I got some other outputs.. when line 6 isn't commented i got 3 1 2 3, i m
using turbo c++ compiler..
On Fri, Dec 6, 2013 at 8:17 AM, pawan yadav pawan1991ya...@gmail.comwrote:
Hi All,
I'm not able to get output of following c program :
#includestdio.h
main()
{
int a[] ={
i think because of operator the program is working differently... also
there is four data value print operation..
On Mon, Dec 9, 2013 at 2:34 AM, Tamanna Afroze afroze...@gmail.com wrote:
I got some other outputs.. when line 6 isn't commented i got 3 1 2 3, i m
using turbo c++ compiler..
Hi All,
I'm not able to get output of following c program :
#includestdio.h
main()
{
int a[] ={ 1,2,3,4,5,6,7};
char c[] = {'a','x','h','o','k'};
//printf(%u %u\n, a[3], a[0]); line 6
printf(%d %d %d %d \n, (a[3]-a[0]));
}
If line 6 is commented,
void dosomething(int num)
{
int mask=~(15-1);
int res=nummask;
printf(%d,res);
}
int main()
{
dosomething(56);
dosomething(64);
dosomething(127);
return 0;
}
please explain the logic behind the output.
Thanks,
Rajesh
--
As arithmetic operator has higher precedence than shifting operator , mask
will have 5th bit zero from left.
So it will deduct 16 from the input if 5th bit is set in the binary
representation of number.
so it will deduct 16 from 56 and 127..but not from 64
But its just the
let me keep it simple
y is integer pointer pointing to address 20 .. x=20
y+7 will point to the addressy + ( size(int) * 7) i.e 50 here
/*y is integer pointer*/
On Tue, Nov 6, 2012 at 11:45 AM, Rahul Kumar Patle
patlerahulku...@gmail.com wrote:
because difference is of 30
*int *x ,int *y;
x=(int *) 50;
y=(int *)20;
coutx-yendl;
why the output is 7.*
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To post to this
its because it is integer pointer subtraction, So subtraction result will
be divided by integer size.
so 30/4 = 7.
2012/11/6 rajesh pandey rajesh.pandey.i...@gmail.com
*int *x ,int *y;
x=(int *) 50;
y=(int *)20;
coutx-yendl;
why the output is 7.*
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anybdy has basic trie implementation (insertion and printing) ?
o/p
1)
main()
{
char *p1=Name;
char *p2;
p2=(char *)malloc(20);
while(*p2++=*p1++);
printf(%s\n,p2);
} ...it's giving me empty string
2)
i=5;
printf(%d,i++ * i++);
o/p and also tell the reason?
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#include stdio.h
#include stdlib.h
#define SIZEOF(arr) (sizeof(arr)/sizeof(arr[0]))
#define PrintInt(expr) printf(%s:%d\n,#expr,(expr))
int main(int argc, char *argv[])
{
/* The powers of 10 */
int pot[] = {
0001,
0010,
0100,
1000
};
0001,
0010,
0100,
these number are represented in octal ..so u r getting decimal of the same.
1000 - this is a decimal value;
preceding number by 0 means you are representing it in octal format
similarly preceding 0x means representing in hexa format.
On Sun, Feb 26, 2012 at 3:38 AM, Ravi
#includestdio.h
int main()
{
int a=1;
switch(a)
{
int b=6;
case 1:
printf(b is %d,b);
break;
default:
printf(b is %d:,b);
compile time error... except case labels nothing else will be executed in
case of switch statements... so definition of b will not be executed.so
compilation error
On Tue, Sep 27, 2011 at 6:28 AM, Ratan success.rata...@gmail.com wrote:
#includestdio.h
int main()
{
int a=1;
b=6 will not execute..
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On ideone.com its showing compiler error and on ubuntu garbage value
of b gets as the result...
On Tue, Sep 27, 2011 at 10:21 AM, gaurav yadav
gauravyadav1...@gmail.com wrote:
b=6 will not execute..
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http://codepad.org/erdnF74M
can anyone explain the output ???
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*MCA final year,*
*Nit Durgapur*
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To
main(){printf(%s,printf(samsung)+fun());}fun(){return electronic;}
The printf is a function which returns the number of printed
characters , and scanf is a function
which returns the number of inputs scanned .
