it violates sequence pt. rule..so output is compiler dependent , but
as there is Lvalue error it would compile fine.
but in prev case pre decrement expects Lvalue but has r-value instead
bcoz of the post increment.
On 7/9/12, md shaukat ali <ali.mdshau...@gmail.com> wrote:
> then atul what would be the output of this prob...
> int a=10;
> int b=a++*a--;
> prinf ("%d",b);
> On Sun, Jul 8, 2012 at 11:52 PM, atul anand <atul.87fri...@gmail.com>
> wrote:
>
>> b=--a--; this will result into compiler error because 1st the post
>> decrement will occur and value will be saved in a temp variable . but
>> you cannot apply pre decrement on temp variable.
>>
>> On 7/8/12, vindhya chhabra <vindhyachha...@gmail.com> wrote:
>> > int a=10;
>> > int b;
>> > b=--a--;
>> > printf("%d %d",a,b);. l value error in this ques..
>> >
>> > On Sun, Jul 8, 2012 at 11:48 PM, vindhya chhabra
>> > <vindhyachha...@gmail.com>wrote:
>> >
>> >> .i think there will be an error in this -l value required, as post
>> >> increment has more precedence than pre increment
>> >>
>> >>
>> >> On Sun, Jul 8, 2012 at 11:44 PM, ashish jain
>> >> <ashishjainco...@gmail.com>wrote:
>> >>
>> >>> I think it should output:
>> >>> 9 9
>> >>>
>> >>>
>> >>> On Sun, Jul 8, 2012 at 11:42 PM, md shaukat ali
>> >>> <ali.mdshau...@gmail.com>wrote:
>> >>>
>> >>>> but i am confused in this problem...
>> >>>> int a=10;
>> >>>> int b;
>> >>>> b=--a--;
>> >>>> printf("%d %d",a,b);..what will output?
>> >>>>
>> >>>>
>> >>>> On Sun, Jul 8, 2012 at 11:39 PM, md shaukat ali
>> >>>> <ali.mdshau...@gmail.com
>> >>>> > wrote:
>> >>>>
>> >>>>> agree with adarsh
>> >>>>>
>> >>>>>
>> >>>>> On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar
>> >>>>> <algog...@gmail.com>wrote:
>> >>>>>
>> >>>>>> Sorry, its 6/6 and not 6/5,
>> >>>>>>
>> >>>>>> regds.
>> >>>>>>
>> >>>>>> On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar
>> >>>>>> <algog...@gmail.com>wrote:
>> >>>>>>
>> >>>>>>> Firstly, this is ambiguous and expressions with multiple
>> >>>>>>> increment/decrement operators will get executed according to the
>> >>>>>>> compiler.
>> >>>>>>>
>> >>>>>>> Even if you consider the normal way, as we(humans) percieve it,
>> >>>>>>> it
>> >>>>>>> will be evaluated as
>> >>>>>>> (++i)/(i++), which is 6/5, which is 1.
>> >>>>>>>
>> >>>>>>> Simple!
>> >>>>>>>
>> >>>>>>>
>> >>>>>>>
>> >>>>>>> On Sun, Jul 8, 2012 at 10:23 PM, rahul sharma <
>> >>>>>>> rahul23111...@gmail.com> wrote:
>> >>>>>>>
>> >>>>>>>> int i=5;
>> >>>>>>>> i=++i/i++;
>> >>>>>>>> print i;
>> >>>>>>>>
>> >>>>>>>>
>> >>>>>>>> i=1
>> >>>>>>>>
>> >>>>>>>> how?
>> >>>>>>>>
>> >>>>>>>> --
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>> >>
>> >>
>> >> --
>> >> Vindhya Chhabra
>> >>
>> >>
>> >>
>> >>
>> >
>> >
>> > --
>> > Vindhya Chhabra
>> >
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