Besides i prefer to write the following #define which is generic and works
in all case, i.e., it doesn't matter whether you pass pointers, float, int,
etc;
*#define swap( a, b )size_t t; t = a; a = b; b = t;*
1 more thing now you don't have to pass any other information i.e., whether
your
Please dont reply to thi post..wrng questionsry for that
On Fri, Oct 26, 2012 at 2:31 AM, rahul sharma wrote:
> address of each byte is printed...each byte is 32 bit long...so 32/4=8 hex
> for each row...but how these are separated with %.2x tin group of
> 2's%.2x means min. 2 width ..if
address of each byte is printed...each byte is 32 bit long...so 32/4=8 hex
for each row...but how these are separated with %.2x tin group of
2's%.2x means min. 2 width ..if <2..then leading zeros...
so it should be ("%.2x",start[i]);
start[i] in 2 but it is more so ignored...print start[i]s
As per o/p below:
00 00 80 3f
01 00 00 00
44 ff 28 00
01 00 00 00
first byte address is first row.
second byte address is second row.
third byte address third row
and so on
but how first row containg 3 values...as 1byte=2hexdigits..si only two
digits must be there.
and u said 44 ff 28 00 is ad
Sorry, I don't understand your question. *%.2x *is only a precision
specifier still.
(%.2x was used for neat formatting only, because you are printing the
values only 1 Byte long and a Byte can occupy at max 2digits in hex)
hex representated by 4 bits.
Yes hex is represented by 4 bits i.e. 1 B
Sorry, about that.
Read it as:
Yes a hex digit is represented by 4 bits but 1 Byte is being read using a
char pointer* and you're printing the values in those Bytes.
On 21 October 2012 01:03, Saurabh Kumar wrote:
> Sorry, I don't understand your question. *%.2x *is only a precision
> specifier s
Actually i have taken form http://www.geeksforgeeks.org/archives/730
Please explain me o/p...as hex representated by 4 bitsthen how cum is
following o/p
00 00 80 3f
01 00 00 00
44 ff 28 00
01 00 00 00
total we have to represent 32 bits and 8 bits in eachplz xplain
On Sun, Oct 21, 20
i - j is giving -3 as output. why is this so ?
On Tue, Oct 16, 2012 at 11:18 PM, iwill wrote:
> how to explain the output if instead of j-i we try to print i-j ?
>
>
> On Saturday, October 13, 2012 3:36:49 PM UTC+5:30, bharat wrote:
>>
>> #include
>> main()
>> {
>> int *i,*j;
>> i=(int*)60;
>>
how to explain the output if instead of j-i we try to print i-j ?
On Saturday, October 13, 2012 3:36:49 PM UTC+5:30, bharat wrote:
>
> #include
> main()
> {
> int *i,*j;
> i=(int*)60;
> j=(int*)71;
> printf("%d",j-i);
> }
>
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Sorry wrng question copiedits a silly question to be asked..sorry guys
On Sun, Oct 7, 2012 at 6:06 PM, rahul sharma wrote:
> #include
> int main()
> {
> int i;
> char ch;
>
> scanf("%c",&ch);
>
> printf("%d",ch);
> // getchar();
> getchar();
> }
>
> when i enter one dig
6/5=1, in integer division, how come you can think it as 2?
On Tuesday, 3 July 2012 13:22:07 UTC+5:30, rahul sharma wrote:
>
> #include
> #include
> int main()
> {
> int i;
> i=5;
> i=++i/i++;
> printf("%d",i);
> getch();
> }
>
> Why o/p is 1 and not 2?? what happened for p
@Kailash: The code violates the sequence point rule, which says that
you may not change the value of a variable more than once between
sequence points. The result is compiler dependent.
Dave
On Jan 21, 5:42 am, Kailash Bagaria
wrote:
> Please explain the output of the following program
>
> #
its right to left
On Wed, Sep 21, 2011 at 7:41 PM, kartik sachan wrote:
> thanks sukran...
> i have one more question evaluation of printf is also undefined in C??
>
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thanks sukran...
i have one more question evaluation of printf is also undefined in C??
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hmmm it is compiler dependent... try out with different compilers... but
even though dey give same results,standard c left this eature as undefined
for efficiency purposes
On Wed, Sep 21, 2011 at 7:16 PM, kartik sachan wrote:
> @sukran if it is undefined in ansi c so it can return any value???
