[algogeeks] C o/p help
#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?
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Re: [algogeeks] C o/p help
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 *argv[1]
hence output will be
o/p = cd
On Thu, Jan 26, 2012 at 10:53 PM, rahul sharma rahul23111...@gmail.comwrote:
#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?
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Re: [algogeeks] C o/p help
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.
(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 *argv[1]
hence output will be
o/p = cd
On Thu, Jan 26, 2012 at 10:53 PM, rahul sharma rahul23111...@gmail.comwrote:
#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?
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Re: [algogeeks] C o/p help
[-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 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 *argv[1]
hence output will be
o/p = cd
On Thu, Jan 26, 2012 at 10:53 PM, rahul sharma
rahul23111...@gmail.comwrote:
#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?
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Re: [algogeeks] C o/p help
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;
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.
(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 *argv[1]
hence output will be
o/p = cd
On Thu, Jan 26, 2012 at 10:53 PM, rahul sharma
rahul23111...@gmail.comwrote:
#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?
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Re: [algogeeks] C o/p help
@ 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 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 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 *argv[1]
hence output will be
o/p = cd
On Thu, Jan 26, 2012 at 10:53 PM, rahul sharma rahul23111...@gmail.com
wrote:
#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?
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