char ch
ch=i++[s];
printf(%c,ch); this will print i[s],then i is incrementrd after
assigning to ch
ch=++i[s];// this will inccrement value at i[s]
My question is what is role of priority which is making them behaving
differentI am not getting y not first i is incremented then i[s]
ch=i++[s]; // in this value is assigned first and then increment will
take place...bcozz you are using post increment.
here i does not have any other option it has to do post increment
before [] comes...but it will not assign value to 'i' ( i.e
incremented 'i' value)
so compiler will do something
use radix sort 4 times. using bucket sort with chaining as the sorting
algorithm.
time complexity= O(4(n+k))
space complexity = O(n)
Ashima
M.Sc.(Tech)Information Systems
4th year
BITS Pilani
Rajasthan
On Sun, Sep 25, 2011 at 6:01 AM, Jasveen Singh jasveen.sing...@gmail.comwrote:
i have a
yes i can do that but how do u want me to proceed if u can give me the code
to tell me how to fetch data from the file and use bucket sort on it
it will be a huge favour and am i right at my approach to use 2d array to
store the values
plz tell me sir plz!!
On Sun, Sep 25, 2011 at 1:08 PM, Ashima
i have a file 'star.txt' containing recordss in the form name score as given
below
*name score*
dave 52.67
steve60.09
and so on till 64 names
i have to make a program to read and sort these names in a rank
just show not to store
all scores are in float and need to rank
oh i'm sorry guys galz,its my mistake..actually i forgot to add one
more field in struct
this is the correct one
typedef struct
{
int bit1:29;
int bit2:4;
}bit;
int main()
{
printf(%d\n,sizeof(bit));
return 0;
}
now what will be the output...i'm sure you will amazed to see the result
The result will depend on alignment : byte alignment or Word alignment.
Sanju
:)
On Wed, Aug 31, 2011 at 2:51 AM, PRATEEK VERMA prateek...@gmail.com wrote:
oh i'm sorry guys galz,its my mistake..actually i forgot to add one
more field in struct
this is the correct one
typedef struct
@ prateek i am still not geeting why now out put is 4
but if we change second bit2:5 then out put again 8
plzz explain throughly
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@kartik:
typedef struct
{
int bit1:29;
int bit2:4;
}bit;
This makes total number of bits 33, which is not on int boundry. (multiple
of 32 bits)
So to make it aligned on int boundry, 31 extra bits are padded at the end
of bit2.
This makes size of struct 29 + 4 + 31 = 64 bits -- 8 bytes.
When bit1
why it(total) should be a multiple of 32 bits..can u clarify it?
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becoz int is of 4btyes so 32 bits required
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@kartik it's wrong output...output will be 8 bytes.I asked this question
just to give more clear picture of bit field
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@prateek:can u pls explain y the o/p will be 8??according to me also o/p
should be 4.
On Tue, Aug 30, 2011 at 11:42 PM, PRATEEK VERMA prateek...@gmail.comwrote:
@kartik it's wrong output...output will be 8 bytes.I asked this question
just to give more clear picture of bit field
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@prateek check this i have complied on ideone
http://ideone.com/UV7MG
output is 4 only
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it will be 00.
Bit fields are allocated within an integer from least-significant to
most-significant bit. In the following code
struct mybitfields
{
unsigned short a : 4;
unsigned short b : 5;
unsigned short c : 7;
} test;
int main( void );
{
test.a = 2;
test.b = 31;
class{
int bit1:1;
}
please explain the meaning of line 2.
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int bit1:1;
above definition instructs compiler that use 1 bit for storing
integer(signed by default) in bit1.
Such notation is used for optimization of memory use.
So if someone stores 1 in bit1 then,it will be treated as -1 by compiler
since we have instructed the compiler to store it in one
Why does the following code not give an error?
#includestdio.h
void fun(char *a) //array decays to pointer here
{
a[0] = 'a';
printf(%s\n,a);
}
int main()
{
char p[]=hello;
fun(p);
getch();
}
if i write something like this in main
char *arr=hello
arr[0] = 'r';
cz when u do char p[]=hello compiler automatically allocates enough
memory(6 charc here) and puts your string there..
but whn u do char*a=hello then its complr dependent where d string will be
storedsome compiler stores it in a special location in data segmnt whch
is (read only )so
Thanks for your reply himanshu, ...I am aware of the above stated fact but
my doubt is that when i pass this char array to a function and accept it as
a char pointer (char *a), and then when write a[0] = 'a', why don't i get an
error?
