@anandregarding problem 2 i agree with you and observed the samebut
that gives me the feel that the compiler in working in the interprating
manner though it doesn't...btw thanks for pointing it out.
and in problem 3 what i understood after reading with what you wrote is that
size
On Tue, Jul 12, 2011 at 2:36 PM, nicks crazy.logic.k...@gmail.com wrote:
and in problem 3 what i understood after reading with what you wrote is
that size of tag will remain sizeof(int) irrespective of the number of
constants in it.correct me if i am wrong !!
Right.
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You received
for the second problem,y should the answer be 3 1?.it should be 3 2
and it is working fine on dev c++ giving 3 2.
On Tue, Jul 12, 2011 at 2:48 PM, Anand Saha anands...@gmail.com wrote:
On Tue, Jul 12, 2011 at 2:36 PM, nicks crazy.logic.k...@gmail.com wrote:
and in problem 3 what i
3 1 is the output which i predicted without running...it is giving 3 3
on gcc and may give 3 2 on dev c++ . output will be compiler dependent
as these types of situations are not defined in standards hence
implementation dependent..i want someone to explain the output with
reference to
Guys plz help me in understanding the following output
*PROBLEM 1.*
*
*
#includestdio.h
main()
{
int scanf=78;
//int printf=45;
int getchar=6;
printf(%d,scanf);
printf(\n%d,getchar);
}
*OUTPUT-*
78
6
in this problem my problem is using printf and scanf as variable
names.they are functions
@sandeep u mean whenever printf will demand for %f then it will print
2.0.is it random behavior or always going to happen ??
anyone else having better idea regarding 1st and 2nd problem...
On Mon, Jul 11, 2011 at 9:51 AM, Sandeep Jain sandeep6...@gmail.com wrote:
TurboC has many flaws, one
problem 3. I think tag is a reference so its size is 4 bytes.
On Tue, Jul 12, 2011 at 1:29 AM, nicks crazy.logic.k...@gmail.com wrote:
Guys plz help me in understanding the following output
*PROBLEM 1.*
*
*
#includestdio.h
main()
{
int scanf=78;
//int printf=45;
int getchar=6;
*OUTPUT-*
78
6
in this problem my problem is using printf and scanf as variable
names.they are functions in the stdio.h then how are they available for
variable name ??...generally what happens is that whenever we use some name
for the function and then use that name for some variable
I'll put it in simple words, when printf is executed, it expects the
arguments to be of the same type and in the same order as they appear in the
format string.
Otherwise, it starts to exhibit random behavior whenever a first mismatch
occurs.
Regards,
Sandeep Jain
On Tue, Jul 12, 2011 at 1:32
Someone please help me in understanding the following output -
Problem *1.*
#includestdio.h
#ifdef getchar //this expression is evaluated to zero.why is so
happening ??getchar is defined as macro in stdio.h.i mean else
part shouldn't be executed which is happening
#undef
*Problem 2:*
I think u are using TurboC, since you have written void main()
Secondly, printf (since based on macros) scans through the format string.
And the it process arguments one by one depending on the format specifiers
given in this string.
I've observed that whenever there is a mismatch
for the first question...it will take #ifdef getchar to be '1' only when it
is defined as a MACRO in your program..if u dont define macro it will not
take it into consideration even if it is defined in header file.
On Mon, Jul 11, 2011 at 12:38 AM, nicks crazy.logic.k...@gmail.com wrote:
probelm 5:It must be giving runtime error not segmentation fault coz it is
an infinite recursion
On Mon, Jul 11, 2011 at 1:09 AM, Kamakshii Aggarwal
kamakshi...@gmail.comwrote:
for the first question...it will take #ifdef getchar to be '1' only when it
is defined as a MACRO in your
@sandeep,kamakshiithanks both...your replies were really helpfuli
understood my fault in 3,4,5...they are clea now..but i am still stuck
with problem 1 and 2
@sandeepwhat if i am using turbo C...though i am using gcc on terminal
in my linux system.
moreover acc. t KR printf
TurboC has many flaws, one of the simplest examples would be
char *p;
scanf(%s, p);
In gcc/g++ this will surely lead to segmentation fault as memory has not
been allocated. Whereas in TC it will execute fine in most of the cases.
Infact this will crash when your code is really large.
