Is this a function call or a variable declaration?

Question

I can not understand the meaning of the following code, please help me, thank you.

In the following code:

FrameDetect::Point FrameDetect::tracer(LabelData *ldata, int x, int y, int &pos, int lbl)
{
    for (int i=7; i>=0; i--)
    {
        int tx(x);
        int ty(y);
        nextPoint(tx, ty, pos);
        if (tx>0 && ty>0 && tx < bimg->width() && ty < bimg->height())
        {
            const int &l( ldata->at(tx, ty) );
            if (bimg->at(tx, ty) == ccolor && (l == 0 || l == lbl))
            {
                return Point(tx, ty);
            }
            if (bimg->at(tx, ty) == bcolor)
            {
                ldata->at(tx, ty) = -1;
            }
        }
        pos = (pos + 1)%8;
    }
    return Point(-1, -1);
}

int tx(x); is function call or variable declaration? Thanks for your help.

Source

It looks like C++. The method is called `tracer` and it's a member of the `FrameDetect` class. It returns a `Point` which is also a member of `FrameDetect`. In answer to their question: It's a function declaration, so, neither. — Olical, Jul 18 '13 at 10:47
Lots of good info here: http://stackoverflow.com/questions/5350125/constructor-of-type-int — sje397, Jul 18 '13 at 11:38

score 2 · Accepted Answer · answered Jul 18 '13 at 10:46

2

The same as

int tx = x;

"An int constructor"

answered Jul 18 '13 at 10:46

Joop Eggen

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Looks like author intended that, but it should be parsed as a function declaration – SpongeBobFan Jul 18 '13 at 11:01
@SpongeBobFan No, see Tadeusz Kopec's answer for why not. – sje397 Jul 18 '13 at 11:41

score 2 · Answer 2 · edited Jul 18 '13 at 11:43

2

It means declare an int type variable named tx. Invoke the constructor tx(x) to initialize tx, its value is x. The code can also written like this:

int tx = x;

edited Jul 18 '13 at 11:43

sje397

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answered Jul 18 '13 at 10:59

Douglas

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1

Fundamental types don't have constructors. – awesoon Jul 18 '13 at 11:25
@soon - I think they can: http://stackoverflow.com/questions/11117450/default-constructor-for-int – sje397 Jul 18 '13 at 11:45
@sje397, constructor is a member function. Does `int` have member functions? :) – awesoon Jul 18 '13 at 11:50
@soon - big debate here: http://stackoverflow.com/questions/5113365/do-built-in-types-have-default-constructors ... I guess it depends on whether you want to beleive SO or Bjarne himself :) – sje397 Jul 18 '13 at 11:53
1

@sje397, I of course believe Bjarne and SO, but in most case I believe the Standard. And I can't find any information about the existence of constructors for fundamental types. But you also right - that discussion is big, and I don't think we should debate it here, so, this is my last comment here. Have a nice day! – awesoon Jul 18 '13 at 12:19

score 2 · Answer 3 · answered Jul 18 '13 at 11:20

2

It is a variable declaration. It can't be parsed as a function declaration, because an expression in parenthesis does not name a type.
It can't be a function call either - the syntax is invalid. You can't write

double sin(2);

answered Jul 18 '13 at 11:20

Tadeusz Kopec for Ukraine

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score 1 · Answer 4 · answered Jul 18 '13 at 11:53

It's a copy constructor. In c++ the confusion arises when you declare a variable, with no parameters. In that situation you omit the brackets

I'll present several examples:

void afunction_thatDoesNothing(int x)
{
   int aFuncDecl();     //1: function declaration
   int aVariable;       //2: default construction of int
   int aValue1 = x;     //3: constructing with x
   int aValue2(x);      //4: constructing with x
   int aFuncDecl2(int); //5: declaration of a function taking an int
}

The only case above where there is a declaration vs initialization ambiguity is case 1 - in your code you've supplied a value typed expression to the constructor (case 4), and it can not be misinterpreted as a declaration.

score 0 · Answer 5 · answered Sep 28 '13 at 21:04

0

int tx(x); explanation.

int x(5); a variable x. and we are initilizing variabe at its creation time.

int x = 5;// in this statement we are assigning 5 to varible x. x in this case already declared. we updating its value just.

answered Sep 28 '13 at 21:04

Muhammad Danish

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Is this a function call or a variable declaration?

5 Answers5