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@rohit: why do you think this should give error???
On Sun, Aug 28, 2011 at 5:17 PM, rohit raman.u...@gmail.com wrote:
Thanks for your reply himanshu, ...I am aware of the above stated fact but
my doubt is that when i pass this char array to a function and accept it as
a char pointer (char
because when u pass an array, the address of memory location that is recievd
in the func is accessible and write allowed also to u..so u dont get an
error
On Sun, Aug 28, 2011 at 5:17 PM, rohit raman.u...@gmail.com wrote:
Thanks for your reply himanshu, ...I am aware of the above stated
wen u pass an array as parameter to a func, it does decay to a pointer, nd
compiler doesn't stop u to change contents of the array through this
pointer.
basically u can change or not the value at some memory location depends on
how u have allocated it not how u r accessing it.
On Sun, Aug 28,
@PRATEEK if i make int bit1:2 then it will store 2 bytes
for example if bit1=4;then out will be 0 or 1
my question is it storing number from least significant bit or most
significant bit???
suppose for 4 0100 so out will be 00 or 01??
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say that you have a structure with some fields of known size and unknown
size.For example, a char, an integer and an integer array.I do not know the
size of the integer array until the user mentions the number of bytes he
needs this integer array to cover in the command line as an argument.Is it
u will be taking pointer in structure and use malloc to allocate memory for
the number of integer u need
On Wed, Aug 24, 2011 at 10:21 AM, Arun Vishwanathan
aaron.nar...@gmail.comwrote:
say that you have a structure with some fields of known size and unknown
size.For example, a char, an
Suppose i create a block of 10 ints, p is a pointer pointing to this
block.
int (*p)[10];
How can i initialize these 10 integer blocks using pointer p.
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#includestdio.h
main()
{
int a[10],i;
int (*p)[10];
p=a;
for( i=0;i10;i++)
{
*((int *)p+i)=i;
}
for( i=0;i10;i++)
{
printf( %d,a[i]);
}
}
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CSE-3
B-Tech 3rd Year
@MNNIT
#includestdio.h
#includeconio.h
struct abc
{
char p[1];
};
int main()
{
struct abc *c =NULL;
char name[]=abcde;
c = (struct abc *) malloc (sizeof(struct abc));
strcpy(c-p,name);
printf(%s,c-p);
getch();
return 0;
}
why it is printing the whole string
this is a case of heap corruption
On Mon, Aug 8, 2011 at 10:40 PM, Mohit Goel mohitgoel291...@gmail.com wrote:
#includestdio.h
#includeconio.h
struct abc
{
char p[1];
};
int main()
{
struct abc *c =NULL;
char name[]=abcde;
c = (struct abc *) malloc (sizeof(struct
when you do maloc or new , space is allocated from heap segment of
your process address space.
Though you have alllocated (sizeof(struct abc) space but strcpy copies
abcde beyond your allocated space and overwrite the contents, we
should avoid this scenario and use strncpy so that it cannot go
Hi All,
In the code below, even though pD is actually pointing to a base
class object, how is the print function being successfully called?
#includeiostream
class base
{
public:
int BVal;
base()
{
BVal = 100;
}
};
class derived : public base
{
public:
How do we declare an array of N pointers to functions that return int??
An array of N pointers to integers is this: int *p[N]
Just can't figure out the case for pointers to function
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This following code produces output as 8 4 4.
8 is fine, as the size of an int plus a pointer is 8, but why the size of p and
q is being shown as 4?
#includestdio.h#includestdlib.hint main(){
struct node
{
int data;
struct node * link;
};
struct
u hv answered ur ques in ur ques itself...:)
c u understand size of struct node as 4+4=8 rite?? now in this 4 is size of
int and 4 is the size of the pointer of the type struct node...
now wat are p and q? dey are also pointers to struct node jst like
link...now whn u understand dat the size of
On Tue, Jul 26, 2011 at 1:43 AM, Nitish Garg nitishgarg1...@gmail.comwrote:
This following code produces output as 8 4 4.
8 is fine, as the size of an int plus a pointer is 8, but why the size of p
and q is being shown as 4?
#includestdio.h#includestdlib.hint main(){
struct node
My bad! Thanks.
