- The definition of (non-virtual) "overriding" in C++ - 5 Updates
- smart pointers-to-char replacing strings - 1 Update
| JiiPee <no@notvalid.com>: Feb 01 10:59PM On 01/02/2015 22:54, Chris Vine wrote: >> You get my point? > No. Read the whole of §10.3 and it is quite clear. > Chris can you post it here please. I dont know where to find it. Again, I am talking about logic here. Yes, its possible the standard has not been written totally logically here... true, and that is another issue. But anyway, thats what I am discussing here. In a court of law so far there is no evidence that "override" belongs only to polynorphism :). |
| Chris Vine <chris@cvine--nospam--.freeserve.co.uk>: Feb 01 11:06PM On Sun, 01 Feb 2015 22:56:39 +0000 > seen the definition still yet. What "override" is defined in > polymorphism has been shown to me, but not what it means in > non-polymorphism situation/case. It has been explained to you concurrently by two people (me and another) that in §10.3/2 of C++11 the word "overrides" is a defined expression. Not an employed expression, but a _defined_ expression. And it is defined and applied exclusively and only in relation to virtual functions. That's the the end of it, from the point of view of C++. Get a good night's sleep as I think everything will become clear in the morning. You are chasing shadows. Chris |
| Paavo Helde <myfirstname@osa.pri.ee>: Feb 01 05:10PM -0600 > "overrides" A's foo, isnt it? am just talking about english language > grammar here, not really C++ language. So it would be natural for me > to say that even non-virtual function overrides. In order to see the "override" effect with the non-virtual functions you need also to use an object of another type (or a pointer or reference to another type). It is natural that an object of another type can have different member functions than the first type, even if they happen to share the same name and parameter list. In this sense, you are not overriding anything, you are just using another type. With virtual functions, the aim is to use pointers or references to the base type and get the functionality of the actual (most derived) type. This is quote another thing (and what OOP is mostly about). That said, I believe most C++ folks will understand what you mean when you are talking about overriding of non-virtual functions, just keep in mind this is not quite correct technically. Cheers Paavo |
| Paavo Helde <myfirstname@osa.pri.ee>: Feb 01 05:22PM -0600 > What am saying is that 10.3/2 there does not *define* "overriding" > *generally* in C++, it only defines its use with virtual functions > there. But how about for non-virtual functions, my question remains. The C++ standard only defines what is there. It does not define what is not there (that would be infinite amont of things!). And I searched through the standard and did not find any other definitions for "override". Is that sufficient to calm you down? For C++ standard, see http://lmgtfy.com/?q=latest+c%2B%2B+standard+download Cheers Paavo |
| JiiPee <no@notvalid.com>: Feb 01 11:32PM On 01/02/2015 23:22, Paavo Helde wrote: > there (that would be infinite amont of things!). And I searched through the > standard and did not find any other definitions for "override". Is that > sufficient to calm you down? there is nothing to calm down, we are just discussing about this matter :) > For C++ standard, see http://lmgtfy.com/?q=latest+c%2B%2B+standard+download ok, thanks. I ll download it |
| Paul <pepstein5@gmail.com>: Feb 01 03:22PM -0800 I am trying to rewrite text-handling code involving char* and const char* in terms of boost::scoped_array<char> Could someone possibly provide a reference to this topic? For example, the following function is wrong, and I don't really know why. boost::scoped_array<char> get(const char *s) { int size = std::strlen(s); boost::scoped_array<char> text(new char[size]); for(int i = 0; i < size; i++) text[i] = s[i]; return text; } I'm sure I just need to read up on it but google searches didn't seem to broach this topic. Many thanks. Paul |
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