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Little Quiz [message #57757]

Thu, 09 December 2021 09:57

mirek
Messages: 14291
Registered: November 2005

Ultimate Member

What does this mean:

DropList dl;
...
if(*~~dl == '$') ...

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Re: Little Quiz [message #57758 is a reply to message #57757]

Thu, 09 December 2021 10:19

Tom1
Messages: 1319
Registered: March 2007

Ultimate Contributor

Hi,

Is it the short version for e.g.:

if(((const char*)dl.Get())[0] == '$')

?

Best regards,

Tom

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Re: Little Quiz [message #57760 is a reply to message #57758]

Thu, 09 December 2021 10:51

mirek
Messages: 14291
Registered: November 2005

Ultimate Member

Tom1 wrote on Thu, 09 December 2021 10:19

Hi,

Is it the short version for e.g.:
if(((const char*)dl.Get())[0] == '$') 
?

Best regards,

Tom

Almost correct.

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Re: Little Quiz [message #57769 is a reply to message #57757]

Fri, 10 December 2021 22:22

Lance
Messages: 656
Registered: March 2007

Contributor

the inner ~ is

DropList::operator ~

which inherits

Value Ctrl::operator ~()(maybe const modifier?)

the outer ~ is

String Value::operator~()(const modifier?)

the * cause a auto type cast be called

String::operatr const char * ()(modifier?);

and then dereference of the returned constant char * returns the first char in the String.

something like

Value t1=~dl;
String t2=~t1;
const char* t3=(const char*)t2;
char t4=*t3;
if(t3=='$')

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Re: Little Quiz [message #57779 is a reply to message #57769]

Mon, 13 December 2021 15:53

mirek
Messages: 14291
Registered: November 2005

Ultimate Member

* is String::operator*

Otherwise solved Smile

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Re: Little Quiz [message #57800 is a reply to message #57779]

Sat, 18 December 2021 05:47

Lance
Messages: 656
Registered: March 2007

Contributor

Didnot realise there is an operator * defined.

BTW, if there is no operator *, will the route suggested by my previous post be taken? IE, will * cause a implicit type case operator by called? I guess so.

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Re: Little Quiz [message #57801 is a reply to message #57800]

Sat, 18 December 2021 09:07

mirek
Messages: 14291
Registered: November 2005

Ultimate Member

It indeed is. I guess that operator* is there from times when compilers were not up to the task.

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