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:
#includestdio.h
#includeconio.h
int main()
{
int i;
i=5;
i=++i/i++;
printf(%d,i);
getch();
}
Why o/p is 1 and not 2?? what happened for
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 kartik.sac...@gmail.comwrote:
ans in both cases is zero
i think if return is written and nthg specified value is given it will
return by default
@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
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
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 kartik.sac...@gmail.comwrote:
@sukran if it is undefined in ansi c so it
thanks sukran...
i have one more question evaluation of printf is also undefined in C??
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group,
its right to left
On Wed, Sep 21, 2011 at 7:41 PM, kartik sachan kartik.sac...@gmail.comwrote:
thanks sukran...
i have one more question evaluation of printf is also undefined in C??
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks
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 agr.bhav...@gmail.com wrote:
another que..
#includestdio.h
main()
{
int a;
int i=10;
printf(%d %d %d\n,i+++i,i,i---i);
@ sukran
if it is giving same ans then there has to be some reason,
On Sep 19, 12:45 pm, sukran dhawan sukrandha...@gmail.com 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
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 abhishek.ma...@gmail.com wrote:
@ sukran
if it is giving
abhishek can you plz explain
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more
@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 dave_and_da...@juno.com
@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 wujinchen...@gmail.com wrote:
@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
@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
@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 wujinchen...@gmail.com 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
+1 dave.
Sanju
:)
On Tue, Sep 6, 2011 at 3:40 PM, Dave dave_and_da...@juno.com 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
thanks a lot guys :)
On Sep 6, 11:48 am, Rajeshwar Patra rajeshwarpa...@gmail.com wrote:
pointer points to read only memory location
and this address is then compared whch evaluates to true
hence the output...
--
You received this message because you are subscribed to the Google Groups
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.
--
*UTKARSH SRIVASTAV
CSE-3
B-Tech 3rd Year
@MNNIT ALLAHABAD*
--
You received this message because you are subscribed to the Google Groups
@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
it will be 4 3
On Aug 31, 9:01 pm, aditi garg aditi.garg.6...@gmail.com 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,
yes u r correct
On Aug 16, 8:22 pm, Sanjay Rajpal sanjay.raj...@live.in 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
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 p.venkatesh...@gmail.com wrote:
yes u r correct
On Aug 16, 8:22 pm, Sanjay Rajpal sanjay.raj...@live.in wrote:
it is compiler dependent
On Aug 1, 5:20 pm, thanu moorthy moorthyth...@gmail.com 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);
}
I think i got it... STRING always return address of S , which then get
summed with 1 and 2.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To view this discussion on the web visit
https://groups.google.com/d/msg/algogeeks/-/kFrZUxzFTxsJ.
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.
--
You received this message because you are subscribed to the Google Groups
compiler dependent and will print some address
On Sun, Aug 14, 2011 at 11:48 AM, sukran dhawan sukrandha...@gmail.comwrote:
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
what is the logic of your concept
On Aug 13, 5:01 pm, Kunal Patil kp101...@gmail.com 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.
--
You received this message because you are subscribed
@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.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to
Thanks a lot
On Aug 12, 11:25 am, Avinash Dharan avinashdha...@gmail.com wrote:
Pointer incrementation and subtraction are done in terms of memory blocks
and not addresses of memory.
For example,
int *p;
p++;
The pointer here jumps to the next integer location and not the next address
in
no one can explain such things correctly because they are compiler
dependent.The result is not deterministic,it may vary from one
platform to another
On Aug 1, 8:22 pm, Prashant Gupta prashantatn...@gmail.com wrote:
Nikhil only explained third rightly. :-/
I think 1,2,4's output are given wrong
@agyat +1
these outputs are not standardit's implementation dependent.
--
Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad
On Mon, Aug 1, 2011 at 8:55 PM, Agyat jalsa.n.sa...@gmail.com wrote:
no one can explain such things correctly because they are
The output for third only correct 1,2 and 4 are wrong i think
printf() is evaluated from right to left and the post inceremet
operator increses after assigning the value hence third is
correct.
also tested them using Dev C++ compiler its working as i said above
On Aug 1, 5:20 pm, thanu
i think there will be warning due to inner structure is not declaring
variable.
Inner structure must be like:
struct yy
{
char s;
struct xx *p;
}a;
On Jul 27, 9:22 pm, Aman Goyal aman.goya...@gmail.com wrote:
#include‹stdio.h›
main()
{
struct xx
{
int x;
struct yy
{
char s;
It is not showing compiler error on Codeblocks on my machine.
On Thu, Jul 28, 2011 at 2:26 PM, prabhat prabhat0...@gmail.com wrote:
i think there will be warning due to inner structure is not declaring
variable.
Inner structure must be like:
struct yy
{
char s;
struct xx *p;
}a;
no error in dev c++.