As for
program 1;
output :1
int main()
{
int i;
i= 1,2,3;
printf(i:%d\n,i);
return 0;
}
program 2;
output: ERROR
int main()
{
int i=1,2,3;
printf(i:%d\n,i);
return 0;
}
pls explain
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@harry: it ll b btr if u go to this link n clr ur doubts link :
http://en.wikipedia.org/wiki/Comma_operator
On Fri, Jul 1, 2011 at 3:56 AM, hary rathor harry.rat...@gmail.com wrote:
program 1;
output :1
int main()
{
int i;
i= 1,2,3;
printf(i:%d\n,i);
can anybody explain that in following code y output is coming to be: 7 6 8
void call(int a,int b,int c)
{
printf(%d %d %d,a,b,c);
}
int main()
{
int a=5;
call(a++,a++,++a);
return 0;
}
On Sat, Jun 11, 2011 at 8:21 PM, PRAMENDRA RATHi rathi
prathi...@gmail.comwrote:
IN second
thanks himanshu finally i got the reason!!
:)
On Sun, Jun 12, 2011 at 5:59 PM, himanshu kansal
himanshukansal...@gmail.com wrote:
@anika:cz on gcc arguemnts r eval frm right to left and assgnment to a pre
increment expression is delayed vry mch
so on eval frm right to left
frst a is
@himanshuwhat abt this ??
#includestdio.H
# include conio.h
int i=2;
main()
{
void add();
add(i++,--i);
printf(\ni=%d \n,i);system(pause);
}
void add(int a ,int b)
{
printf(\na=%d b=%d,a,b);
}
*OUTPUT -*
a=1 b=1
i=2
acc. to ur logic output should be -
a=1 b=2
i=2
On Sun, Jun 12,
void call(int a,int b,int c)
{
printf(%d %d %d,a,b,c);
}
int main()
{
int a=5;
call(a++,a++,++a);
return 0;
}
On Sat, Jun 11, 2011 at 8:21 PM, PRAMENDRA RATHi rathi
prathi...@gmail.com wrote:
IN second program:
in function value are always push in the stack from
@umesh...thanks :) this explanation seems to be satisfactory but failed to
understand from where @anika and @himanshu were getting 7 6 8i ran
it...output comes out to be 7 6 6 !!
On Sun, Jun 12, 2011 at 6:15 AM, UMESH KUMAR kumar.umesh...@gmail.comwrote:
void call(int a,int b,int c)
{
with GCC the above code gives a = 1 and b = 2
On Sun, Jun 12, 2011 at 6:39 PM, nicks crazy.logic.k...@gmail.com wrote:
@himanshuwhat abt this ??
#includestdio.H
# include conio.h
int i=2;
main()
{
void add();
add(i++,--i);
printf(\ni=%d \n,i);system(pause);
}
void add(int a
no it's
a=1 b=1
i=2
i ran it on gcc (linux ubuntu 11.04)
On Sun, Jun 12, 2011 at 6:26 AM, sanjay ahuja sanjayahuja.i...@gmail.comwrote:
with GCC the above code gives a = 1 and b = 2
On Sun, Jun 12, 2011 at 6:39 PM, nicks crazy.logic.k...@gmail.com wrote:
@himanshuwhat abt this ??
but program by anika is giving *7 6 8* on gcc.but *7 6 6 *on
dev-cpp...i am wondering if the output is compiler dependent !!
On Sun, Jun 12, 2011 at 6:33 AM, nicks crazy.logic.k...@gmail.com wrote:
no it's
a=1 b=1
i=2
i ran it on gcc (linux ubuntu 11.04)
On Sun, Jun 12, 2011
Hello friends..plz help me in understanding the following C Output
first one is --
#includestdio.h
#includeconio.h
main()
{
int a=5;
printf(a=%d\n,a);
printf(%a=%d,a);
getch();
}
*OUTPUT - *
a=5
0x1.2ff380p-1021=4199082
and the other one is --
#includestdio.H
# include conio.h
int i=2;
main()
In 1st program, 2nd printf requires one more argument. And basically
%a is used for printing a double value in hex. see man 3 printf.