So after printing samsung it returns 7. fun() is returning a pointer
to the constant
int main(){printf(%s,printf(samsung)+fun());return 0;}fun(){return
electronic;}
On Tue, Sep 13, 2011 at 11:28 PM, kumar raja rajkumar.cs...@gmail.comwrote:
main(){printf(%s,printf(samsung)+fun());}fun(){return electronic;}
The printf is a function which returns the number of printed
@Kumar: +1
@Kumar Rajeshwar: ExitFailure is outputted because main is expected to
return something which is not done in your case. Just add return 0; at the
end of main to get expected output.
On Tue, Sep 13, 2011 at 11:33 PM, Ishan Aggarwal
ishan.aggarwal.1...@gmail.com wrote:
int main()
#includestdio.h
#includestring.h
int main()
{
typedef union
{
int a;
char b[10];
float c;
}
Union;
Union x,y = {100};
x.a = 50;
strcpy(x.b,hello);
x.c = 21.50;
printf(Union x : %d %s %f \n,x.a,x.b,x.c );
printf(Union y :%d %s%f \n,y.a,y.b,y.c);
return 0;
}
Someone pls explain
U should know that union's elements share memory alloted through the largest
data type like in this case 10 bytes is alloted and is shared by all the
union elements.
So in union x :- last element modified is c, so that's why it is printing
garbage values for int and char[] elements...
actually if
+1 to sagar.
@vartika: if you don't know the concept of Union then understanding this
program will be quite lenghty, because it will involve lot of calculation.
try to check some simple examples, like:
union a
{
int a;
char c;
};
int main()
{
union a xy;
xy.a=65;
printf(%c,xy.c);
#includestdio.h
int main()
{
char arr[5] = geeks;
printf(%s, arr);
getchar();
return 0;
}
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macro has not been terminated and it has so many errors
On Thu, Aug 11, 2011 at 11:36 PM, manvir siyo manis...@gmail.com wrote:
output of the program?
#define prn(a) printf(%d,a)
#define print(a,b,c) prn(a), prn(b), prn(c)
#define max(a,b) (a
main()
{
int x=1, y=2;
this is the question :
#includestdio.h
#define prn(a) printf(%d,a)
#define print(a,b,c) prn(a), prn(b), prn(c)
#define max(a,b) (ab?a:b)
int main()
{
int x=1, y=2;
print(max(x++,y),x,y);
print(max(x++,y),x,y);
return 0;
}
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Answer=: 32
On Fri, Aug 12, 2011 at 8:05 PM, Avenged nitee...@gmail.com wrote:
this is the question :
#includestdio.h
#define prn(a) printf(%d,a)
#define print(a,b,c) prn(a), prn(b), prn(c)
#define max(a,b) (ab?a:b)
int main()
{
int x=1, y=2;
print(max(x++,y),x,y);
output of the program?
#define prn(a) printf(%d,a)
#define print(a,b,c) prn(a), prn(b), prn(c)
#define max(a,b) (a
main()
{
int x=1, y=2;
print(max(x++,y),x,y);
print(max(x++,y),x,y);
}
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this code has so many errors
On Thu, Aug 11, 2011 at 11:36 PM, manvir siyo manis...@gmail.com wrote:
output of the program?
#define prn(a) printf(%d,a)
#define print(a,b,c) prn(a), prn(b), prn(c)
#define max(a,b) (a
main()
{
int x=1, y=2;
print(max(x++,y),x,y);
print(max(x++,y),x,y);
this is the question in de shaw written test
On Thu, Aug 11, 2011 at 11:48 PM, Anika Jain anika.jai...@gmail.com wrote:
this code has so many errors
On Thu, Aug 11, 2011 at 11:36 PM, manvir siyo manis...@gmail.com wrote:
output of the program?
#define prn(a) printf(%d,a)
#define
@sandeep: so the statement becomes if(ch=0) since printf returns integer
0...whats does this mean now actually?0 is ascii for NULL and so ch is
assinged to null? I am slightly confused..
On Tue, Aug 9, 2011 at 7:04 PM, SANDEEP CHUGH sandeep.aa...@gmail.comwrote:
@all sorry i give wrong
#includestdio.h
#includeconio.h
int main()
{
struct value
{
int bit1:1;
int bit3:4;
int bit4:4;
}bit={1,2,2};
printf(%d %d %d\n,bit.bit1,bit.bit3,bit.bit4);
getche();
return 0;
}
the above code gives output : -1 2 2
any idea why???