>
@sukran if it is undefined in ansi c so it can return any value???
at complilaton.but all time it is giving me same value
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IT is not the case ... I already said acc to ANSI c result is undefined
!! not 0 by dault
On Wed, Sep 21, 2011 at 6:50 PM, kartik sachan wrote:
> ans in both cases is zero
> i think if return is written and nthg specified value is given it will
> return by default zero
>
> --
> You re
ans in both cases is zero
i think if return is written and nthg specified value is given it will
return by default zero
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@Wujin: Okay. b is but an integer, yet it is being printed as a long
long. Since there are no implicit type conversions in procedure calls
such as the printf statement, b is taken as a long long, not of an
int. So b and the next 32-bits are printed. The format that these are
printed in is a figment
@Wujin: Okay. But b is an integer and it is being printed as a long
long. Since there are no implicit type conversions in procedure calls
such as the printf statement, b is taken as a long long, not of an
int. So b and the next 32-bits are printed.
Dave
On Sep 19, 8:18 pm, wujin chen wrote:
> @D
@Dave printf("a=%x, b=%llx",a,b,c); i think c will be ignored~~ , and the
output is a=9,b=10
2011/9/19 Dave
> @Wujin: What do you expect the output to be? How does it differ from
> what you actually get?
>
> Dave
>
> On Sep 18, 8:47 am, wujin chen wrote:
> > usigned long long x = 0x12345678;
thanks siddharam,but i am form computers .I have given my interview.I
heard that C-dot asks questions from networking.But since networking
was not in our syllaybus i was too worried as i had only 2-3 hrs to
read about that .I took an idea about networking,but wasn't confident
that i will be able
abhishek can you plz explain
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For more opti
According to K and R c book , when the return value is specified and yet we
fail to return a value, it is undefined. It is compiler dependent.
So i feel there is no meaning in thinking abt it
On Mon, Sep 19, 2011 at 4:38 PM, abhishek wrote:
> @ sukran
> if it is giving same ans then there has
@ sukran
if it is giving same ans then there has to be some reason,
On Sep 19, 12:45 pm, sukran dhawan wrote:
> common guys its undefined acc to standard c
>
> On Mon, Sep 19, 2011 at 12:36 PM, Siddhartha Banerjee <
>
>
>
>
>
>
>
> thefourrup...@gmail.com> wrote:
> > on running,every ti
output is
18 9 0
0
0
8
the same was expected.
printf uses stack when it has multiple arguments to print
On Sep 18, 1:19 pm, Bhavesh agrawal wrote:
> another que..
>
> #include
> main()
> {
> int a;
> int i=10;
> printf("%d %d %d\n",i+++i,i,i---i);
> printf("%d\n",i---i);
> a
@Wujin: What do you expect the output to be? How does it differ from
what you actually get?
Dave
On Sep 18, 8:47 am, wujin chen wrote:
> usigned long long x = 0x12345678;
> int a = 0x09;
> int b = 0x10;
> printf("a=%x, b=%llx",a,b,c);
>
> the result is: a=9,b=123456780010
>
> i wonder why~~
it's running..
http://ideone.com/XjGIo
wat is does is that through the system function u can run commands as
u do in command prompt.. nd dis is d command to show the files n all
as ur question asked.
try it on dev C++.
On Sep 17, 9:45 pm, teja bala wrote:
> @ pooja
>
> in borland c++ 4.5 compiler
@ pooja
in borland c++ 4.5 compiler version its giving linker error,
wat actually ur code does?
On Sat, Sep 17, 2011 at 9:27 PM, pooja wrote:
> i tried it for Windows.
>
> #include
>
> using namespace std;
> int main()
> {
> char ch;
> int i=system("dir /s |more");
> ch=getchar();
i tried it for Windows.
#include
using namespace std;
int main()
{
char ch;
int i=system("dir /s |more");
ch=getchar();
}
On Sep 17, 6:46 pm, teja bala wrote:
> you have to print the list of all the files in a directory and all
> its sub directories?
--
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This is not a C puzzle. It depends on operating system.
On Sep 17, 3:46 pm, teja bala wrote:
> you have to print the list of all the files in a directory and all
> its sub directories?
