ok..thnxi got it.your r ryt n i m ryt too:)..thnx
On Fri, Nov 16, 2012 at 11:54 PM, Neeraj Gangwar wrote:
> Ignore last to last mail. Sorry. Do show expanded content in last mail.
> On 16 Nov 2012 23:49, "Neeraj Gangwar" wrote:
>
>> Yes, it would be like copying the code in the other
Ignore last to last mail. Sorry. Do show expanded content in last mail.
On 16 Nov 2012 23:49, "Neeraj Gangwar" wrote:
> Yes, it would be like copying the code in the other file. You have to find
> a way to do it in Dev-C++.
> In linux it's simple. Just use *gcc file1.c file2.c *in terminal (as to
Ignore last to last mail. Sorry. Expand previous mail.
*Neeraj Gangwar*
B.Tech. IV Year
Electronics and Communication IDD
Indian Institute of Technology Roorkee
Contact No. : +91 9897073730
On Fri, Nov 16, 2012 at 11:49 PM, Neeraj Gangwar wrote:
> Yes, it would be like copying the code in the o
Yes, it would be like copying the code in the other file. You have to find
a way to do it in Dev-C++.
In linux it's simple. Just use *gcc file1.c file2.c *in terminal (as told
earlier).
If you are still confused, Think it this way
If you are compiling only one file in which you have declared varia
Think it this way
If you are compiling only one file in which you have declared variable as
intern, where would compiler find its actual definition because you are *not
compiling *the second file.
*file1.c : file in which variable is defined*
*file2.c : file in which variable is declared as exte
but with adding it willl copy aalll the codewe we dont need to copy..if
we declare int i in file 1...and include in file 2..then i can use it in
file 2 with its extern declaration...m i ryt?
On Fri, Nov 16, 2012 at 2:42 PM, Neeraj Gangwar wrote:
> For Dev-C++, you have to include one file in
For Dev-C++, you have to include one file in another.
So either add *#include "file1.c" *in file2.c and compile file2.c or
add *#include
"file2.c" *in file1.c and compile file1.c.
Hope this helps.
*Neeraj Gangwar*
B.Tech. IV Year
Electronics and Communication IDD
Indian Institute of Technology R
@rahulsharma
file1.c
#include
extern int i;// defintion provided in this file itself
extern int j; // definition provided in file2
void next()
{
++i;
other();
}
int main()
{
++i;
printf("%d\n",j);
printf("%d\n",i);
next();
}
int i=3;
// end of file1.c
file2.c
extern int i; // dec
That's why you are getting the error. You have to compile both the files
together. Search on google. I don't use dev c++.
*Neeraj Gangwar*
B.Tech. IV Year
Electronics and Communication IDD
Indian Institute of Technology Roorkee
Contact No. : +91 9897073730
On Thu, Nov 15, 2012 at 11:32 PM, rahul
No...individually...dev cpp..how to compile both together???
On Thu, Nov 15, 2012 at 9:26 PM, Neeraj Gangwar wrote:
> Which compiler are you using ? Are you compiling both the files together ?
>
> *Neeraj Gangwar*
> B.Tech. IV Year
> Electronics and Communication IDD
> Indian Institute of Techno
Which compiler are you using ? Are you compiling both the files together ?
*Neeraj Gangwar*
B.Tech. IV Year
Electronics and Communication IDD
Indian Institute of Technology Roorkee
Contact No. : +91 9897073730
On Thu, Nov 15, 2012 at 9:10 PM, rahul sharma wrote:
> but how can i use extern..if i
but how can i use extern..if i simply declare a variable in file1 as int j
and try to use in file2 with extern then it shows that j nit defined..how
cum file2 knows in which file j is definedfor e.g if i use extern in
file it means that this variable/fxn is defined somewhr else.then what are
th
@rahul it will compile perfectly well . note that you have declared j in
file 1 as extern and used it and have not provided its definition any where
so getting compile error.
as far as functions are concerned they are external by defaullt as
specified by @shobhit
i am attaching your corrected code
can nyone provide me dummy code of how exactly to use extern in c..
in dev environment
when i declare int i in one fyl
and try use use with extern int i in another then it doesnt compile..plz
coment
On Wed, Oct 24, 2012 at 9:58 PM, rahul sharma wrote:
> Then why its not running?
>
>
> On Wed, Oc
Then why its not running?