--
You
I think it is because bit1 is only 1 bit fwide and whn u initialize it to
1,since MSB is 1 it treats it as a negative integer...Plz correct me if i am
wrong...
On Tue, Aug 9, 2011 at 12:24 PM, Rohit Srivastava access2ro...@gmail.comwrote:
#includestdio.h
#includeconio.h
int main()
{
Int bit3:4 will be read as lower order 4 bits of bit3 and this will be
treated as int (signed). Thus lower order bit of bit3 which is 2, are
0010 which is 2
try with
1) int bit3:2, output will be -2
2) unsigned int bit3:2, output will be 2.
I hope it is cleared now
On 8/9/11, Rohit Srivastava
hey guys thanks got it!!
On Tue, Aug 9, 2011 at 12:49 PM, sanjay ahuja sanjayahuja.i...@gmail.comwrote:
Int bit3:4 will be read as lower order 4 bits of bit3 and this will be
treated as int (signed). Thus lower order bit of bit3 which is 2, are
0010 which is 2
try with
1) int bit3:2,
On a little-endian machine, bit structure will be represented as:
0x00 00 00 45
which is bit.bit4 = 0010 (2 in decimal) bit.bit3 = 0010 (2 in decimal)
bit.bit1 = 1 (1 in decimal)
since bit.bit1 exists at the rightmost position, while displaying data as
integer, compiler just repeats the same
i dnt get it.. any better explanation..
On Tue, Aug 9, 2011 at 1:13 PM, dinesh bansal bansal...@gmail.com wrote:
On a little-endian machine, bit structure will be represented as:
0x00 00 00 45
which is bit.bit4 = 0010 (2 in decimal) bit.bit3 = 0010 (2 in decimal)
bit.bit1 = 1 (1 in
#includestdio.h
int main()
{
char ch;
if((ch=printf()))
printf(it matters);
else
printf(it doesn't matter);
return 0;
}
what will b the output??
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it doesnt matter
On Tue, Aug 9, 2011 at 10:20 PM, tech rascal techrascal...@gmail.comwrote:
#includestdio.h
int main()
{
char ch;
if((ch=printf()))
printf(it matters);
else
printf(it doesn't matter);
return 0;
}
what will b the output??
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it matters..
On Tue, Aug 9, 2011 at 10:20 PM, tech rascal techrascal...@gmail.comwrote:
what will b the output??
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explanation??
On Tue, Aug 9, 2011 at 10:24 PM, aditi garg aditi.garg.6...@gmail.comwrote:
it doesnt matter
On Tue, Aug 9, 2011 at 10:20 PM, tech rascal techrascal...@gmail.comwrote:
#includestdio.h
int main()
{
char ch;
if((ch=printf()))
printf(it matters);
else
printf(it doesn't
On 9 August 2011 22:20, tech rascal techrascal...@gmail.com wrote:
#includestdio.h
int main()
{
char ch;
if((ch=printf()))
printf(it matters);
else
printf(it doesn't matter);
return 0;
}
It doesn't matter
what will b the output??
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it doesn't matter
On Tue, Aug 9, 2011 at 10:24 PM, Jayanthi shravan
jayanthisra...@gmail.com wrote:
it matters..
On Tue, Aug 9, 2011 at 10:20 PM, tech rascal techrascal...@gmail.com
wrote:
what will b the output??
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printf returns the numbers of character printed and here it prints nothing
i.e. 0 characters hence returns 0
and hence it doesnt matter gets printed
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Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad
On Tue, Aug 9, 2011 at 10:25 PM, tech rascal
o/p:
It doesn't matter
Reason:
printf() returns the number of characters printed to screen.
since printf() will return 0, hence the *else* is selected.
