrctors, and give your member variables an initial value at the time they are created.
In this case, it is initializing m_classID to -1 and m_userData to NULL.
It is not quite equivalent to assigning in the body of the constructor, because the latter first creates the member variables, then assigns to them. With the initialization, the initial value is provided at the time of creation, so in the case of complex objects, it can be more efficient.
回答7:
It isn't precisely an operator. It's a part of the syntax for a constructor.
What it is saying is that following it will be a list of member variables and their initial values.
Constant members have to be initialized this way. Non-constants can be initialized here too, as long as it can be done with a single expression. If it takes more code than that to initialize a member, you have to put actual code between the {}'s to do it.
A lot of people like to put pretty much all their constructor code in the initilizer list. I have one co-worker who regularly writes classes with several screens of initilizers, and then puts "{}" for the constructor code.
回答8:
Its the start of an initialiser list which sets member variables during the construction of the object. Your example "MyClass(m_classID = -1, m_userdata = 0);" is not possible as you have not defined the correct constructor and you would not be able to access the member variables in the parameter list anyway... you could have something like:
MyClass( int classId = -1, void* userData = 0 ) : m_classID(classId), m_userdata(userData) {}
The initialiser list is considered better than:
MyClass( int classId = -1, void* userData = 0 ) {
m_classID = classId;
m_userdata = userData;
}
Google for more info.
回答9:
In this case: Yes, ist is equivalent because only primitive types are concerned.
If the members are classes (structs) then you should prefer the initialization list. This is because otherwise the objects are default constructed and then assigned.
来源：https://stackoverflow.com/questions/1272680/what-does-a-colon-following-a-c-constructor-name-do