one question

[ogedezvoltare]

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) : ifaces(i)

the code above is equivalent to:

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) {
iface=i;
}

or to

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) {
std::map <int, struct net_interface> &ifaces=1
}
thx

[weaknessforcats]

These are equivalent:

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) : ifaces(i)
 
the code above is equivalent to:
 
std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) {
iface=i;
}
 

The difference is in the first case ifaces is intialized when the member is created using a constructor, or equivalent. Whereas the second case takes an already initialized ifaces and uses the assignment operator to change the existing contents.

The first form is preferred.

This all assumes crouter is a constructor.If it's not, then the first case won't compile.

This code:

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) {
std::map <int, struct net_interface> &ifaces=1
}
 

uses a local variable ifaces which is set to 1. This won't compile as a map reference and a 1 are incompatible.

[ogedezvoltare]

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) {
std::map <int, struct net_interface> &ifaces=1
}
thx

this is actually :
std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) {
std::map <int, struct net_interface> &ifaces=i
}
sorry for the typing error

[ogedezvoltare]

These are equivalent:

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) : ifaces(i)
 
the code above is equivalent to:
 
std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) {
iface=i;
}
 

The difference is in the first case ifaces is intialized when the member is created using a constructor, or equivalent. Whereas the second case takes an already initialized ifaces and uses the assignment operator to change the existing contents.

The first form is preferred.

This all assumes crouter is a constructor.If it's not, then the first case won't compile.

This code:

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) {
std::map <int, struct net_interface> &ifaces=1
}
 

uses a local variable ifaces which is set to 1. This won't compile as a map reference and a 1 are incompatible.

What i don't understand is that ifaces is a reference and it has to be initializated with an std::map <int, struct net_interface> variable from what i knew.

PS:sorry for the typing error

[weaknessforcats]

Yes, references must be initialized at the time the refercne is created. That means this code:

std::map <int, struct net_interface> &ifaces;
 

won't compile as it stands. I took your examples a fer-instance.

With functions, the refernce argument is bound to the variable used to make the call. With class reference members, you must use the initializer list in the constructor to bind the reference to the constructor argument.

As with all things passed around a program, be sure the object of your reference is not deleted unexpectedly or you will crash. If this is an issue, consider using a handle (see the Bridge Design Pattern).

[ogedezvoltare]

OK so

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) : ifaces(i)

the code above is equivalent to:

crouter(std::map <int, struct net_interface> &i) {

std::map <int, struct net_interface> &ifaces=i
}
thx

[weaknessforcats]

No, the code is not equivalent.

This code:

crouter(std::map <int, struct net_interface> &i) {
 
std::map <int, struct net_interface> &ifaces=i
}
 

creates a local variable ifaces that is a reference and initializes it with i.
This is fine. However, this code:

std::map <int, struct net_interface> &ifaces;
crouter(std::map <int, struct net_interface> &i) : ifaces(i)
 

creates a reference ifaces without initializing it. This won't compile unless the reference is a member variable of crouter. Like this:

class crouter
{
    private:
    std::map <int, struct net_interface> &ifaces;
    public:
    crouter(std::map <int, struct net_interface> &i);
};
crouter::crouter(std::map <int, struct net_interface> &i)  : ifaces(i)
{
}