Classes and variables scope in C++

Question

class car {
    int speed; 
    double position;

    public:
       car(int v,double d);
       int getspeed();
};

int car::getspeed() {
return speed;
}

car::car(int s, double x){
   speed=s;
   position=x;
}

Even though i specified different variables for car(int s,v), why does it work? i though it should give me a compile time error?

this code:which var it uses?

class car {
    int speed; 
    double position;

    public:
       car(int speed,double time);
       int getspeed();
};

int car::getspeed() {
   return speed;
}

car::car(int speed, double position){
   speed=speed;
   position=position;
}

I think the global variable might be used, or is it something you can't say certain about

actually i tired to do this.speed, and the compiler gave me an error, that "this" is not defined. I remember i was able to do "this" with out including any special headers, it seems now i can't. — Rave
– Rave, Commented Feb 1, 2012 at 21:45
@Rave: this is always a valid symbol in member function scope. Always. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Feb 1, 2012 at 22:04

user1073719 · Accepted Answer · 2012-02-01 21:52:03Z

7

car::car(int speed, double position){
   speed=speed;
   position=position;
}

In this function definition, it does nothing with the class member car::speed, and car::position, because you declared the local int speed and double position in the function parameter list, they hide the class member variables. To do it properly, you need explicitly say so:

car::car(int speed, double position){
   this->speed=speed;
   this->position=position;
}

answered Feb 1, 2012 at 21:52

user1073719

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Comments

gruszczy · Accepted Answer · 2012-02-01 22:35:39Z

4

The names of parameteres are not important. Types of parameteres create the signature. The signature is the same, so there is no compile error.

In second example speed in constructor will shadow speed atribute. Therefore you will assign parameter value to parameter variable. You need:

this->speed = speed;

And this is not guesswork ;-).

edited Feb 1, 2012 at 22:35

answered Feb 1, 2012 at 21:42

gruszczy

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1 Comment

Mark Smith Over a year ago

It took me a minute to figure out what your comment was... I've been in C# too much lately. I made the same mistake in a comment above.

Bo Persson · Accepted Answer · 2012-02-01 22:01:11Z

2

This constructor doesn't work

car::car(int speed, double position){
   speed=speed;
   position=position;
}

because it assigns the parameters to themselves.

This version does work because of the slightly odd scoping rules of a class

car::car(int speed, double position) : speed(speed), position(position)
{  }

answered Feb 1, 2012 at 22:01

Bo Persson

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Comments

Michael Chinen · Accepted Answer · 2012-02-01 21:43:44Z

1

The compiler doesn't care about your variable names in the method declaration, just the signature, which is

car::car(int,double)

for both your constructor declaration and your implementation, so it knows to match these up when linking. This is possible because you cannot have two methods in the class with the same signature. (You can do this with subclasses, but the result is an override).

answered Feb 1, 2012 at 21:43

Michael Chinen

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Comments

vmpstr · Accepted Answer · 2012-02-01 21:45:20Z

1

In the second case, I believe the closest scope variable is used. So, first it checks the local function scope, and finds both speed and position, so the search stops there. In effect, the second constructor isn't actually assigning obje

answered Feb 1, 2012 at 21:45

vmpstr

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Comments

bames53 · Accepted Answer · 2012-02-01 21:58:00Z

1

In your code:

car::car(int speed, double position){
   speed=speed;
   position=position;
}

you're just assigning each variable's value to itself. You can however do:

car::car(int speed, double position)
  :speed(speed)
  ,position(position)
{}

in addition to explicitly accessing the member variables via this->

answered Feb 1, 2012 at 21:58

bames53

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1 Comment

Rave Over a year ago

if i have a private member int *p; and in the constructor i do this; this->p=new int; to assign the value i end up doing this *p=value; is there a way to do that with "this"

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