On Wed, Oct 24, 2012 at 6:50 PM, SHOBHIT GUPTA
wrote:
> http://www.geeksforgeeks.org/archives/840
>
> By default, the declaration and definition of a C function have “extern”
> prepended with them. It means even though we don’t use extern with the
> declaration/definitio
http://www.geeksforgeeks.org/archives/840
By default, the declaration and definition of a C function have “extern”
prepended with them. It means even though we don’t use extern with the
declaration/definition of C functions, it is present there. For example,
when we write.
int foo(int arg1, c
Pleaase reply with sol as asp
Fille 1:
#include
extern int i;
extern int j;
void next(void);
int main()
{
++i;
printf("%d",i);
next();
getchar();
}
int i=3;
void next()
{
++i;
printf("%d",i);
printf("%d",j);
other();
}
File 2:
extern int i;
void other()
{
++i;
printf
#include
int main()
{
int i;
char ch;
scanf("%c",&ch);
printf("%d",ch);
// getchar();
getchar();
}
when i enter one digit no. it showswhen 2 digit it halts...y so??? we
can store 2 digit number like 65 in 8 bit char???plz tell
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++i/i++=> 6/6
++i * i++ = 36.00
http://ideone.com/j4n0Q
On Tue, Jul 10, 2012 at 3:13 PM, rahul sharma wrote:
> yeahu r ryt
>
>
> On Tue, Jul 10, 2012 at 10:01 AM, Firoz Khursheed <
> firozkhursh...@gmail.com> wrote:
>
>> Well, when i compiled the code the output ie i is alway i=2,
>>
>> http
he is right.
On Tue, Jul 10, 2012 at 3:13 PM, rahul sharma wrote:
> yeahu r ryt
>
>
> On Tue, Jul 10, 2012 at 10:01 AM, Firoz Khursheed <
> firozkhursh...@gmail.com> wrote:
>
>> Well, when i compiled the code the output ie i is alway i=2,
>>
>> http://ideone.com/AFljo
>> http://ideone.com/87w
yeahu r ryt
On Tue, Jul 10, 2012 at 10:01 AM, Firoz Khursheed
wrote:
> Well, when i compiled the code the output ie i is alway i=2,
>
> http://ideone.com/AFljo
> http://ideone.com/87waz
>
> This expression is ambiguous, and compiler dependent.
> --
> You received this message because you are
Well, when i compiled the code the output ie i is alway i=2,
http://ideone.com/AFljo
http://ideone.com/87waz
This expression is ambiguous, and compiler dependent.
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"Algorithm Geeks" group.
To post to this group, send
i dont think it will work like u said...7/5i think it will go as
6/6=1..explain nyone???
On Mon, Jul 9, 2012 at 6:38 PM, Firoz Khursheed wrote:
> int i=5;
> i=++i/i++;
> print i;
>
>
> i=1
> coz ++ operator in c has preference from right to left, therefor first
> (i++ is ca;cu;ated) i=5 is u
int i=5;
i=++i/i++;
print i;
i=1
coz ++ operator in c has preference from right to left, therefor first
(i++ is ca;cu;ated) i=5 is used then it's incremented ie i=6 now. Now at
this point of time ++i is calculated, which makes i=7;
finally / operator is performed and i=7/5 is calculated, which m
what about post increment??
On Sun, Jul 8, 2012 at 10:37 PM, mitaksh gupta wrote:
> the o/p will be 2 not 1 because of the post-increment operator.
>
> On Sun, Jul 8, 2012 at 10:23 PM, rahul sharma wrote:
>
>> int i=5;
>> i=++i/i++;
>> print i;
>>
>>
>> i=1
>>
>> how?
>> --
>> You received
the o/p will be 2 not 1 because of the post-increment operator.
On Sun, Jul 8, 2012 at 10:23 PM, rahul sharma wrote:
> int i=5;
> i=++i/i++;
> print i;
>
>
> i=1
>
> how?
>
> --
> You received this message because you are subscribed to the Google Groups
> "Algorithm Geeks" group.
> To post to thi
it violates sequence pt. rule..so output is compiler dependent , but
as there is Lvalue error it would compile fine.
but in prev case pre decrement expects Lvalue but has r-value instead
bcoz of the post increment.
On 7/9/12, md shaukat ali wrote:
> then atul what would be the output of this pr
then atul what would be the output of this prob...
int a=10;
int b=a++*a--;
prinf ("%d",b);
On Sun, Jul 8, 2012 at 11:52 PM, atul anand wrote:
> b=--a--; this will result into compiler error because 1st the post
> decrement will occur and value will be saved in a temp variable . but
> you cannot
b=--a--; this will result into compiler error because 1st the post
decrement will occur and value will be saved in a temp variable . but
you cannot apply pre decrement on temp variable.
On 7/8/12, vindhya chhabra wrote:
> int a=10;
> int b;
> b=--a--;
> printf("%d %d",a,b);. l value error in this
int a=10;
int b;
b=--a--;
printf("%d %d",a,b);. l value error in this ques..
