@programming love:I can understand what u say but my doubt is that for the
first output which is 2, according to your example, p has address 10 which
it points to.And as u say int * tends to dereference 2 bytes so that wud be
now 10 and 11.finally char * takes only 1 byte so why is value at 11
I believe you are assuming little endian right ?
On Mon, Aug 15, 2011 at 2:14 PM, programming love
love.for.programm...@gmail.com wrote:
The internal representation of array is this:
suppose that the address starts from decimal number 10 and integer occupies
2 bytes
10- 0002 ( num 2 in
#includestdio.h
#define fun(arg) do\
{\
if(arg)\
printf(have fun...,\n);\
}while(i--)
main()
{
int i=6;
fun(i5);
}
give the answer and please give the reason for this
#includestdio.h
#define fun(a,b) a##b
main()
{
int a, b, ab;
a = 1, b = 2, ab
#includestdio.h
main()
{
int arr[3]={2,3,4};
char *p;
p=arr;
p=(char *)((int *)(p));
printf(%d,*p);
p=(char *)((int *)(p+1));
printf(%d,*p);
}
it is giving 2,0 why it is giving 0 ..??
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I think dis is bec int occupies 4 bytes while char occupies 1 byte so in the
memory when we save as int dey are saved as 2000 3000 4000
now whn u take a char pointer pointing to dis array and u increment it by 1
dey will move only by 1 byte and thus u get 0...
u can verify the result by removing
@nitin--
http://www.crazyengineers.com/forum/computer-science-engineering/43458-what-use-c-language.html
On Mon, Aug 15, 2011 at 1:51 PM, hary rathor harry.rat...@gmail.com wrote:
main()
{
int i=6;
do
{
if(i5)
printf(have fun...,\n);
I think while assignment the type conversion will take place. Whether you
write it or not.
check this out: http://ideone.com/y36vj
^^ Just giving off warnings, but it is working.
On 15 August 2011 14:14, aditi garg aditi.garg.6...@gmail.com wrote:
I think dis is bec int occupies 4 bytes while
The internal representation of array is this:
suppose that the address starts from decimal number 10 and integer occupies
2 bytes
10- 0002 ( num 2 in hex)
12- 0003 ( num 3 in hex)
14- 0004 ( num 4 in hex)
Now p points to address 10 and is type char. (Even after type casts) p+1
will increment
#includestdio.h
main()
{
short int i=0;
for(i=5 i=-1;++i;i0)
printf(%u\n,i);
printf(\n);
return 0;
}
o/p is 1.
4294967295 hw???
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its an infinite loop
On Tue, Aug 2, 2011 at 12:21 AM, jagrati verma
jagrativermamn...@gmail.comwrote:
#includestdio.h
main()
{
short int i=0;
for(i=5 i=-1;++i;i0)
printf(%u\n,i);
printf(\n);
return 0;
}
o/p is 1.
it is nt an infinite loop it ends at 4294967295
On Tue, Aug 2, 2011 at 12:53 AM, vijay goswami vjrockks...@gmail.comwrote:
its an infinite loop
On Tue, Aug 2, 2011 at 12:21 AM, jagrati verma
jagrativermamn...@gmail.com wrote:
#includestdio.h
main()
{
short int i=0;
for(i=5
u save its output in a file and note nos. from 32676 to a large value
4294967295 are missing in this..
On Tue, Aug 2, 2011 at 12:59 AM, jagrati verma
jagrativermamn...@gmail.comwrote:
it is nt an infinite loop it ends at 4294967295
On Tue, Aug 2, 2011 at 12:53 AM, vijay goswami
Hi,
I was using a different ID for browsing and commenting on this group
but since yesterday, I am getting a notification that the moderator
has banned me from using this group. I wanted to know the reason why I
was banned and whether this was in error because I don't remember
doing anything
only reason why any of the moderators must have banned you is either for
advertising kind of mails, or abusive mails, spamming..
I hope I get a reply to this instead of this ID also getting banned oh
come on there must have been something that got you banned, anyway you are
not banned
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