On 9 August 2011 22:25, siddharth srivastava akssps...@gmail.com wrote:
On 9 August 2011 22:20, tech rascal techrascal...@gmail.com wrote:
it doesn't matter
On Tue, Aug 9, 2011 at 10:25 PM, siddharth srivastava
akssps...@gmail.comwrote:
On 9 August 2011 22:20, tech rascal techrascal...@gmail.com wrote:
#includestdio.h
int main()
{
char ch;
if((ch=printf()))
printf(it matters);
else
printf(it doesn't matter);
return 0;
my bad..that I didn't notice = (assumed it to be ==)
On 9 August 2011 22:28, SANDEEP CHUGH sandeep.aa...@gmail.com wrote:
it doesn't matter
On Tue, Aug 9, 2011 at 10:25 PM, siddharth srivastava akssps...@gmail.com
wrote:
On 9 August 2011 22:20, tech rascal techrascal...@gmail.com wrote:
@all sorry i give wrong explanation by mistake..:P :P
printf() returns no if characters.. in this case returns 0 . which is
assigned to ch
so in ch 0 is stored and 0 is the ascii value of null character
when we using ch in --- if (ch) -- it will reduce to if(0) -- as 0 is the
ascii value of
#includestdio.h
int main()
{
typedef static int *i;
static int j;
i a = j;
printf(%d, *a);
getchar();
return 0;
}
what is the error in the code?
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kamakshi...@gmail.com
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using typedef u define the static keyword as int
and the statement static int j thus becomes
int int j
hence the compiler gives the error multiple storage class in declaration
specifiers
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Third Year Student
Computer Science and Engineering
MNNIT Allahabad
On Mon, Aug 8, 2011
typedef doesn't declare an instance of a variable, it declares a type (type
alias actually),
static is a qualifier you apply to an instance, not a type, so you can use
static when you use the type, but not when you define the type.
Aseem
On Mon, Aug 8, 2011 at 12:29 AM, Kamakshii Aggarwal
got it...thanks
On Mon, Aug 8, 2011 at 12:39 AM, aseem garg ase.as...@gmail.com wrote:
typedef doesn't declare an instance of a variable, it declares a type (type
alias actually),
static is a qualifier you apply to an instance, not a type, so you can use
static when you use the type, but
any1 plz tell me y output of float type of dis prgrm is zero?
#includestdio.h
#define INT 1
main()
{
struct a
{
int type;
};
struct c
{
int type;
int e;
};
struct b
{
out put is :
10.375001
float value =10.4
integer value=1
1 1 1 10.37500
what us the problem ?
On Sat, Aug 6, 2011 at 2:19 AM, SHIVAM AGRAWAL shivi...@gmail.com wrote:
any1 plz tell me y output of float type of dis prgrm is zero?
#includestdio.h
#define INT 1
main()
{
struct a
output wud be:
0.00
float value= 00.0
integer value=1
1 1 1 0.00
On Sat, Aug 6, 2011 at 2:49 PM, SHIVAM AGRAWAL shivi...@gmail.com wrote:
any1 plz tell me y output of float type of dis prgrm is zero?
#includestdio.h
#define INT 1
main()
{
struct a
{
program output comes - hi
why??
#includestdio.h
main()
{
float a=0.7;
if(.7 a)
printf(hi\n);
else
printf(hello\n);
}
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the output will be hi
even if u have putted equal to sign there this is becoz 0.7 do not exactly
stored like 0.7 in a but some thing lesser as describe in ASCII format
that's why it is awlays lesser than 0.7
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.7 in float is taken like .699 i blv ?
On Fri, Aug 5, 2011 at 8:20 PM, kartik sachan kartik.sac...@gmail.comwrote:
the output will be hi
even if u have putted equal to sign there this is becoz 0.7 do not exactly
stored like 0.7 in a but some thing lesser as describe in ASCII format
coz by default all real values are converted to double and it is larger than
float.ie real nos are not stored accurately inside computer memory.so
since double occupies more memory it prints hi
On Fri, Aug 5, 2011 at 8:24 PM, Kamakshii Aggarwal kamakshi...@gmail.comwrote:
the reason is dat 0.7
Thanks to all
On Fri, Aug 5, 2011 at 8:53 PM, sukran dhawan sukrandha...@gmail.comwrote:
coz by default all real values are converted to double and it is larger
than float.ie real nos are not stored accurately inside computer memory.so
since double occupies more memory it prints hi
On Fri,
ha ha
its not like it stores slightly less than actual :P :P
try this
float a=0.5,b=1.0,c=0.7;
if(0.5==a)
printf(yes a=0.5\n);
if(1.0==b)
printf(yes b=1.0\n);
if(0.7!=c)
printf(Yes c!=0.7\n);
o/p will be
yes a=0.5
yes b=1.0
yes c!=0.7
check it out i will explain it later...
On Fri, Aug 5,
# define swap(a,b) temp=a; a=b; b=temp;
main( )
{
int i, j, temp;
i=5;
j=10;
temp=0;
if( i j)
swap( i, j );
printf( “%d %d %d”, i, j, temp);
}
On compiling i got ans 10, 0, 0.explain..
