5 replies to this topic

[irancplusplus]

hi
we have a class like this:

class int2

{

public:

	int2(int a)

	{

	}

};

are these two lines:

int2 s(10);

int2 s = 10;

completely equivalent in any C++ compiler & any compiler options?

Edited by irancplusplus, 06 February 2012 - 07:04 AM.

[untitled_1]

those two lines are the same thing

[DRK]

Initialization by assignment operator = works only for 1-parameter constructor (like in your example). For constructor with more parameters only initialization by brackets () is possible.

[notes]

Also constructors initialization list accepts only bracket version.

class int2 {

    int a;

    public :

    int2(int b) : a(b)  

    {    


    } 


};

In this case while creating new int2 object first compiler will make operations on initialization list then everything in braces { } .

[DRK]

EDIT: Never mind.

[kernelcoder]

class int2
{
};

If you write above code, compilers (as far as I know any compiler) will generate 3 things for you -- i) default constructor, ii) copy constructor and iii) assignment operator (overloading). So the generated code will be something like following...

class int2
{
public:
int2(){} // This is the reason you are able to write 'int2 i2;'
int2(const int2 &other) {} // This is the reason you are able to write 'int2 i1; int2 i2(i1); int2 i2=i1; '
int2 operator=(const int2& other){} // This is the reason you are able to write 'int2 i1; int2 i2; i2 = i1; '
};

However, if you define any constructor of your own, compiler will not generate default constructor. If you write copy constructor or copy assignment operator of your own, compiler will not generate them.

So, in your case, yes, 'int2 s(10);' and 'int2 s = 10;' are completely equivalent.