On Sat, Jun 11, 2011 at 5:29 PM, nicks crazy.logic.k...@gmail.com wrote:
Hello friends..plz help me in understanding the following C Output
first one is --
IN second program:
in function value are always push in the stack from left.
so first value is --i that will make i=1 and pass to function
and
after that i++ will be passed.that's means i will be passed and after that
value will be incremented.
so 1 will pass and after passing value. i will
IN second program:
in function value are always push in the stack from right.
so first value is --i that will make i=1 and value 1 will be passed to
function
and
after that i++ that's means i will be passed.
so 1 will be passed and after passing value. i will changed to 2.
if u want to know why
because compiler have know about g funtion while evaluating f
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int main()
{
g();
f();
}
inline int f(){
g();
return g()+1;
}
inline int g()
{
return 1;
}
this code works but if removed inline than gives run time error can any one
explain
--
SOURABH JAKHAR,(CSE)(3 year)
ROOM NO 167 ,
TILAK,HOSTEL
'MNNIT ALLAHABAD
The Law of Win says, Let's not do it your
Check this is working fine..
int g()
{
return 1;
}
int f(){
g();
return g()+1;
}
int main()
{
g();
f();
}
On Tue, Jun 7, 2011 at 12:29 PM, sourabh jakhar sourabhjak...@gmail.comwrote:
int main()
{
g();
f();
}
inline int f(){
g();
return g()+1;
}
inline int g()
{
return 1;
}
Because in C++ we need to declare the function before using it. With keyword
inline...the complete code of the function is replaced wherever it is
called.
I hope now its clear why the above code is not working.
On Wed, Jun 8, 2011 at 9:32 AM, nicks crazy.logic.k...@gmail.com wrote:
Check this
a=++b*++b;
if b=3 initially, then a is coming out to be 25.why
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you may want to read: http://c-faq.com/expr/seqpoints.html
On Wed, Jun 1, 2011 at 5:19 PM, himanshu kansal
himanshukansal...@gmail.com wrote:
a=++b*++b;
if b=3 initially, then a is coming out to be 25.why
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This will be same as:
b=b+1;
b=b+1;
a=b*b;
Basically, all prefix increment and decrement operators will be executed
first. Similarly all postfix operators will be executed at last.
Anuj Agarwal
Engineering is the art of making what you want from things you can get.
On Wed, Jun 1, 2011 at 5:27
#includestdio.h
int main(void)
{
float a=0.08;
if(a0.08)
printf(Hello\n);
else
printf(Hii\n);
return 0;
}
The o/p is: *Hello * why
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To
a 0.08f will make it compare to float. by default 0.08 is
considered as double.
On Sun, May 29, 2011 at 9:51 PM, Ankit Agarwal ankitgeniu...@gmail.com wrote:
#includestdio.h
int main(void)
{
float a=0.08;
if(a0.08)
printf(Hello\n);
else
printf(Hii\n);
and, I read it long time back that.. the value of 0.8 alone will be stored
as 0.7995 (not sure on the number of 9's but.. the last digit in the
precision will be a 5)
that could be a reason.
may be what vishal said is correct.
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you may want to check how the floats and doubles are stored into
memory using ieee notation.
i tried to print 0.08 and 0.08f in hex format and got the following result.
vishal@ubuntu:~/progs/c\ 10:03:56 AM $ cat fl.c
#include stdio.h
int main()
{
float f=0.08;
if (f 0.08f)
when u store 0.08 on the float variable a then it does not store exactly
0.08 on 'a'
actually it store 0.7999
that's why after comparition it make less quantity with 0.08
i.e 'if ' condition is true and print it HELLO
On Sun, May 29, 2011 at 9:21 AM, Ankit Agarwal
Thank u guys :) :)
On Mon, May 30, 2011 at 10:40 AM, arjoo kumar 2009ar...@gmail.com wrote:
when u store 0.08 on the float variable a then it does not store exactly
0.08 on 'a'
actually it store 0.7999
that's why after comparition it make less quantity with 0.08
i.e 'if ' condition is
#includestdio.h
#includestdio.h
#includeconio.h
#define PRINTX printf(%d\n,x)
main()
{
int x=2,y,z;
x*=3 + 2; PRINTX;
x*= y = z = 4; PRINTX;
x = y == z; PRINTX;
x == ( y = z ); PRINTX;
getch();
}
can anyone explain the output
--
SOURABH JAKHAR,(CSE)(3 year)
The Law of Win
output is
10
40
1
1
10 because x = x*(3+2) = 2 * (3+2)
40 because x = x*4 = 10 * 4
1 because x = y == z which is x = 4 == 4 which become x = 1
1 becase value of x is not changed in 4th expression.