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+1 dave.
Sanju
:)
On Tue, Sep 6, 2011 at 3:40 PM, Dave wrote:
> @Srivastav: Yeah. You need more parens:
>
> printf("%d",(int)(3.14*6.25*6.25));
>
> Without the extra parens, the 3.14 is cast to an int, but then
> implicitly recast to a double for the multiplications. With the
> parens, the p
@Srivastav: Yeah. You need more parens:
printf("%d",(int)(3.14*6.25*6.25));
Without the extra parens, the 3.14 is cast to an int, but then
implicitly recast to a double for the multiplications. With the
parens, the product is formed in type double, and then the result is
cast into integer.
Sorry
printf behaves abnormally when it sees arguments not matching with its
datatype
@dave
printf("%d",(int)3.14*6.25*6.25); is also giving 0.
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@Sisavknesh: The product of those integers is 122.65625, which is 122
and 21/32. It turns out that the low order 42 bits of the mantissa are
zero. Since numbers on PCs are stored little-endian, and because you
asked for an integer format conversion, the first 32 of those 42 bits
are treated as an i
thanks a lot guys :)
On Sep 6, 11:48 am, Rajeshwar Patra wrote:
> pointer points to read only memory location
> and this address is then compared whch evaluates to true
> hence the output...
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@nitesh good explanation brother
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For mo
@teja ...nitesh hv correctly explained it
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output is :
-9-11-13-15
first j=19222 then
find(19222)=find(1922)-2
find(1922)=find(192)-2
find(192)=find(19)-2
find(19)=find(1)-9
find(1) returns 0 then
find(19)=0-9=-9
find(192)=-9-2=-11
find(1922)=-11-2=-13
find(19222)=-13-2=-15
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@neha could u xplain it?
On Sun, Sep 4, 2011 at 12:05 PM, Neha Gupta wrote:
> o/p : -9-11-13-15
>
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o/p : -9-11-13-15
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but o/p is -15 -13 -11 -9
On Sun, Sep 4, 2011 at 3:38 AM, vikas wrote:
> f(19222)
> |
> |
> f(1922) -11 -2 =-13
> |
> |
> f(192) = -9 -2 = -11
> |
> |
> f(19) = f(1) - 9 = -9
> |
> |
> f(1) = 0
>
> output -9 -11 -13
>
>
> On Sep 3, 10:29 pm, teja bala wrote:
> > Find the output of the
f(19222)
|
|
f(1922) -11 -2 =-13
|
|
f(192) = -9 -2 = -11
|
|
f(19) = f(1) - 9 = -9
|
|
f(1) = 0
output -9 -11 -13
On Sep 3, 10:29 pm, teja bala wrote:
> Find the output of the following code - plzzz xplain the o/p
> int find(int j)
> {
> if(j>1)
> {
> j=find(j/10)-(j%10);
> printf
because your are trying to edit an unmodifiable object which C-
compiler will not allow
On Sep 3, 6:18 pm, priya ramesh
wrote:
> ok but y does the compiler say lvalue required if you perform a++??
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it will be 4 3
On Aug 31, 9:01 pm, aditi garg wrote:
> sorry my mistake...
> it wud be 4 3
>
> second output is quite obvious as it will count till it get string termination
and in first one it is a pointer to character so it will be of size 4.
>
>
>
>
>
>
>
> On Wed, Aug 31, 2011 at 9:30 PM,
yes i agree since in the question they dint mention so we can solve it the
other way also .
On Sun, Aug 28, 2011 at 7:31 PM, saurabh singh wrote:
> Well if you go on using system calls functions etc the job can easily be
> done in many ways but the real idea behind the quine problem is the good
Well if you go on using system calls functions etc the job can easily be
done in many ways but the real idea behind the quine problem is the good old
spirit of a challenge (not merely for an interview) So the best way remains
the store everything way...
On Sun, Aug 28, 2011 at 3:11 PM, aditya kuma
one more method : use system funtion .
eg system("cat filename") //system funtion executes the command
On Sun, Aug 28, 2011 at 1:51 PM, UTKARSH SRIVASTAV
wrote:
> good logic
>
>
> On Sun, Aug 28, 2011 at 12:14 AM, Piyush Grover > wrote:
>
>> yeah you can do that by opening the file and printing
good logic
On Sun, Aug 28, 2011 at 12:14 AM, Piyush Grover
wrote:
> yeah you can do that by opening the file and printing it but as far as I
> know, interviewer adds the constraint of not using the file method.
