.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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