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thnx all...i got it :)
On Thu, Jul 14, 2011 at 2:13 AM, Anika Jain anika.jai...@gmail.com wrote:
As in base 10 we say 552.5 is same as 5.525 *10^2 , here 2 is the exponent
of base 10.. so similarly in binary nos. for base 2 here if i make
101.00110011.. to 1.0100110011.. *2^2 my exponent
binary equivalent of 5.2 is
101.0011001100110011001100110011(nonterminating)..
now it is actually stored in normalised frorm in 32 bits..
like this
--1 bit for sign8 bits for exponent-23 bits for
fraction--
this is from higher order byte to lower order for little endian..
Shouldn't the value of 1100 be -64
On Wed, Jul 13, 2011 at 4:53 PM, Anika Jain anika.jai...@gmail.com wrote:
binary equivalent of 5.2 is
101.0011001100110011001100110011(nonterminating)..
now it is actually stored in normalised frorm in 32 bits..
like this
--1 bit for sign8
sorry its 0100 not 1100 coz 5.2 is a positive no. so sign bit is 0
On Wed, Jul 13, 2011 at 7:58 PM, Piyush Kapoor pkjee2...@gmail.com wrote:
Shouldn't the value of 1100 be -64
On Wed, Jul 13, 2011 at 4:53 PM, Anika Jain anika.jai...@gmail.comwrote:
binary equivalent of 5.2
why is the order of numbers reversed?
On Wed, Jul 13, 2011 at 8:07 PM, Anika Jain anika.jai...@gmail.com wrote:
sorry its 0100 not 1100 coz 5.2 is a positive no. so sign bit is 0
On Wed, Jul 13, 2011 at 7:58 PM, Piyush Kapoor pkjee2...@gmail.comwrote:
Shouldn't the value of
it is not unsigned its signed but positive thts y it is 0..
the order isnt reversed.. look at the code, the code 1st prints lowest order
byte then next and then next and then highest soo output is 102 102 -90 64
On Wed, Jul 13, 2011 at 8:08 PM, Piyush Kapoor pkjee2...@gmail.com wrote:
why is
I think it is machine dependent and current output is for a little endian
machine
and output should be reverse for a big endian machine
On Wed, Jul 13, 2011 at 10:18 PM, Anika Jain anika.jai...@gmail.com wrote:
it is not unsigned its signed but positive thts y it is 0..
the order isnt
ya thts ryt, for big endian it will be 64 -90 102 102
On Wed, Jul 13, 2011 at 10:22 PM, sunny agrawal sunny816.i...@gmail.comwrote:
I think it is machine dependent and current output is for a little endian
machine
and output should be reverse for a big endian machine
On Wed, Jul 13, 2011 at
Will u guyz pls tell me frm where do u study terms like endian ,it is a
pity i had to google it :( :(
On Wed, Jul 13, 2011 at 10:30 PM, Anika Jain anika.jai...@gmail.com wrote:
ya thts ryt, for big endian it will be 64 -90 102 102
On Wed, Jul 13, 2011 at 10:22 PM, sunny agrawal
computer organization carl hamacher. computer system architecture by Morris
Mano. Computer organization and architecture by william stallings.. if u
dont have these refer wikipedia.
On Wed, Jul 13, 2011 at 10:59 PM, Piyush Kapoor pkjee2...@gmail.com wrote:
Will u guyz pls tell me frm where do u
thanks,i hv just entered 2nd year,so most probably i will study this
year
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@anika:can u please explain the meaning of this line..
so i here get an exponent of 2 as 2 here.. now in exponent 8 bits this
exponent is stored as 127+exponent so here it becomes 1001..
On Wed, Jul 13, 2011 at 11:44 PM, Piyush Kapoor pkjee2...@gmail.com wrote:
thanks,i hv just entered 2nd
As in base 10 we say 552.5 is same as 5.525 *10^2 , here 2 is the exponent
of base 10.. so similarly in binary nos. for base 2 here if i make
101.00110011.. to 1.0100110011.. *2^2 my exponent is 2.. now in storage of
floats exponent is stored as exponent +127 i.e here 2 +127 so we get
1001
#includestdio.h
main(){int a=10,b;
a=5?b=10:b=20;printf
http://www.opengroup.org/onlinepubs/009695399/functions/printf.html(%d\n,b);}
y this is asking for lvalue
while this(below) not?