>
>
> On Sun, Aug 28, 2011 at 12:35 PM, rahul sharma wrote:
>
>> this logic is ok...bu
yeah you can do that by opening the file and printing it but as far as I
know, interviewer adds the constraint of not using the file method.
On Sun, Aug 28, 2011 at 12:35 PM, rahul sharma wrote:
> this logic is ok...but we have pre defined everything in char f...if i
> add one or two more stateme
this logic is ok...but we have pre defined everything in char f...if i
add one or two more statements then it will require corresponding
change in char *f...can i open the same file with f open n prin t it
out?
On Aug 28, 11:31 am, Piyush Grover wrote:
> char*f="char*f=%c%s%c;main(){printf(f,
got it...thnkx
On Aug 28, 11:31 am, Piyush Grover wrote:
> char*f="char*f=%c%s%c;main(){printf(f,34,f,34,10);}%c";
>
> f is a global pointer to the char array which contains the string
> "char*f=%c%s%c;main(){printf(f,34,f,34,10);}%c"
>
> Now in main function you are printing this string with arg
char*f="char*f=%c%s%c;main(){printf(f,34,f,34,10);}%c";
f is a global pointer to the char array which contains the string
"char*f=%c%s%c;main(){printf(f,34,f,34,10);}%c"
Now in main function you are printing this string with arguments 34,f,34,10.
ASCII value of " is 34.
->So in f, the first %c is
i m not able to understand hw it wokrs..elaborate it
On Aug 28, 11:16 am, rahul sharma wrote:
> plz expalin char*f=" "
>
> On Aug 28, 11:12 am, Piyush Grover wrote:
>
>
>
>
>
>
>
> > it's a quine problem.
>
> > char*f="char*f=%c%s%c;
> > main(){
> > printf(f,34,f,34,10);}%c";
>
>
plz expalin char*f=""
On Aug 28, 11:12 am, Piyush Grover wrote:
> it's a quine problem.
>
> char*f="char*f=%c%s%c;
> main(){
> printf(f,34,f,34,10);}%c";
>
> main()
> {
> printf(f,34,f,34,10);
>
> }
>
> I have used whitespaces to make it understand.
>
> On Sun, Aug 28, 2011 at 11:39
@Rajesh: abs() can be done without conditional operations. There
probably are many ways. The first two that come to mind are:
abs(x) = (x >> 31) & x | ~(x >> 31) & ~x
abs(x) = (x >> 31) ^ x + (x >> 31) & 1
Dave
On Aug 26, 2:56 am, rajesh singarapu wrote:
> abs function itself has a condition??
abs function itself has a condition??
I think it is not that good solution.
On Wed, Aug 24, 2011 at 6:33 PM, priyanka raju wrote:
>
>
> int a,b,max,min;
> max=(a+b+abs(a-b))/2;
> min=(a+b-abs(a-b))/2;
>
> --
> cheers
> priyanka
>
> --
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C-Dot procedure (For Computers):
- First shortlist on the basis of cgpa/ %age for interviews.
- Only one round of interview, based on your resume. Discussion of projects
and questions related to mainly OS, Networking and Programming.
There is nothing as selection of "Toppers" only. If you are goo
On Thu, Aug 25, 2011 at 2:04 AM, Ninad Page wrote:
> If you stick to one convention its not that difficult to code without
> bothering about endian's.In your case if you want to have a variable data
> size for int(As far as I can guess),develop ur own integer interface using
> raw bytes like you
Using getchar() after the first scanf ll be much better...!!
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On Aug 24, 11:41 pm, Arun Vishwanathan wrote:
> @don: so if a 16bit value is put into a 32 bit field and i need to read the
> value, do i need to read last 16 bits only somehow ?
>
>
>
>
>
>
As long as you are saving those 16 bits in a variable (an integer, a
bit-field or otherwise), you don't
On Aug 24, 6:00 pm, Sanjay Rajpal wrote:
> after first scanf, place a call to fflush(stdin).
