Yes, you are right. Pre increment simply increments and returns
*reference*to same object.
somewhat like:
int& operator++(int &n){
n = n+1;
return n;
}
On 31 October 2012 02:08, rahul sharma <rahul23111...@gmail.com> wrote:
> @saurabh..thnx....plz provide me code snippet for pre increment also..it
> will help a lot..i can guess it will do n++ and return itself..plz give
> code snippet...
>
>
> On Mon, Oct 29, 2012 at 10:50 AM, Saurabh Kumar <srbh.ku...@gmail.com>wrote:
>
>> Post increment returns temp object because it has to return the old value
>> of object not the new incremented one:
>> Simply put, the code for post increment somewhat goes like this:
>>
>> int operator++ (int &n){
>> int temp = n;
>> n = n+1;
>> return temp;
>> }
>>
>> that's also the reason you cannot use it as a lvalue in an exprression,
>> as you would be assigning nowhere.
>>
>> On 27 October 2012 20:09, rahul sharma <rahul23111...@gmail.com> wrote:
>>
>>> But y post returns temp. object
>>>
>>>
>>> On Fri, Oct 26, 2012 at 8:18 PM, Saurabh Kumar <srbh.ku...@gmail.com>wrote:
>>>
>>>> i++: Post increment can't be a lvalue because Post increment internally
>>>> returns a temporary object (NOT the location of i) hence you cannot assign
>>>> anything to it. The result of i++ can be used only as a rvalue.
>>>> ++i: whereas, in Pre-increment value gets incremented and the same
>>>> location is returned back to you, hence can be used as lvalue in an
>>>> expression.(C++ allows this)
>>>> Both can be used as rvalues though.
>>>>
>>>> On 26 October 2012 18:54, rahul sharma <rahul23111...@gmail.com> wrote:
>>>>
>>>>> why pre inc. is l value and post is r value..please explain
>>>>>
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