@rahul According to C specification, half filled array will be filled with
value 0. In your example you are setting str[0] as 'g' and str[1] as 'k'.
So the compiler sets str[2....9] as 0. So you string str becomes
{'g', 'k', '\0', '\0', '\0', '\0', '\0', '\0', '\0', '\0'}
Confusion is arising from the fact that you have created an array of 10
elements.
To answer you original question "gk" is inherently {'g', 'k', '\0'} and has
size 3 while {'g', 'k'} has size 2.
Regards,
Sachin
On Saturday, October 6, 2012 9:34:30 PM UTC+5:30, rahul sharma wrote:
>
> #include<stdio.h>
>
>
> int main()
> {
> char str[10]={'g','k'};
> char str1[10]="gh";
> int i;
> for(i=0;str1[i]!=NULL;i++)
> printf("%c",str[i]);
> getchar();
> }
>
> NUll is there in character array also...make clear me...
>
> On Sat, Oct 6, 2012 at 9:22 PM, rahul sharma <[email protected]<javascript:>
> > wrote:
>
>> int main()
>> {
>> char str[10]={'g','k'};
>> char str1[10]="gh";
>>
>>
>> printf("%s",str);
>> printf("%s",str1);
>> getchar();
>> }
>> then how does this work???
>> str printing gk...then NULL is automatically appended in this also...plz
>> tell
>>
>>
>> On Sat, Oct 6, 2012 at 6:33 PM, Rathish Kannan
>> <[email protected]<javascript:>
>> > wrote:
>>
>>> For string, C appends '\0' internally. hence sizeof(str) returned the
>>> value 3.
>>> str1 is char array with two character. hence sizeof(str1) returned the
>>> value 2.
>>>
>>> -- RK :)
>>>
>>>
>>> On Sat, Oct 6, 2012 at 5:53 PM, rahul sharma
>>> <[email protected]<javascript:>
>>> > wrote:
>>>
>>>> char str[]="ab";
>>>> char str1[]={'a','b'};
>>>>
>>>> sizeof(str) ...o/p is 3
>>>> sizeof(str1)....o/p is 2..
>>>>
>>>> Why so????
>>>> plz explain...
>>>>
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>>
>
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