>
> you will get the result.
>
> Sanju
> :)
>
>
>
Flushing the input stream is undefined.
>
>
>
>
> On Wed, Aug 24, 2011 at 5:58 AM, Mehnaaz wrote:
> > #include
> > #define max 30
> > int no_p;
> > i
@don: so if a 16bit value is put into a 32 bit field and i need to read the
value, do i need to read last 16 bits only somehow ?
On Wed, Aug 24, 2011 at 8:22 PM, Don wrote:
> It is most common to use 4 bytes to store an integer value, even if
> the full range will not be used. There is no proble
It is most common to use 4 bytes to store an integer value, even if
the full range will not be used. There is no problem putting a 16-bit
value into a 32-bit field. The only case where this is not true is
when memory is extremely limited and you need to pack as much into
every word as possible. Do
thanks guys!
On Wed, Aug 24, 2011 at 6:37 PM, Don wrote:
> Yes, the memory provided by malloc will not be in the structure. Only
> the pointer to that memory will be in the structure. The size of a
> struct is defined at compile time, so it can't be dynamically sized at
> run time.
>
> struct ju
Yes, the memory provided by malloc will not be in the structure. Only
the pointer to that memory will be in the structure. The size of a
struct is defined at compile time, so it can't be dynamically sized at
run time.
struct junk
{
int size;
int *data;
};
Somewhere in the code:
struct junk m
that was actually a mistake , i should have taken the value before i
declared it BUT as for arrays goes, we know we can access the a[10]
even though we may just declare a[5] ! may be its the same reason i
was able to read in "for" loop besides here having the size of
column matters more than th
See if we use dynamic memory allocation then still the size of pointer will
be 4 bytes only
Mean that int* pointer still have the size equals to pointer ... malloc only
returns new alloted memory which is now only *pointed *by that pointer
check this out :- http://www.ideone.com/20ayq
On Wed
@ Mehnaaz :the variable no_p will be equal to 0,since it is an
external one
so the declaration char p[no_p][max] is equivalent to p[0][max];
and size of p is zero.
then how you can insert anything into it
as you are doing in for loop?
no_p may receive some non zero value afterwards but array p is
If you are working in C++, stl has a vector container class which will
do this. Otherwise, declare an integer pointer in the struct and use
malloc to allocate memory for it. Then you can use it like an array.
Don
On Aug 23, 11:51 pm, Arun Vishwanathan wrote:
> say that you have a structure with s
The problem is that scanf("%c" will take whatever character is next,
even if it is not the one you think you typed. When you typed your
first entry, the number "3" and then pressed "Return", the "\n"
character is left in the buffer, so your scanf gets that character
instead of the one you want. The
@ vikas ..the space actually worked !!..is there any
explanation ??..thanks a ton!
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i hav executed it in dev..
it waits for second scanf
On Wed, Aug 24, 2011 at 7:05 PM, Mehnaaz wrote:
> @lakshay and sanju.after placing fflush(stdin), i am continuing to get
> the same output...is there anything else i can do..? and will you also
> please execute and see
>
> --
> You received th
@lakshay and sanju.after placing fflush(stdin), i am continuing to get
the same output...is there anything else i can do..? and will you also
please execute and see
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nice solution priyanka :)
Sanju
:)
On Wed, Aug 24, 2011 at 6:03 AM, priyanka raju wrote:
>
>
> int a,b,max,min;
> max=(a+b+abs(a-b))/2;
> min=(a+b-abs(a-b))/2;
>
>
> --
> cheers
> priyanka
>
> --
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int a,b,max,min;
max=(a+b+abs(a-b))/2;
min=(a+b-abs(a-b))/2;
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cheers
priyanka
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@Abhishek:
int c = (a - b) >> 31;
max = c & b | ~c & a;
Explanation: c = 0 if a >= b, else c = all one bits.
Then c & b = 0 if c = 0, but c & b = b if c = all ones, i.e. if the
max is b,
and ~c & a = a if c = 0, i.e., if the max is a, but ~c & a = 0 if c =
all ones.
Dave
On Aug 23, 8:07 am, Abh
can any1 tell me wat type of question are asked in TCS nd INFOSYS interview.
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algog
Moreover they also asked what is .NET as if they heard it for the first
timecome on...can you believe it...!!!