#includestdio.h
main(){int a=10,b;
a=5?b=10:(b=20);printf
this is something like this thats why it is giving lvalue required.
u r going to store the the value over value which is not a variable at left
hand side.
main(){int a=10,b;
(a=5?b=10:b)=20;printf
http://www.opengroup.org/onlinepubs/009695399/functions/printf.html(%d\n,b);}
On Mon, Jul 11, 2011
remember in C , the false statement cannot be used for assignment, if u
write it without braces.It is allowed in C++ but not in C
In C compiler would treat it as what aditya has explained.hence the error.
On Mon, Jul 11, 2011 at 12:59 PM, aditya pratap contacttoadity...@gmail.com
wrote:
this
@vaibhav.i don't think there is any rule like you mentioned.if it is
plz mention it's source
i think the explanation by aditya is correct it's just due to the precedence
order which is
=
?:
=
in descending order
hence first = is evaluated then ?: followed by =...due to which lvalue
Can anybody give a full explanation
On Sat, Jul 9, 2011 at 10:49 PM, sunny agrawal sunny816.i...@gmail.comwrote:
try to find out the binary representation of float value 5.2
On Sat, Jul 9, 2011 at 10:46 PM, Sangeeta sangeeta15...@gmail.com wrote:
int main(){
int i;
float a=5.2;
char
Sunny is right.
Try to observe the binary representation, and you shall get your answers.
Regards,
Sandeep Jain
On Mon, Jul 11, 2011 at 5:54 PM, Piyush Kapoor pkjee2...@gmail.com wrote:
Can anybody give a full explanation
On Sat, Jul 9, 2011 at 10:49 PM, sunny agrawal
On Mon, Jul 11, 2011 at 5:54 PM, Piyush Kapoor pkjee2...@gmail.com wrote:
Can anybody give a full explanation
http://ideone.com/K1QmV
On Sat, Jul 9, 2011 at 10:49 PM, sunny agrawal sunny816.i...@gmail.comwrote:
try to find out the binary representation of float value 5.2
On Sat, Jul 9,
class a
{
int x;
public:
a()
{
}
a(int i){x=i;coutin a xendl;}
a(a obj){coutin copy cons of aendl;}
};
a obj1=14; //error no matching call to a::a(a)
why.
and just adding a const in the constructor saves me from error...but
how
--
use a(int arg)
{
x = arg;
}
ur call will work...:)
On Sun, Jul 10, 2011 at 11:46 PM, himanshu kansal
himanshukansal...@gmail.com wrote:
class a
{
int x;
public:
a()
{
}
a(int i){x=i;coutin a xendl;}
a(a obj){coutin copy cons of aendl;}
my badadd const in copy construcori think...that compiler expect...
On Sun, Jul 10, 2011 at 11:48 PM, rahul rahulr...@gmail.com wrote:
use a(int arg)
{
x = arg;
}
ur call will work...:)
On Sun, Jul 10, 2011 at 11:46 PM, himanshu kansal
himanshukansal...@gmail.com wrote:
a obj3(obj1);but this statement works fine.so it means it is calling
copy constt. perfectly...
On Sun, Jul 10, 2011 at 11:49 PM, rahul rahulr...@gmail.com wrote:
my badadd const in copy construcori think...that compiler expect...
On Sun, Jul 10, 2011 at 11:48 PM, rahul
The reason is... that when u write
a obj1=14;
it is same as writing a obj1 = a(14);
So first a temporary object is created using the constructor
a(int i)
And this temporary object is passed in the copy constructor. BUT since it is
temp object it must be referred by a const alias.
Regards,
thanku sir...sir 1 more thngcn u gv a link or some pdf for studying
virtual inheritance elaborating the vptr mechanism more clearly...
On Sun, Jul 10, 2011 at 11:56 PM, Sandeep Jain sandeep6...@gmail.comwrote:
The reason is... that when u write
a obj1=14;
it is same as writing a obj1 =
http://www.parashift.com/c++-faq-lite/virtual-functions.html
Its one of my favorite sites... :)
Regards,
Sandeep Jain
On Mon, Jul 11, 2011 at 12:02 AM, himanshu kansal
himanshukansal...@gmail.com wrote:
thanku sir...sir 1 more thngcn u gv a link or some pdf for studying
virtual
int main(){
int i;
float a=5.2;
char *ptr;
ptr=(char *)a;
for(i=0;i=3;i++)
printf(%d ,*ptr++);
}
output:
102 102 -90 64.explain?