On Wed, May 25, 2011 at 3:35 PM, sourabh jakhar sourabhjak...@gmail.comwrote:
#includestdio.h
1st print : x = x * (3 + 2) - 10
2nd print : x = x * 4 - 40 : while evaluating the expression, the rightmost
result will be used
3rd print : x = (4 == 4) - 1
4th print : x == ( y = z ) evaluates to false; prev val of x, 1 is printed
On Wed, May 25, 2011 at 3:35 PM, sourabh jakhar
ok i got it is the precedence of *= is less than of +
On Wed, May 25, 2011 at 3:52 PM, Shachindra A C sachindr...@gmail.comwrote:
1st print : x = x * (3 + 2) - 10
2nd print : x = x * 4 - 40 : while evaluating the expression, the
rightmost result will be used
3rd print : x = (4 == 4) - 1
#includestdio.h
void main()
{
long x;
float t;
scanf(%f,t);
printf(%d\n,t);
x=90;
printf(%f\n,x);
{
x=1;
printf(%f\n,x);
{
x=30;
printf(%f\n,x);
}
printf(%f\n,x);
}
x==9;
printf(%f\n,x);
#define SIZE 10
int main()
{
int arr[SIZE];
printf(size of array is:%d\n,sizeof(arr));
return 0;
}
when we call the function and pass the name of the ARRAY ,then PUSH the base
Address of the ARRAY in the Stack of the Calling
Function not PUSH the
Is it that the first case its a pointer to a pointer. So size of pointer to
a pointer is 4.
In next case , Its the size of array. So its 40. Is the explanation correct?
On Wed, Dec 15, 2010 at 9:43 AM, rahul rahulr...@gmail.com wrote:
you would like to read peter ven der linden.(deep C
#define SIZE 10
void size(int arr[SIZE])
{
printf(size of array is:%d\n,sizeof(arr));
}
int main()
{
int arr[SIZE];
size(arr);
return 0;
}
the out put should be 40 considering 4 byte integer...
but out put is only 4... how this is possible...
When you pass an array as argument, it gets broken into a pointer. So you
get size as 4 for 32 bit machine.
On Tue, Dec 14, 2010 at 10:59 PM, Divesh Dixit
dixit.coolfrog.div...@gmail.com wrote:
#define SIZE 10
void size(int arr[SIZE])
{
printf(size of array is:%d\n,sizeof(arr));
When u r passing an array to a function u only pass the base address
nt the total array...bt when sizeof is applied in main() u hv the
whole array. Thats why in the first case the output is 4(only one
address that is capable of holdin one integer)bt in the 2nd case the
output is 40(as u hv 10
you would like to read peter ven der linden.(deep C secrets).
On Tue, Dec 14, 2010 at 11:19 PM, Saurabh Koar saurabhkoar...@gmail.comwrote:
When u r passing an array to a function u only pass the base address
nt the total array...bt when sizeof is applied in main() u hv the
whole array. Thats
@Divya
o/p
c prints b
1 - prints a
6, 2 --- printf returns no of characters printed.
6=5 (length of a) + 1 (for \n)
2=1 (length of b) + 1 (for \n)
Hope it's
printf returns number of successfully printed characters.
so printf(1\n) will return 6
and printf(c\n) will return 2
due to the extra '\n' character which is also being printed
I hope the output is now clear
Anurag Sharma
On Fri, Jun 11, 2010 at 12:26 AM, divya sweetdivya@gmail.com
The return value of printf is the number of characters it prints successfully.
So the rightmost printf is going to return 2 (one for char c and one
for new line), similarly second one returning 6.
Regards,
Chetan.
On 11 June 2010 00:56, divya sweetdivya@gmail.com wrote:
#include stdio.h
It will be c
1
6
2
6 and 2 because of newline
On Fri, Jun 11, 2010 at 9:26 AM, Rohit Saraf rohit.kumar.sa...@gmail.comwrote:
c
1
6,2
u might be expecting 5,1 if u are forgetting the newline character :)
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Rohit Saraf
Second Year
Explanation:
The prototype for printf as per ANSI C is:
*int printf( const char *format,…)*
the return value is integer and returns the number of characters
successfully read by printf.
Also,in case of printf(),the evaluation of expressions passed on as
arguments is done from right to
#include stdio.h
main()
{
int a = 1;
char b='c';
int i,j;
printf(%d,%d,printf(%d\n,a),printf(%c\n,b));
wat shd b the o/p of this..plzz explain y?
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c
1
6,2
u might be expecting 5,1 if u are forgetting the newline character :)
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Rohit Saraf
Second Year Undergraduate,
Dept. of Computer Science and Engineering
IIT Bombay
http://www.cse.iitb.ac.in/~rohitfeb14
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