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hint:MACRo is just substituted as it is
On Sat, Aug 6, 2011 at 10:23 AM, ranjane ranjane...@gmail.com wrote:
# define swap(a,b) temp=a; a=b; b=temp;
main( )
{
int i, j, temp;
i=5;
j=10;
temp=0;
if( i j)
swap( i, j );
printf( “%d %d %d”, i, j, temp);
}
On compiling i got ans
Try compiling with gcc -E filename.c.That will clear everything
On Sat, Aug 6, 2011 at 10:25 AM, saurabh singh saurab...@gmail.com wrote:
hint:MACRo is just substituted as it is
On Sat, Aug 6, 2011 at 10:23 AM, ranjane ranjane...@gmail.com wrote:
# define swap(a,b) temp=a; a=b;
@saurabh:ya got it..thanks
On Sat, Aug 6, 2011 at 10:26 AM, saurabh singh saurab...@gmail.com wrote:
Try compiling with gcc -E filename.c.That will clear everything
On Sat, Aug 6, 2011 at 10:25 AM, saurabh singh saurab...@gmail.com wrote:
hint:MACRo is just substituted as it is
On
when swap is called ,as it is a macro ,before compilation its code is
replaced and the code before compilation bcomes:
# define swap(a,b) temp=a; a=b; b=temp;
main( )
{
int i, j, temp;
i=5;
j=10;
temp=0;
if( i j)
//code replaced here
temp=a; //not executed (false)
a=b;
#include stdio.h
void foo(const char **p) { }
int main(int argc, char **argv)
{
foo(argv);
return 0;
}
why the above code giving me warning can body explain me??
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the warning comes is as
warning: passing argument 1 of ‘foo’ from incompatible pointer type
foo expects a constant char** but it is not the case...hence warning is
displayed...
remove the const keyword.you'll see no warning !!
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*@AMOL *sending a normal pointer to a const pointer doesn't give any warning
why this is giving??
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int b=2;b=b++;
b=???
Plz explain..
Aseem
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3.it is same as b=b+1;
On Sun, Jul 31, 2011 at 1:32 AM, aseem garg ase.as...@gmail.com wrote:
int b=2;b=b++;
b=???
Plz explain..
Aseem
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@Kamakshi: Run karke dekh leti pehle. :-/
Aseem
On Sun, Jul 31, 2011 at 1:34 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
3.it is same as b=b+1;
On Sun, Jul 31, 2011 at 1:32 AM, aseem garg ase.as...@gmail.com wrote:
int b=2;b=b++;
b=???
Plz explain..
Aseem
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it would be undefined...
On Sun, Jul 31, 2011 at 1:34 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
3.it is same as b=b+1;
On Sun, Jul 31, 2011 at 1:32 AM, aseem garg ase.as...@gmail.com wrote:
int b=2;b=b++;
b=???
Plz explain..
Aseem
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oh sorry,mane galat pad lia..mane sirf b++ dekha...
it is compiler dependent.value of b is changing twice b/w two serial points
On Sun, Jul 31, 2011 at 1:37 AM, aseem garg ase.as...@gmail.com wrote:
@Kamakshi: Run karke dekh leti pehle. :-/
Aseem
On Sun, Jul 31, 2011 at 1:34 AM, Kamakshii
@Aditi: Run karke dekh leti pehle. :-/
Aseem
On Sun, Jul 31, 2011 at 1:38 AM, aditi garg aditi.garg.6...@gmail.comwrote:
it would be undefined...
On Sun, Jul 31, 2011 at 1:34 AM, Kamakshii Aggarwal kamakshi...@gmail.com
wrote:
3.it is same as b=b+1;
On Sun, Jul 31, 2011 at 1:32 AM,
concept of sequence points.search wiki
On Sun, Jul 31, 2011 at 1:38 AM, aditi garg aditi.garg.6...@gmail.comwrote:
it would be undefined...
On Sun, Jul 31, 2011 at 1:34 AM, Kamakshii Aggarwal kamakshi...@gmail.com
wrote:
3.it is same as b=b+1;
On Sun, Jul 31, 2011 at 1:32 AM, aseem
@aseem output is 3 only...check it once again
http://ideone.com/mWGc5
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Third Year Student
Computer Science and Engineering
MNNIT Allahabad
On Sun, Jul 31, 2011 at 1:37 AM, aseem garg ase.as...@gmail.com wrote:
@Kamakshi: Run karke dekh leti pehle. :-/
Aseem
On Sun,
@aseem its 3
different on yours ?