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try to find out the binary representation of float value 5.2
On Sat, Jul 9, 2011 at 10:46 PM, Sangeeta sangeeta15...@gmail.com wrote:
int main(){
int i;
float a=5.2;
char *ptr;
ptr=(char *)a;
for(i=0;i=3;i++)
printf(%d ,*ptr++);
}
output:
102 102 -90 64.explain?
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#Iincludestdio.h
#includestring.h
main()
{
char str[]=S\061AB;
printf(\n%d,strlen(str));
}
output:4
why?
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because \061 is considered as a single char in ur string..
On Mon, Jul 4, 2011 at 12:52 PM, Sangeeta sangeeta15...@gmail.com wrote:
#Iincludestdio.h
#includestring.h
main()
{
char str[]=S\061AB;
printf(\n%d,strlen(str));
}
output:4
why?
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Continuous memory allocation is used for 2-d array.
So a[2][5] will assign 10 continuous memory spaces.
Hello will be stored on the 1st 5 spaces. Hi will be saved on the next 2
consecutive spaces itself.
There is no null character saved for the 1st string so while printing it
prints until it finds
If you try to visualize the internal representation.
You've allocated 10 bytes.
| h | e | l | l | o |
| h | i |\0 |\0 |\0 |
Since these are stored in linear form, so the actual representation would be
| h | e | l | l | o | h | i |\0 |\0 |\0 |
Now a[0] points to 'h' in the first row, and printf
thanx i got it :)
On Mon, Jul 4, 2011 at 2:05 PM, Sandeep Jain sandeep6...@gmail.com wrote:
If you try to visualize the internal representation.
You've allocated 10 bytes.
| h | e | l | l | o |
| h | i |\0 |\0 |\0 |
Since these are stored in linear form, so the actual representation would
ya i got it.. thanx
On Thu, Jun 23, 2011 at 6:01 PM, rajeev bharshetty rajeevr...@gmail.comwrote:
It is because of the ! operator in front of a , !a is always 0(in case a is
value other than 0) so it takes it as integer and hence the size is 4 bytes
whereas only a you will get 12 . The !
int main()
{
long double a;
signed char b;
printf(%d\n,sizeof(!a+b));
return 0;
}
why is it 4?? shouldnt it be 12 as per sizeof long double in gccc
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i thing that if u negate any value by ! the result will be always a int..
so !a will be a integer value and !a+b due to typecast will be also int..
you can verified this fact by seeing the result of
sizeof (!a+a) = 12
sizeof (!b) =4
sizeof (!a) =4
sizeof(!(a+b)) =4
int main()
{
char a;
printf(%d %d,sizeof(a),sizeof('a'));
return 0;
}
Output:
1 4
why does the expression sizeof('a') evaluates it as an integer,not as a
character??
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@piyush Because of its interpretation as ascii .The character's ascii value
is taken into account which is an integer .
On Thu, Jun 23, 2011 at 6:32 PM, piyush kapoor pkjee2...@gmail.com wrote:
int main()
{
char a;
printf(%d %d,sizeof(a),sizeof('a'));
return 0;
}
Output:
1 4
You didn't understand my question...
Why is sizeof() interpreting a constant character literal as ascii???
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int main()
{
int a,b, c;
scanf(%d%*d%d,a,b,c);
printf(%d %d %d,a,b,c);
}
output: 25 35 garbage
how is it happening??
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An asterisk indicates that the data is to be retrieved from the use
but ignored, i.e. it is not stored in the corresponding
argument...hence the third value entered gets stored for b and for c
the output comes to garbage value
One beautiful application of such type of implementation is in
thanx .. can u explain me how this is used in finding sum of 2 vars without
using + ??
On Thu, Jun 23, 2011 at 7:20 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
An asterisk indicates that the data is to be retrieved from the use
but ignored, i.e. it is not stored in the corresponding
sorry by mistake i added it in scanf situation..
actually this type of specifier can be used with printf statement for
finding the sum...
look at the code below
main()
{
int a=9;
int b=3;
printf(%d\n,printf(%*s%*s,a,,b,));
system(pause);
}
On 6/23/11,
@anika can u will please tell me the source of all dese questions,,
actually i am new to this all
nd getting to learn a lot..
thanks in anticipation
On Thu, Jun 23, 2011 at 7:52 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
sorry by mistake i added it in scanf situation..
actually
i mean how it working actually?