On Thu, Jul 28, 2011 at 2:43 PM, Piyush Kapoor pkjee2...@gmail.com wrote:
It is not showing compiler error on Codeblocks on my machine.
On Thu, Jul 28, 2011 at 2:26 PM, prabhat prabhat0...@gmail.com wrote:
i think there will be warning due to inner structure is
No error in gcc 4.3.2 too
On Thu, Jul 28, 2011 at 2:47 PM, Rohit Srivastava access2ro...@gmail.comwrote:
no error in dev c++.
On Thu, Jul 28, 2011 at 2:43 PM, Piyush Kapoor pkjee2...@gmail.comwrote:
It is not showing compiler error on Codeblocks on my machine.
On Thu, Jul 28,
Only a warning in Dev C++ due to no variable declaration of inner structure
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To view this discussion on the web visit
https://groups.google.com/d/msg/algogeeks/-/HtevjjeGDZoJ.
To post to this
hello is a string literal and this is what the C standard says about
string literals:
The multibyte character sequence is then used to initialize an array of
static storage duration and length just
sufficient to contain the sequence.
So memory will remain reserved for hello for the entire
No it will not cause a memory leak as its a string literal and it behaves
more like a read only entity.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To view this discussion on the web visit
Passing a pointer across functions does not create any problem and doesn't
cause memory leak. but passing an array ,for eg
int *fun()
{
int a[12];
return a;
}
this creates memory leak as the call stack no more holds the array. So, in
essence, though int *a and int a[] behave quite similarly
ideone.com is also great
On Jul 26, 10:03 pm, Ram CEG honest...@gmail.com wrote:
ya it would ignore \ alone..
On Tue, Jul 26, 2011 at 10:27 PM, aditi garg aditi.garg.6...@gmail.comwrote:
ya thats wat my doubt was...if its not a recognised escape sequence thn how
is it interpreted??
Hey, I guess , there is no defined order in which the function values are
evaluated .. Plz check this link ...
http://stackoverflow.com/questions/376278/parameter-evaluation-order-before-a-function-calling-in-c/376288#376288...
I think we should clear our concept on this part However, you
@T3rminal- Agreed. Thus the expressions like : a[i] = i++; , i=i++ - --i;
are irrelevant
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To view this discussion on the web visit
https://groups.google.com/d/msg/algogeeks/-/-lYHVFcDjXYJ.
To
+1 to t3rminal
On Jun 12, 11:38 pm, T3rminal piyush@gmail.com wrote:
@all
Stop guessing and making your own standards. C standards haven't
defined anything (though in gcc arguments are processed from left to
right) about processing arguments in a function call. And sentence
like
@nicks -- ur code for 1st problem is giving me 1 not -1 on gcc..which
compiler are you using
On Fri, Jul 22, 2011 at 11:58 AM, sumit sumitispar...@gmail.com wrote:
+1 to t3rminal
On Jun 12, 11:38 pm, T3rminal piyush@gmail.com wrote:
@all
Stop guessing and making your own standards. C
its showing same output...
On Jul 20, 6:40 pm, chetan kapoor chetankapoor...@gmail.com wrote:
but its showing output
On Wed, Jul 20, 2011 at 6:53 PM, mohit verma mohit89m...@gmail.com wrote:
hey guys...
1. char c='a';
while(c=='a')
{
printf(%c,c);
c=getchar();
}
guys just give input a stream of characters as: a a a a a a a a a
a a a ---(press enter)
for second case input as: c d e f g h i j b -(press enter)
and see the difference. (I know that printf() is line buffered. So
why?)
1.printf() shows 'a' one time only.
2.here it shows
In first case first character input is 'a' and second is space so loop
breaks
in second loop runs till b is not read hence all characters including spaces
are found in output
On Wed, Jul 20, 2011 at 7:29 PM, mohit mohit89m...@gmail.com wrote:
guys just give input a stream of characters
ohh ...got it.
On Jul 20, 7:06 pm, sunny agrawal sunny816.i...@gmail.com wrote:
In first case first character input is 'a' and second is space so loop
breaks
in second loop runs till b is not read hence all characters including spaces
are found in output
On Wed, Jul 20, 2011 at
@Nicks : Kindly share the source from where you are practicing these c
output questions ..
Thank You :)
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this
The keyword inline means the code is substituted for the declared inline
function call in its place itself and this is done at compile time and
mostly an optional one left to the compiler to substitute or not.
since you have used inline in the snippet it means i checks for the function
PROBLEM 2: First all the expressions are evauated from right to left in
printf() and then results printed with tha final value of i.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To view this discussion on the web visit
For 1st problem learn the concepts of linking.Scanf hasn't been required to
link yet in the main.