On Sun, Jul 31, 2011 at 1:37 AM, aseem garg ase.as...@gmail.com wrote:
@Kamakshi: Run karke dekh leti pehle. :-/
Aseem
On Sun, Jul 31, 2011 at 1:34 AM, Kamakshii Aggarwal kamakshi...@gmail.com
wrote:
3.it is same as b=b+1;
On Sun, Jul 31, 2011 at
Yes, it's UB.
For such ques, try to run it on ideone.
It will give a warning if any operation is not standard.
http://ideone.com/bJBGl
http://ideone.com/bJBGl
On Sun, Jul 31, 2011 at 1:38 AM, aditi garg aditi.garg.6...@gmail.comwrote:
it would be undefined...
On Sun, Jul 31, 2011 at 1:34
I am getting 2. .
Aseem
On Sun, Jul 31, 2011 at 1:41 AM, Neeraj Gupta neeraj.gupta...@gmail.comwrote:
Yes, it's UB.
For such ques, try to run it on ideone.
It will give a warning if any operation is not standard.
http://ideone.com/bJBGl
http://ideone.com/bJBGl
On Sun, Jul 31, 2011 at
On Dev CPP.
Aseem
On Sun, Jul 31, 2011 at 1:42 AM, aseem garg ase.as...@gmail.com wrote:
I am getting 2. .
Aseem
On Sun, Jul 31, 2011 at 1:41 AM, Neeraj Gupta
neeraj.gupta...@gmail.comwrote:
Yes, it's UB.
For such ques, try to run it on ideone.
It will give a warning if any
yup it is undefined and the warning is justified
b=b++ will assign 2 to b then increment b hence 3 is printed
--
Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad
On Sun, Jul 31, 2011 at 1:41 AM, Neeraj Gupta neeraj.gupta...@gmail.comwrote:
Yes, it's UB.
For
I am getting b=2.
Maybe coz we are assigning 2 to b then incrementing b. But still acc to me
it should be 3.
:-\
No warnings in dev cpp.
On Sun, Jul 31, 2011 at 1:43 AM, Amol Sharma amolsharm...@gmail.com wrote:
yup it is undefined and the warning is justified
b=b++ will assign 2 to b then
@sanchit:it is undefined..
On Sun, Jul 31, 2011 at 1:47 AM, Sanchit Manchanda sanchit...@gmail.comwrote:
I am getting b=2.
Maybe coz we are assigning 2 to b then incrementing b. But still acc to me
it should be 3.
:-\
No warnings in dev cpp.
On Sun, Jul 31, 2011 at 1:43 AM, Amol Sharma
I dint get any error, not even warning. it might depend from compiler to
compiler. but yeah it should be undefined.
On Sun, Jul 31, 2011 at 1:49 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
@sanchit:it is undefined..
On Sun, Jul 31, 2011 at 1:47 AM, Sanchit Manchanda
bt y it sud b undefind?
On Sun, Jul 31, 2011 at 1:51 AM, Sanchit Manchanda sanchit...@gmail.comwrote:
I dint get any error, not even warning. it might depend from compiler to
compiler. but yeah it should be undefined.
On Sun, Jul 31, 2011 at 1:49 AM, Kamakshii Aggarwal kamakshi...@gmail.com
@gaurav:refer the link below
http://c-faq.com/expr/seqpoints.html
On Sun, Jul 31, 2011 at 1:53 AM, gaurav gupta grvgupta...@gmail.com wrote:
bt y it sud b undefind?
On Sun, Jul 31, 2011 at 1:51 AM, Sanchit Manchanda
sanchit...@gmail.comwrote:
I dint get any error, not even warning. it
thanx kamakshi
On Sun, Jul 31, 2011 at 1:56 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
@gaurav:refer the link below
http://c-faq.com/expr/seqpoints.html
On Sun, Jul 31, 2011 at 1:53 AM, gaurav gupta grvgupta...@gmail.comwrote:
bt y it sud b undefind?