On Thu, Jun 23, 2011 at 8:06 PM, Anika Jain anika.jai...@gmail.com wrote:
hey ya its working :) but whats the logic behind it??
On Thu, Jun 23, 2011 at 7:52 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
sorry by mistake i added it in scanf situation..
there is no as such logic behind it..its just the format specifier...
u must be knowing printf returns the number of values it has printed(u can
check that)
now, in printf if u write like *printf(%7s,a), *it will create 7
columns for the output and print a in the last column and the returned
In C, the type of a character constant like 'a' is actually an int, with
size of 4. In C++, the type is char, with size of 1. This is one of many
small differences between the two languages.
On Thu, Jun 23, 2011 at 6:50 PM, piyush kapoor pkjee2...@gmail.com wrote:
You didn't understand my
@ Piyush Could u provide the link to some source , because i am still
unclear about the above concept .
Regards
Rajeev N B
On Thu, Jun 23, 2011 at 8:32 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
there is no as such logic behind it..its just the format specifier...
u must be knowing
@rajeev
http://www.cplusplus.com/reference/clibrary/cstdio/scanf/
http://www.cplusplus.com/reference/clibrary/cstdio/printf/
On Thu, Jun 23, 2011 at 9:39 PM, rajeev bharshetty rajeevr...@gmail.comwrote:
@ Piyush Could u provide the link to some source , because i am still
unclear about the
or u cud consult ANSI C by Balaguruswamy in chapter of Console I/Ps and O/Ps
On 6/23/11, harshit pahuja hpahuja.mn...@gmail.com wrote:
@rajeev
http://www.cplusplus.com/reference/clibrary/cstdio/scanf/
http://www.cplusplus.com/reference/clibrary/cstdio/printf/
On Thu, Jun 23, 2011 at 9:39
Thanks a lot :)
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*CSE-IT-BHU*
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@anika: the negation operation returns 0 or 1 which is an integerso u
get this output
On Thu, Jun 23, 2011 at 10:29 PM, piyush kapoor pkjee2...@gmail.com wrote:
Thanks a lot :)
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*Piyush Kapoor,*
*CSE-IT-BHU*
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@ harshit: i find such questions from books like let us c, test ur c, dennis
ritchie and from test papers of companies that visit campus
On Thu, Jun 23, 2011 at 9:15 AM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
or u cud consult ANSI C by Balaguruswamy in chapter of Console I/Ps and
O/Ps
@anika -thankx :)
On Thu, Jun 23, 2011 at 11:57 AM, Anika Jain anika.jai...@gmail.com wrote:
@ harshit: i find such questions from books like let us c, test ur c,
dennis ritchie and from test papers of companies that visit campus
On Thu, Jun 23, 2011 at 9:15 AM, Piyush Sinha
It is because of the ! operator in front of a , !a is always 0(in case a is
value other than 0) so it takes it as integer and hence the size is 4 bytes
whereas only a you will get 12 . The ! changes the data type here .
Hope it clears the doubt :)
-
Suppose I want to copy an integer array to another array pointed by a
void pointer of different size.
Some thing like this is done probably in realloc function.
The problem is it is not working for me ... here's the code
#include stdio.h
int main(){
int *p,i=0;
void *x;
@Amit Jaspal...there is a fundamental error in the code related to
pointersu have incremented the pointers p and x..so before the printf
loop, pointer p (as well as x) is pointing to 4th location...I hope I am
clear...
this can be done by creating two more pointers that will initialised as
Usually it is a bad practice to increment the pointer. Rather, one
should use the indexing variable to dereference the pointer at some
index. One more bug in your code was, you are allocating the space for
only 3 ints for p and storing 4 ints. Here is the code with some
modification.
#include
are bhai just you use memcpy function in string.h you need worry about
any ...
is it sufficient ?
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does the check x==x+1 always return false for integer x?
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yaeh its correct it always return the false statement according to the
operator rules
On Sat, Feb 12, 2011 at 6:16 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
according to me,it always return false..
On Sat, Feb 12, 2011 at 5:24 PM, snehal jain learner@gmail.comwrote:
does the
When we can convert Derived* to Base* then why can't we convert Derived** to
Base** ??
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