to clear ur these concepts , i think u should refer to Programming in C ,
Schaum series, Byron S Gottfried, and then Ritchie book. They are clearly
given there.
Sanju
:)
On Mon, Aug 22, 2011 at 11:25 PM, Vijay Khandar wrote:
> Thanks ...got it..
>
>
> On Tue, Aug 23, 2011 at 11:30 AM,
Thanks ...got it..
On Tue, Aug 23, 2011 at 11:30 AM, binayakranjan das wrote:
> In this case = has right to left associativity and as such < has no
> associativity.but,the parsing occurs from left to right.so first (x is checked which evaluates to 0 then (0 that is what is assigned to i
In this case = has right to left associativity and as such < has no
associativity.but,the parsing occurs from left to right.so first (x wrote:
> main()
> {
> int x=10,y=10,z=5;
> int i=x pf("\n%d",i);
>
> }
>
> o/p is 1 .pls any1 explain me hw is it printing?
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+1 to 16.
whats the confusion here?
On 21 August 2011 16:44, sagar pareek wrote:
> Arey yaar just see the my post...
> i explained it step by step :)
>
>
> On Sun, Aug 21, 2011 at 4:43 PM, Sanjay Rajpal wrote:
>
>> 16
>>
>> 8*(8-1*(8-1-1))
>>
>>
>> Sanju
>> :)
>>
>>
>>
>> On Sun, Aug 21, 2011
Arey yaar just see the my post...
i explained it step by step :)
On Sun, Aug 21, 2011 at 4:43 PM, Sanjay Rajpal wrote:
> 16
>
> 8*(8-1*(8-1-1))
>
>
> Sanju
> :)
>
>
>
> On Sun, Aug 21, 2011 at 4:10 AM, Anjul Sharma wrote:
>
>> is 336 the answer??
>>
>> On Aug 21, 11:10 am, SuDhir mIsHra wrote:
16
8*(8-1*(8-1-1))
Sanju
:)
On Sun, Aug 21, 2011 at 4:10 AM, Anjul Sharma wrote:
> is 336 the answer??
>
> On Aug 21, 11:10 am, SuDhir mIsHra wrote:
> > #include
> > #define FUNC1(i) (i*(i-1))
> > #define FUNC2(i) (i==0?1:i*FUNC1(i-1))
> > main()
> > {
> > int i=8;
> >
> > p
is 336 the answer??
On Aug 21, 11:10 am, SuDhir mIsHra wrote:
> #include
> #define FUNC1(i) (i*(i-1))
> #define FUNC2(i) (i==0?1:i*FUNC1(i-1))
> main()
> {
> int i=8;
>
> printf("\n%d",FUNC2(i));
>
>
>
>
>
>
>
> }
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but these questions don't have an algo
On Sun, Aug 7, 2011 at 2:08 PM, Gary Drocella wrote:
> I thought this was algogeeks, not company question geeks.
>
> On Aug 7, 4:27 pm, UTKARSH SRIVASTAV wrote:
> > please these questions are compiler dependent and have no standard
> > answers...th
I thought this was algogeeks, not company question geeks.
On Aug 7, 4:27 pm, UTKARSH SRIVASTAV wrote:
> please these questions are compiler dependent and have no standard
> answers...these are rarely asked by companies
>
>
>
>
>
>
>
>
>
> On Sun, Aug 7, 2011 at 1:23 PM, Gary Drocella wro
please these questions are compiler dependent and have no standard
answers...these are rarely asked by companies
On Sun, Aug 7, 2011 at 1:23 PM, Gary Drocella wrote:
> @puneet The provided faq is garbage, if you want to learn about the
> semantics of the C programming
> language, then re
@puneet The provided faq is garbage, if you want to learn about the
semantics of the C programming
language, then refer to this original ISO spec here
http://www.open-std.org/jtc1/sc22/wg14/www/docs/n1124.pdf
I also suggest that for all programming languages (OCaml, Ruby, lua
script, etc)
It is de
Also guys, this link:
http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=expr
discusses erroneous expns like
a[i]=i++;
But if u run this code..this gives no error on GNU compiler..
So, are we really referring to a reliable document here..?
http://c-faq.com/~scs/cgi-bin/faqcat.cgi?sec=expr
I really do
@ Amit: Well, the link you have posted refers that i++*i++ is not an
invalid expression, it just runs differently on different OS's because
every OS has a different implementation on how to solve such
expressions that use the same address space(Address of the integer i).
As far as i know the value
What Amit told is exactly correct. But I would like to know the
expression evaluation order of this in gcc and turboc
Arun
On Aug 3, 6:15 pm, Arun Vishwanathan wrote:
> @amit:+1
>
> On Wed, Aug 3, 2011 at 3:14 PM, amit karmakar
> wrote:
>
>
>
>
>
>
>
> > You are wrong.
> > The above program in
@amit:+1
On Wed, Aug 3, 2011 at 3:14 PM, amit karmakar wrote:
> You are wrong.
> The above program invokes undefined behavior. Read the standard
> language draft to know about sequence points, side effects and
> undefined behavior.
>
> Between a previous and next sequence point a variable's value
You are wrong.
The above program invokes undefined behavior. Read the standard
language draft to know about sequence points, side effects and
undefined behavior.
Between a previous and next sequence point a variable's value cannot
be modified twice.
c-faq should be quite useful
http://c-faq.com/~s
x is a pointer which can be changed from out side and stores the
pointers for int*
On Jun 8, 10:58 am, Vishal Thanki wrote:
> Following declaration makes the "x" as a volatile pointer to an integer.
>
> int *volatile x;
>
> But what does following means?
>
> int **volatile x;
>
> ~Vishal
--
You
hi pramod,
thanks for the direction. i didnt know abt function objects before.definitely helped me.
On 3/2/06, pramod <[EMAIL PROTECTED]> wrote:
you can have function objects, i.e., objects which behave likefunctions with overloading the () operator. Define a base class like"base_functor" and for e
you can have function objects, i.e., objects which behave like
functions with overloading the () operator. Define a base class like
"base_functor" and for each different function pointer, define a new
class which will store this function pointer and whose operator () will
take the same parameters
Arun wrote:
> icant use void *, because,i wudnt know how to typecast back.
> for eg: if map[a]=char * (*fptr1)(char*,int*)
> map[b]=float (*fptr2)(obj1)
> so on
> if my string is b then map[b] will give some void ptr, how do i know to
> typecast it to float (*fptr1)(ob
icant use void *, because,i wudnt know how to typecast back.
for eg: if map[a]=char * (*fptr1)(char*,int*)
map[b]=float (*fptr2)(obj1)
so on
if my string is b then map[b] will give some void ptr, how do i know to typecast it to float (*fptr1)(obj1)?
i have about 10 dif
I couldn't ensure that what problem be with you.
If you need some pointer point to many kind of types or
User-defined type, you could need void* within your declaration.
But using it be careful, since any type of pointer
could be accept by complier .
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