On Sun, Jul 31, 2011 at
#includestdio.h
# define MAXROW 3
#define MAXCOL 4
int main()
{
int (*p)[MAXCOL];
p=(int(*)[MAXCOL])malloc(MAXROW*sizeof(*p));
printf(%d %d,sizeof(p),sizeof(*p));
system(pause);
return 0;
}
THE O/P IS 4 16.
I am not getting the reason.plss help
--
You received this message
The system on which u executed this will allocate 4 bytes for int type.
hence sizeof(p) gives 4 bytes and sizeof(*p) will contain 4 such integer
pointers( int *p[4] is the declaration) which gives 4*4=16 bytes
Regards
Ramya
*Try to learn something about everything and everything about
...@gmail.com
*Sender: * algogeeks@googlegroups.com
*Date: *Wed, 27 Jul 2011 19:09:20 +0530
*To: *algogeeks@googlegroups.com
*ReplyTo: * algogeeks@googlegroups.com
*Subject: *[algogeeks] output
#includestdio.h
int main(int argc,char *argv[])
{
const char *s=;
char str[]=hello;
s=str;
while(*s
kamakshi...@gmail.com
*Sender: * algogeeks@googlegroups.com
*Date: *Wed, 27 Jul 2011 19:09:20 +0530
*To: *algogeeks@googlegroups.com
*ReplyTo: * algogeeks@googlegroups.com
*Subject: *[algogeeks] output
#includestdio.h
int main(int argc,char *argv[])
{
const char *s=;
char str[]=hello;
s
sry sry o/p ll be 0 1 0in the 2nd printf the value ll be evaluated
from rite to left...
so in the 1st printf i's vlue ll be 0 and in the 2nd printf stmt rite
expression is evaluated first and then i value ll be modified in it..
and with hte help of it left one is evaluated...crct me if i'm
yes I understand, thanks to both of you
On Sun, Jul 24, 2011 at 10:10 PM, rajeev bharshetty rajeevr...@gmail.comwrote:
value 200 exceeds the char range so -56 , to print 200 try unsigned char .
On Sun, Jul 24, 2011 at 10:06 PM, Prashant Gupta prashantatn...@gmail.com
wrote:
-56
On Sun,
it is printed in %d format nd not as %ch...
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//BE COOL// kavi
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1.
#includestdio.h
main()
{
int i=1;
printf(\n%d,i^=1%2);
printf(\n%d %d,i^=1%2,i=1%2);
return 0;
}
output 3 3
hey shudnt the output be 3 2
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run on gcc compiler it would be
0
1 1
On Mon, Jul 25, 2011 at 10:35 PM, geek forgeek geekhori...@gmail.comwrote:
1.
#includestdio.h
main()
{
int i=1;
printf(\n%d,i^=1%2);
printf(\n%d %d,i^=1%2,i=1%2);
return 0;
}
output 3 3
hey shudnt the output be 3 2
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You received this
@ Deoki : Can u please explain how did dis outcome come?
On Mon, Jul 25, 2011 at 10:49 PM, Deoki Nandan deok...@gmail.com wrote:
run on gcc compiler it would be
0
1 1
On Mon, Jul 25, 2011 at 10:35 PM, geek forgeek geekhori...@gmail.comwrote:
1.
#includestdio.h
main()
{
int i=1;
@Deoki Remove the first printf from the program ..
You will get 3 3 . The first printf should be removed .
And about the output as 3 3 it is beacuse of the right to left eval of
printf so i has value 3 after evaluation.
On Mon, Jul 25, 2011 at 10:49 PM, Deoki Nandan deok...@gmail.com wrote:
run
ma mistake
code is
#includestdio.h
main()
{
int i=1;
printf(\n%d %d,i^=1%2,i=1%2);
return 0;
}
On Mon, Jul 25, 2011 at 10:19 AM, Deoki Nandan deok...@gmail.com wrote:
run on gcc compiler it would be
0
1 1
On Mon, Jul 25, 2011 at 10:35 PM, geek forgeek geekhori...@gmail.comwrote:
1.
yeah output
0
1 1
*is dis because of side effect? *
* *
*
*
*Muthuraj R.
4TH Year BE.**
Information Science Dept*
*PESIT, Bengaluru .
*
On Mon, Jul 25, 2011 at 10:49 PM, Deoki Nandan deok...@gmail.com wrote:
run on gcc compiler it would be
0
1 1
On Mon, Jul 25, 2011 at 10:35 PM, geek
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