thats what i thought
On Mon, Aug 8, 2011 at 12:18 AM, Kamakshii Aggarwal
wrote:
> fun(&p) hona tha..
> but i wrote(*p)
>
>
> On Mon, Aug 8, 2011 at 12:17 AM, aseem garg wrote:
>
>> What was the mistake?
>> Aseem
>>
>>
>>
>> On Mon, Aug 8, 2011 at 12:15 AM, Kamakshii Aggarwal <
>> kamakshi...@gma
fun(&p) hona tha..
but i wrote(*p)
On Mon, Aug 8, 2011 at 12:17 AM, aseem garg wrote:
> What was the mistake?
> Aseem
>
>
>
> On Mon, Aug 8, 2011 at 12:15 AM, Kamakshii Aggarwal > wrote:
>
>> i got my mistake...sorry guys
>>
>>
>> On Mon, Aug 8, 2011 at 12:14 AM, Kamakshii Aggarwal <
>> kamaks
What was the mistake?
Aseem
On Mon, Aug 8, 2011 at 12:15 AM, Kamakshii Aggarwal
wrote:
> i got my mistake...sorry guys
>
>
> On Mon, Aug 8, 2011 at 12:14 AM, Kamakshii Aggarwal > wrote:
>
>> what is the problem with the following code
>>
>> void fun(int **p)
>> {
>> static int q = 10;
>>
>>
i got my mistake...sorry guys
On Mon, Aug 8, 2011 at 12:14 AM, Kamakshii Aggarwal
wrote:
> what is the problem with the following code
>
> void fun(int **p)
> {
> static int q = 10;
>
> *p = &q;
> }
>
> int main()
> {
> int r = 20;
> int *p = &r;
> fun(*p);
>
> printf("%d", *p);
> g
no, its GNU gcc compiler
On Mon, Feb 7, 2011 at 10:27 PM, jalaj jaiswal wrote:
> this error will be given by g++ compiler , i guess.. i also got d same
>
> On Mon, Feb 7, 2011 at 10:20 PM, tech rascal wrote:
>
>> but gcc compiler is giving error saying- can't convert int* to char* in
>> assignme
this error will be given by g++ compiler , i guess.. i also got d same
On Mon, Feb 7, 2011 at 10:20 PM, tech rascal wrote:
> but gcc compiler is giving error saying- can't convert int* to char* in
> assignment..so hw r u getting 566 as answer??
> ( I am using codeblocks on ubuntu )
>
>
> On
but gcc compiler is giving error saying- can't convert int* to char* in
assignment..so hw r u getting 566 as answer??
( I am using codeblocks on ubuntu )
On Mon, Feb 7, 2011 at 10:10 PM, jalaj jaiswal wrote:
> @ dave .. you r correct :)
>
>
> On Mon, Feb 7, 2011 at 9:41 PM, Dave wrote:
>
>>
@ dave .. you r correct :)
On Mon, Feb 7, 2011 at 9:41 PM, Dave wrote:
> @Jalaj: The results depend on the "endianness" of the addressing
> scheme on your hardware. (See http://en.wikipedia.org/wiki/Endianness.)
> The language standard does not specify a required endianness. You must
> be using
@Jalaj: The results depend on the "endianness" of the addressing
scheme on your hardware. (See http://en.wikipedia.org/wiki/Endianness.)
The language standard does not specify a required endianness. You must
be using a little-endian hardware, but the results would be different
on a big-endian syste
For more of these tricks one can see the first 10 lectures
from{exclude 1st one}
http://see.stanford.edu/see/lecturelist.aspx?coll=2d712634-2bf1-4b55-9a3a-ca9d470755ee
On 2/7/11, ranjane wrote:
> thank u jalaj
>
> On Feb 6, 10:03 pm, jalaj jaiswal wrote:
>> the logic is :-
>> int is stored in 32
thank u jalaj
On Feb 6, 10:03 pm, jalaj jaiswal wrote:
> the logic is :-
> int is stored in 32 bits in our systems
>
> 300 is 0001 00101100
> as ptr is character pointer, it points to lower 8 bits
> and when *++ptr=2 gets executed then 0001 changes to 0010(equal to
>
oh... why did u put %u.
i did not even notice that :P
--
Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
--
You received this message because you are subs
reason for that o/p is ...
coz range of short int is -32768 to 32767
and value of i start with i=0
and each time it will increment by 1
so corresponding value of i will be
1
2
3
.
.
.
32766
32767
-32768
-32767
.
.
.
.
-2
-1
but
coz in printf u r using %u so it will print...
1
2
3
.
.
.
3
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