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
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 y.neeraj2...@gmail.comwrote:
For Dev-C++, you have
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
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
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 y.neeraj2...@gmail.comwrote:
Yes, it would be like
Ignore last to last mail. Sorry. Do show expanded content in last mail.
On 16 Nov 2012 23:49, Neeraj Gangwar y.neeraj2...@gmail.com 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
ok..thnxi got it.your r ryt n i m ryt too:)..thnx
On Fri, Nov 16, 2012 at 11:54 PM, Neeraj Gangwar y.neeraj2...@gmail.comwrote:
Ignore last to last mail. Sorry. Do show expanded content in last mail.
On 16 Nov 2012 23:49, Neeraj Gangwar y.neeraj2...@gmail.com wrote:
Yes, it would
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 rahul23111...@gmail.comwrote:
but how
No...individually...dev cpp..how to compile both together???
On Thu, Nov 15, 2012 at 9:26 PM, Neeraj Gangwar y.neeraj2...@gmail.comwrote:
Which compiler are you using ? Are you compiling both the files together ?
*Neeraj Gangwar*
B.Tech. IV Year
Electronics and Communication IDD
Indian
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,
@rahulsharma
file1.c
#includestdio.h
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; //
@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
Pleaase reply with sol as asp
Fille 1:
#includestdio.h
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;
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,
Then why its not running?
On Wed, Oct 24, 2012 at 6:50 PM, SHOBHIT GUPTA
shobhitgupta1...@gmail.comwrote:
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
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 rahul23111...@gmail.comwrote:
Then why its not
#includestdio.h
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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You received this message because
yeahu r ryt
On Tue, Jul 10, 2012 at 10:01 AM, Firoz Khursheed
firozkhursh...@gmail.comwrote:
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
he is right.
On Tue, Jul 10, 2012 at 3:13 PM, rahul sharma rahul23111...@gmail.comwrote:
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
++i/i++= 6/6
++i * i++ = 36.00
http://ideone.com/j4n0Q
On Tue, Jul 10, 2012 at 3:13 PM, rahul sharma rahul23111...@gmail.comwrote:
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
what about post increment??
On Sun, Jul 8, 2012 at 10:37 PM, mitaksh gupta mitak...@gmail.com 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 rahul23111...@gmail.comwrote:
int i=5;
i=++i/i++;
print i;
i=1
how?
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
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 firozkhursh...@gmail.comwrote:
int i=5;
i=++i/i++;
print i;
i=1
coz ++ operator in c has preference from right to left, therefor first
(i++ is
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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int i=5;
i=++i/i++;
print i;
i=1
how?
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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
Sorry, its 6/6 and not 6/5,
regds.
On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar algog...@gmail.com 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)
agree with adarsh
On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar algog...@gmail.com wrote:
Sorry, its 6/6 and not 6/5,
regds.
On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar algog...@gmail.com wrote:
Firstly, this is ambiguous and expressions with multiple
increment/decrement operators
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 ali.mdshau...@gmail.comwrote:
agree with adarsh
On Sun, Jul 8, 2012 at 10:39 PM, adarsh kumar algog...@gmail.com wrote:
Sorry, its 6/6 and
I think it should output:
9 9
On Sun, Jul 8, 2012 at 11:42 PM, md shaukat ali ali.mdshau...@gmail.comwrote:
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
ali.mdshau...@gmail.comwrote:
.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 ashishjainco...@gmail.comwrote:
I think it should output:
9 9
On Sun, Jul 8, 2012 at 11:42 PM, md shaukat ali
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
vindhyachha...@gmail.comwrote:
.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
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 vindhyachha...@gmail.com wrote:
int a=10;
int b;
b=--a--;
printf(%d %d,a,b);. l
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 atul.87fri...@gmail.com 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
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 ali.mdshau...@gmail.com wrote:
then atul what would
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 rahul23111...@gmail.comwrote:
int i=5;
i=++i/i++;
print i;
i=1
how?
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To
#includestdio.h
#includeconio.h
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
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
btw your compiler has sizeof(int)=4;
thats why o/p = fg
On Thu, Jan 26, 2012 at 11:09 PM, atul anand atul.87fri...@gmail.comwrote:
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.
[-1] in end is same as -1 ??
On Thu, Jan 26, 2012 at 11:11 PM, atul anand atul.87fri...@gmail.comwrote:
btw your compiler has sizeof(int)=4;
thats why o/p = fg
On Thu, Jan 26, 2012 at 11:09 PM, atul anand atul.87fri...@gmail.comwrote:
output depends on sizeof(int)so it may be
think in terms of pointers...
they are same :-
p[-1] = *(p - 1)
On Thu, Jan 26, 2012 at 11:15 PM, rahul sharma rahul23111...@gmail.comwrote:
[-1] in end is same as -1 ??
On Thu, Jan 26, 2012 at 11:11 PM, atul anand atul.87fri...@gmail.comwrote:
btw your compiler has sizeof(int)=4;
@ atul...got nw..thnx
On Thu, Jan 26, 2012 at 11:26 PM, atul anand atul.87fri...@gmail.comwrote:
think in terms of pointers...
they are same :-
p[-1] = *(p - 1)
On Thu, Jan 26, 2012 at 11:15 PM, rahul sharma rahul23111...@gmail.comwrote:
[-1] in end is same as -1 ??
On Thu, Jan
#includestdio.h
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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