So during compling the program doesn't give any special treatment to scanf.
check this,
#includestdio.h
int charu(int x);
int charu(int x)
{
return x;
}
main()
{
int charu=67;
int scanf=78;
//int
But this works, where I have both the variable name and function name used:
#includestdio.hint charu(int x);int charu(int x)
{
return x;
}
main(){*
int a = charu(45);
int charu = a;*
int scanf=78;
//int printf=45;
int getchar=6;
@all
Stop guessing and making your own standards. C standards haven't
defined anything (though in gcc arguments are processed from left to
right) about processing arguments in a function call. And sentence
like assgnment to a preincrement expression is delayed vry mch
have no meaning .
Also,Please note that using relational/comparison operators with floating
point numbers/real numbers is not an advisable coding practice.
On Mon, May 30, 2011 at 8:12 PM, Dave dave_and_da...@juno.com wrote:
@vishal: Floats and doubles are stored in different formats, so
looking at the first 8
ya i have seen many questions like this in this group...its all
because of sequence points the output varies from compiler to compiler
On Jun 4, 9:10 pm, nicks crazy.logic.k...@gmail.com wrote:
i agree this type of expressions are not defined and are depend on the
machine..
it's
statement starts executing from the right side!!! 1st it reads it
from left !! while reading b increment two times and new value of b is
thn 5 nw it gets executed!!!
am i right??
On 6/4/11, hary rathor harry.rat...@gmail.com wrote:
in 1st pass
both ++b
++b= 4 then ++b= 5;
at second pass
Order of evaluation of operands is not defined in C and secondly the result
will be compiler implementation dependent. A compiler may store the sub
expressions operand values in a temporary storage or can make use of
variable value as an operand.
*In each case, either of the ++b may be executed
i agree this type of expressions are not defined and are depend on the
machine..
it's something like a[i]=i++;
the result may vary from machine to machine !
On Sat, Jun 4, 2011 at 5:00 PM, Rohit Sindhu rohit.sindhu.spec...@gmail.com
wrote:
1. If the operand values is to be stored in
An ANSI-compliant compiler is not required to generate an error for
undefined code. The code syntax is correct. ANSI doesn't say what the
compiler must do for undefined code, which is why it is undefined. The
compiler can do anything. It might do what you expect, or it might
not.
Don
On Jun 1,
in 1st pass
both ++b
++b= 4 then ++b= 5;
at second pass a= 5*5; because ++b is 5
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
That may be true, but it is not guaranteed. Having multiple side
affects between sequence points is undefined by the ANSI standard.
Therefore an ANSI-compliant compiler could produce an executable which
causes monkeys to fly out of your nose.
Don
On Jun 1, 11:27 am, anuj agarwal
if it is undefined by standard den y dont compilers follow it
On Wed, Jun 1, 2011 at 11:59 PM, Don dondod...@gmail.com wrote:
That may be true, but it is not guaranteed. Having multiple side
affects between sequence points is undefined by the ANSI standard.
Therefore an ANSI-compliant
@vishal: Floats and doubles are stored in different formats, so
looking at the first 8 hex digits of the two numbers isn't really
helpful.
The expression f 0.8 is evaluated as (double)f 0.8, so it would be
more useful to print all 16 hex digits of (double)0.8f and 0.8. Then
it would be easy to
conversion frm int to float and float to int is very poor in case of
printf funtion
so when we give the input to t then that value is displayed for all
the printf function
since %f modifier doesn't accept x (int) as
a successful argument so it takes the
latest float value..
On May 22, 2:41 pm,
one of my frnd askd me this question...
On 11 June 2010 21:34, Raj N rajn...@gmail.com wrote:
@kirubakaran: How can it be 1,1 ? No of characters read in a is 5+ 1 for
'\n' so its 6 and for the next one 1+1=2
On Fri, Jun 11, 2010 at 9:09 AM, kirubakaran kirubakaran1...@gmail.comwrote:
The only gurantee in this is that the outer printf result will come first.
Why ? because comma operator does not act as sequence point.
-Arun prasath N
On Sat, Jun 12, 2010 at 3:44 PM, divya jain sweetdivya@gmail.comwrote:
thanks to all for explanantions :)
On 12 June 2010 15:43, divya
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 left
Output will be
1,1
bcoz printf returns number of characters or integers printed
On Jun 11, 12:26 am, divya sweetdivya@gmail.com wrote:
#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
@kirubakaran: How can it be 1,1 ? No of characters read in a is 5+ 1 for
'\n' so its 6 and for the next one 1+1=2
On Fri, Jun 11, 2010 at 9:09 AM, kirubakaran kirubakaran1...@gmail.comwrote:
Output will be
1,1
bcoz printf returns number of characters or integers printed
On Jun 11, 12:26
70 matches
Mail list logo
