Chose C as language not c++

On Mon, Jan 2, 2012 at 4:54 PM, mandeep <[email protected]> wrote:

> hi..take a look at this now  http://codepad.org/jHR8O5FT
>
>
> On Mon, Jan 2, 2012 at 9:04 PM, Oussama Bounaim <[email protected]>wrote:
>
>> Take a look at http://stackoverflow.com/q/8702511/759916.
>>
>>
>> On Mon, Jan 2, 2012 at 10:35 AM, Ernest Galbrun <
>> [email protected]> wrote:
>>
>>> It is more liquely a  wchar_t  (wide character) which is the default
>>> value for a litteral char. It is 32 bits large with gcc (it is
>>> compiler-dependent).
>>> --
>>> *Ernest Galbrun*
>>>
>>>
>>>
>>> On Mon, Jan 2, 2012 at 10:26, Carlos Guia <[email protected]>wrote:
>>>
>>>> I guess the compiler treats 'a' as an integral constant and ends up
>>>> using int to represent it.
>>>>
>>>> Carlos Guía
>>>>
>>>>
>>>>
>>>> On Mon, Jan 2, 2012 at 2:51 AM, Shoubhik <[email protected]> wrote:
>>>>
>>>>>    #include<stdio.h>
>>>>>
>>>>>    int main()
>>>>>    {
>>>>>
>>>>>            char ch;
>>>>>            fflush(stdin);
>>>>>            ch=getchar();
>>>>>            printf("ch= %d a=%d char=%d",
>>>>> sizeof(ch),sizeof('a'),sizeof(char));
>>>>>
>>>>>
>>>>>    }
>>>>>
>>>>> I type in 'a' (without quotes) as input , and the output I got in my
>>>>> ***gcc version 4.5.1*** is :
>>>>>
>>>>> ch= 1 a=4 char=1
>>>>>
>>>>> My question is :
>>>>>
>>>>> If sizeof(c) is 1 , then how can sizeof('a') be 4 ?
>>>>>
>>>>> --
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>>
>>
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>> Find me on Twitter @obounaim <https://www.twitter.com/obounaim>.
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