Can somebody explain this short code

[Tony]

Note this code is autogenerated
What I don't understand is this
How is it possible to call function dsFrsvoyage_ttFrsvoyage when the
signature doesn't match.
The function dsFrsvoyage_ttFrsvoyage doesn't any argument but you can call
it with an argument.


function dsFrsvoyage()
{
this.ttFrsvoyage = new Array(1);
this.ttFrsvoyage[0] = new dsFrsvoyage_ttFrsvoyage(1);
}

function dsFrsvoyage_ttFrsvoyage()
{
this.Vesselid = "";
this.Arrtime = null;
this.Deptime = null;
this.Objecttype = "";
this.Dnggoods = false;
this.prevport = "";
this.port = "";
this.nextport = "";
this.VoyNum = "";
this.done = false;
this.sent = "";
this.todestination = "";
this.fromdestination = "";
this.Remark = "";
this.errstat = "";
this.cancelled = false;
this.Sendstat = "";
this.Objectid = 0;
this.UID = "";
this.UIDDATE = null;
this.UIDTIME = "";
this.UIDPGM = "";
this.Z_Seq = 0;
this.MISC1 = "";
this.MISC2 = "";
this.MISC3 = "";
this.UIDsign = "";
this.UIDupdate = null;
this.UIDuppgm = "";
this.UIDuptime = "";
}

//Tony

[Tony Mountifield]

In article <knkp6o$sov$1...@dont-email.me>,

Yes, JS does not do any argument prototype checking, and you don't need
to name the parameters in the function definition.

The arguments are placed in an automagically-created object called arguments,
and then the elements are placed in any named parameters there might be.

function foo(a,b) {
// a contains arguments[0]
// b contains arguments[1]
// further arguments are just in arguments[2] onwards.
}

foo(11,22,33,44); // a=arguments[0]=11, b=arguments[1]=22, arguments[2]=33, arguments[3]=44

Since in your example, the generated function doesn't use arguments[],
the 1 passed in is ignored.

> //Tony

// Another Tony :)
--
Tony Mountifield
Work: to...@softins.co.uk - http://www.softins.co.uk
Play: to...@mountifield.org - http://tony.mountifield.org