On Sun, Jul 8, 2012 at 11:48 PM, vindhya chhabra
wrote:
> .i think there will be an error in this -l value required, as post
> increment has more precedence than pre increment
>
>
> On Sun, Jul 8, 2012 at 11:44 PM, ashis
.i think there will be an error in this -l value required, as post
increment has more precedence than pre increment
On Sun, Jul 8, 2012 at 11:44 PM, ashish jain wrote:
> I think it should output:
> 9 9
>
>
> On Sun, Jul 8, 2012 at 11:42 PM, md shaukat ali
> wrote:
>
>> but i am confused in this
I think it should output:
9 9
On Sun, Jul 8, 2012 at 11:42 PM, md shaukat ali wrote:
> but i am confused in this problem...
> int a=10;
> int b;
> b=--a--;
> printf("%d %d",a,b);..what will output?
>
>
> On Sun, Jul 8, 2012 at 11:39 PM, md shaukat ali
> wrote:
>
>> agree with adarsh
>>
>>
>> On
but i am confused in this problem...
int a=10;
int b;
b=--a--;
printf("%d %d",a,b);..what will output?
On Sun, Jul 8, 2012 at 11:39 PM, md shaukat ali wrote:
> agree with adarsh
>
>
> On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar wrote:
>
>> Sorry, its 6/6 and not 6/5,
>>
>> regds.
>>
>> On Sun,
agree with adarsh
On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar wrote:
> Sorry, its 6/6 and not 6/5,
>
> regds.
>
> On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar wrote:
>
>> Firstly, this is ambiguous and expressions with multiple
>> increment/decrement operators will get executed according to
Sorry, its 6/6 and not 6/5,
regds.
On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar wrote:
> Firstly, this is ambiguous and expressions with multiple
> increment/decrement operators will get executed according to the compiler.
>
> Even if you consider the normal way, as we(humans) percieve it, it
Firstly, this is ambiguous and expressions with multiple
increment/decrement operators will get executed according to the compiler.
Even if you consider the normal way, as we(humans) percieve it, it will be
evaluated as
(++i)/(i++), which is 6/5, which is 1.
Simple!
On Sun, Jul 8, 2012 at 10:2
int i=5;
i=++i/i++;
print i;
i=1
how?
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F
@ atul...got nw..thnx
On Thu, Jan 26, 2012 at 11:26 PM, atul anand wrote:
> think in terms of pointers...
>
> they are same :-
>
> p[-1] = *(p - 1)
>
>
> On Thu, Jan 26, 2012 at 11:15 PM, rahul sharma wrote:
>
>> [-1] in end is same as -1 ??
>>
>>
>> On Thu, Jan 26, 2012 at 11:11 PM, atul ana
think in terms of pointers...
they are same :-
p[-1] = *(p - 1)
On Thu, Jan 26, 2012 at 11:15 PM, rahul sharma wrote:
> [-1] in end is same as -1 ??
>
>
> On Thu, Jan 26, 2012 at 11:11 PM, atul anand wrote:
>
>> btw your compiler has sizeof(int)=4;
>> thats why o/p = fg
>>
>> On Thu, Jan 26
[-1] in end is same as -1 ??
On Thu, Jan 26, 2012 at 11:11 PM, atul anand wrote:
> btw your compiler has sizeof(int)=4;
> thats why o/p = fg
>
> On Thu, Jan 26, 2012 at 11:09 PM, atul anand wrote:
>
>> output depends on sizeof(int)so it may be different if you run on
>> different compiler
btw your compiler has sizeof(int)=4;
thats why o/p = fg
On Thu, Jan 26, 2012 at 11:09 PM, atul anand wrote:
> output depends on sizeof(int)so it may be different if you run on
> different compilers.
>
> considering *sizeof(int) = 2;*
>
> argv[] is array of pointers.
> (p+=sizeof(int))[-1];
output depends on sizeof(int)so it may be different if you run on
different compilers.
considering *sizeof(int) = 2;*
argv[] is array of pointers.
(p+=sizeof(int))[-1];
p=p+2 // 2=sizeof(int);
now p will be pointing at index *argv[2];
then you are doing
p=p-1;
i.e p will point to *ar
#include
#include
void fun(char **);
int main()
{
char *argv[]={"ab","cd","de","fg"};
fun(argv);
getch();
return 0;
}
void fun(char **p)
{
char *t;
t=(p+=sizeof(int))[-1];
printf("%s\n",t);
}
o/p: fg
can nyone xplain
the 2nd statement in fun?
--
You receive
#include
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
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