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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