On Tue, Aug 23, 2011 at 4:18 PM, Dheeraj Sharma wrote:
> yeah..the interview was total nakli :D .. they always..take the toppers..
> :) i mean..among the top ten who are sitting..
>
yeah..the interview was total nakli :D .. they always..take the toppers.. :)
i mean..among the top ten who are sitting..
@Sachin:- they also asked , what is Drishti ? :D
On Tue, Aug 23, 2011 at 1:54 PM, Sanjay Rajpal wrote:
> 7.7 LPA @ NIT Kurukshetra.
>
>
> Sanju
> :)
>
>
>
> On Mon, Aug 22, 20
7.7 LPA @ NIT Kurukshetra.
Sanju
:)
On Mon, Aug 22, 2011 at 10:47 PM, siddharam suresh
wrote:
> how much they are offering ?
> Thank you,
> Siddharam
>
>
>
> On Tue, Aug 23, 2011 at 11:12 AM, ranjith kumar > wrote:
>
>>
>>
>> They shortlist candidates based on cgpa and select the highest cgp
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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how much they are offering ?
Thank you,
Siddharam
On Tue, Aug 23, 2011 at 11:12 AM, ranjith kumar
wrote:
>
>
> They shortlist candidates based on cgpa and select the highest cgpa
> candidate.
>
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They shortlist candidates based on cgpa and select the highest cgpa
candidate.
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alg
+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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Thnks:)
On Sat, Aug 20, 2011 at 1:00 AM, sachin sabbarwal
wrote:
> they ask very easy questions in interviews.
> i had faced their interview.
> first of all they asked me if i had done any project work.
> then they asked about the platform used and why i had chosen it.
> then few easy questions a
they ask very easy questions in interviews.
i had faced their interview.
first of all they asked me if i had done any project work.
then they asked about the platform used and why i had chosen it.
then few easy questions about algo of the project.
they questioned me on bubble sort.
then i was aske
No the warning in gcc is
ibm1.c: In function ‘main’:
ibm1.c:7:19: warning: initialization discards qualifiers from pointer target
type
On Wed, Aug 17, 2011 at 11:50 AM, venkat wrote:
> yes u r correct
>
> On Aug 16, 8:22 pm, Sanjay Rajpal wrote:
> > This is becuase "Hello" is a constant strin
yes u r correct
On Aug 16, 8:22 pm, Sanjay Rajpal wrote:
> This is becuase "Hello" is a constant string and constant strings get stored
> in *Data Area, not in stack for the function you called. *Thats why pointer
> to constant string will be returned and program will not produce any error.
>
> S
it is compiler dependent
On Aug 1, 5:20 pm, thanu moorthy wrote:
> Please help me...
>
> How can the following output be obtained :
>
> 1.main()
>
> {
>
> int i=1;
>
> printf("%d\t%d\t%d\t",i,i++,i);
>
> }
>
> output: 2 1 2
>
> 2.main()
>
> {
>
> int i=1;
>
> printf("%d\t%d\t%d\t",i,++i,i);
>
what is the logic of your concept
On Aug 13, 5:01 pm, Kunal Patil wrote:
> @rohit: Cast pointer to an integer into an int to get what you are
> expecting.
>
> for e.g.
>
> printf("%d",(int)&a[4]-(int)&a[0]);
>
> This will give 16.
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compiler dependent and will print some address
On Sun, Aug 14, 2011 at 11:48 AM, sukran dhawan wrote:
> try out with Microsoft VC++
>
>
> On Sun, Aug 14, 2011 at 11:47 AM, Brijesh Upadhyay <
> brijeshupadhyay...@gmail.com> wrote:
>
>> I think i got it... "STRING" always return address of S , whi
try out with Microsoft VC++
On Sun, Aug 14, 2011 at 11:47 AM, Brijesh Upadhyay <
brijeshupadhyay...@gmail.com> wrote:
> I think i got it... "STRING" always return address of S , which then get
> summed with 1 and 2.
>
> --
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I think i got it... "STRING" always return address of S , which then get
summed with 1 and 2.
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@rohit: Cast pointer to an integer into an int to get what you are
expecting.
for e.g.
printf("%d",(int)&a[4]-(int)&a[0]);
This will give 16.
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