Assignment initialization of a constructor

Question

I am trying out different versions of calling the constructor, and I came up with this

#include<iostream>
#include<string>
using namespace std;
class game{

    public:
        float version;
        string name;
        game()
        {
            name = "game";
            version = 1.0;
        }
        game(float v,string n)
        {
            version = v;
            name = n;
        }
        game(float v)
        {
            version = v;
            name="any";
        }
};
int main()
{
    game lol1(1.0,"league of legends"); //functional form
    game lol2 = 2.0;    //assignment form
    game lol3{3.0,"league2"}; //uniform initialization
    game *pt = &lol1;
    cout<<pt->name<<endl;
    return 0;
}

Every statement compiles, but if I write

 game lol2 = 2.0,"league of legends2"; //code 2

I get an error:

expected unqualified-id before string constant

But the following code works fine:

game lol2 = {2.0,"league of legends2"}; //code 3

I am not getting what exactly the issue is with the second code. Any ideas?

You can condense your constructors down to a single one using default parameters. — Sebastian Lenartowicz
– Sebastian Lenartowicz, Commented Sep 14, 2016 at 16:26
And for amusement, add a constructor that only takes a string .... — Andre Kostur
– Andre Kostur, Commented Sep 14, 2016 at 22:14

Cheers and hth. - Alf · Accepted Answer · 2016-09-14 16:48:30Z

11

What you call "assignment form" is a copy initialization.

It works as if a temporary object is constructed from a single argument specified as initializer, and is then passed to the class' copy constructor or move constructor.

So, the code

game lol2 = 2.0,"league of legends2"; //code 2

… is just syntactically invalid.

Tip: Instead of three constructors, where the second adds a first argument, and the third adds a second argument, you can just use default argument values:

class game{

public:
    float version;
    string name;
    game(float v = 1.0,string n = "game")
        : version( v ), name( n )
    {}
};

The : syntax is a constructor member initializer list.

It can sometimes be more efficient, sometimes necessary, and anyway is usually more concise and conventional.

edited Sep 14, 2016 at 16:48

answered Sep 14, 2016 at 16:27

Cheers and hth. - Alf

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4 Comments

Striezel Over a year ago

Just for completeness: It should be game(float v = 1.0,string n = "game"), not game(float v = 1.0,string n = "game), because the code won't compile with the missing quote.

Cheers and hth. - Alf Over a year ago

@Striezel: Thanks, fixed.

Anil Over a year ago

@Cheersandhth.-Alf ,Just little doubt, as u told "It works as if a temporary object is constructed from a single argument specified as initializer, so according the promblem ,A new object is created with version = 2.0 with code 2?

Cheers and hth. - Alf Over a year ago

@SubSea: The temporary is in practice always optimized away. That's called "elision" of the temporary and the copy or move constructor call. But still, since the construction of a temporary plus implicit call of copy or move constructor call is the spec, a copy or move constructor must be accessible and must not be declared explicit.

NathanOliver · Accepted Answer · 2016-09-14 16:44:16Z

4

You are not using the correct grammar to initialize a object from multiple values. When you do

game lol2 = 2.0,"league of legends2";

The grammer expects a variable name after the comma like

type name1 = value1, name2 = value2;

So you get an error as you have

type name1 = value1, value2;
                    ^ missing variable declaration here

When you have multiple variables that you need to construct with you can only use the folowing forms

type name = {value1, value2, ..., valuen};
type name{value1, value2, ..., valuen};
auto name = type{value1, value2, ..., valuen};
type name(value1, value2, ..., valuen);
auto name = type(value1, value2, ..., valuen);

edited Sep 14, 2016 at 16:44

answered Sep 14, 2016 at 16:29

NathanOliver

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5 Comments

Cheers and hth. - Alf Over a year ago

Are you sure it's the comma operator in this context? Do you get a complaint about the type of argument? Test this by adding a parenthesis around (then you do get comma operator for sure).

NathanOliver Over a year ago

@Cheersandhth.-Alf Oops. It is just bad grammar. fixing

Anil Over a year ago

@NathanOliver ,there is no error like missing variable declartion here ,but i got your point of constructin objects with multiple variables.

NathanOliver Over a year ago

@SubSea Yeah that is not what you will actually get. I was just putting it in human readable language.

Anil Over a year ago

@NathanOliver .okay.i got you..:)

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