Re: [algogeeks] Please explain the output
#includestdio.h
#define power(a) #a
int main()
{
printf(%d,*power(432));
return 0;
}
the printf statement, after preprocessing, will look like
printf(%d,*432);
so, when u print the value at the first position of the string, 52, which is
the ascii value of 4, will be printed.
On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla vaibhav200...@gmail.comwrote:
#includestdio.h
#define power(a) #a
int main()
{
printf(%d,*power(432));
return 0;
}
ans is 52 on gcc. Explain plss
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Re: [algogeeks] Please explain the output
printf(%d,*power(432)) will expand as
*printf(%d, *432)*
432 represents here a string and *432 is pointing to the first string
literal i.e 4 whose ascii value is 52..hence the output is 52
On Thu, Jun 23, 2011 at 4:02 PM, Shachindra A C sachindr...@gmail.comwrote:
#includestdio.h
#define power(a) #a
int main()
{
printf(%d,*power(432));
return 0;
}
the printf statement, after preprocessing, will look like
printf(%d,*432);
so, when u print the value at the first position of the string, 52, which
is the ascii value of 4, will be printed.
On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla
vaibhav200...@gmail.comwrote:
#includestdio.h
#define power(a) #a
int main()
{
printf(%d,*power(432));
return 0;
}
ans is 52 on gcc. Explain plss
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Vaibhav Shukla
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Re: [algogeeks] Please explain the output
hmm i got it.thnx
On Thu, Jun 23, 2011 at 4:05 PM, Piyush Sinha ecstasy.piy...@gmail.comwrote:
printf(%d,*power(432)) will expand as
*printf(%d, *432)*
432 represents here a string and *432 is pointing to the first string
literal i.e 4 whose ascii value is 52..hence the output is 52
On Thu, Jun 23, 2011 at 4:02 PM, Shachindra A C sachindr...@gmail.comwrote:
#includestdio.h
#define power(a) #a
int main()
{
printf(%d,*power(432));
return 0;
}
the printf statement, after preprocessing, will look like
printf(%d,*432);
so, when u print the value at the first position of the string, 52, which
is the ascii value of 4, will be printed.
On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla
vaibhav200...@gmail.comwrote:
#includestdio.h
#define power(a) #a
int main()
{
printf(%d,*power(432));
return 0;
}
ans is 52 on gcc. Explain plss
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Vaibhav Shukla
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best wishes!!
Vaibhav Shukla
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Re: [algogeeks] Please explain the output
#a is the replacement sequence which is substituted in the printf statement
The statements
#define power(a) #a
printf(%d,power(a));
is substituted as
printf(%d,a);
it is replaced with the string literal a . then *power(a) is converted as
value at that string literal address.
Hope this solves the problem :)
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I Blog @ www.opensourcemania.co.cc
On Thu, Jun 23, 2011 at 3:40 PM, vaibhav shukla vaibhav200...@gmail.comwrote:
#includestdio.h
#define power(a) #a
int main()
{
printf(%d,*power(432));
return 0;
}
ans is 52 on gcc. Explain plss
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Vaibhav Shukla
DU-MCA
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Re: [algogeeks] please explain the output
COOL BRO THIS IS A GOOD SOLN
On Tue, Apr 5, 2011 at 4:10 PM, Azhar Hussain azhar...@gmail.com wrote:
Few Important things about macros, before I explain the output
1. Macros are replaced in passes.
2. Macros are not recursive.
regarding the output remember the rule for expansion
A parameter in the replacement list, *UNLESS* preceded by a # or ##
preprocessing token or followed by a ## preprocessing token, is replaced by
the corresponding argument after all macros contained therein have been
expanded.
In other words, macros are replaced inside out unless # or ## exists
printf(%s,g(f(1,2))); is replaced as #f(1,2) --- f(1,2) according to
the replacement rule.
printf(\t%s,h(f(1,2))); As this does not replace with # or ## directly,
inside out expansion leads to h(1,2) -- g('1,2) -- 1,2
for the first pass
printf(%s, f(1,2)); --- g(a) #a
printf(\t%s, h(1,2));
second pass
printf(%s, f(1,2)); not processed(exhausted)
printf(\t%s, g(1,2)); -- h(a) g(a)
Third pass
printf(%s, f(1,2)); not processed(exhausted)
printf(\t%s, 1,2); -- g(a) #a
Hope this answers your question.
-
Azhar.
On Tue, Apr 5, 2011 at 3:22 PM, Vandana Bachani vandana@gmail.comwrote:
Hi Arvind,
These are preprocessor specific operators. Check out
http://msdn.microsoft.com/en-us/library/wy090hkc(v=vs.80).aspx
-Vandana
On Tue, Apr 5, 2011 at 12:45 PM, Arvind akk5...@gmail.com wrote:
#includestdio.h
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
int main()
{
printf(%s,g(f(1,2)));
printf(\t%s,h(f(1,2)));
return 0;
}
i have run this program in gcc compiler and getting : f(1,2) 12 as
output.
can anyone explain the reason for getting this output?
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Re: [algogeeks] please explain the output
nice explanation -- 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 options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] please explain the output
u can see the pre-processed file using gcc -E prog_name.cand @
bottom u can see what actually the code is doing.
On Tue, Apr 5, 2011 at 12:45 PM, Arvind akk5...@gmail.com wrote:
#includestdio.h
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
int main()
{
printf(%s,g(f(1,2)));
printf(\t%s,h(f(1,2)));
return 0;
}
i have run this program in gcc compiler and getting : f(1,2) 12 as
output.
can anyone explain the reason for getting this output?
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CoEP
BTech IT
8149198343
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Re: [algogeeks] please explain the output
thx pratik
On Sat, Apr 9, 2011 at 8:13 PM, Pratik Kathalkar
dancewithpra...@gmail.comwrote:
u can see the pre-processed file using gcc -E prog_name.cand @
bottom u can see what actually the code is doing.
On Tue, Apr 5, 2011 at 12:45 PM, Arvind akk5...@gmail.com wrote:
#includestdio.h
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
int main()
{
printf(%s,g(f(1,2)));
printf(\t%s,h(f(1,2)));
return 0;
}
i have run this program in gcc compiler and getting : f(1,2) 12 as
output.
can anyone explain the reason for getting this output?
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CoEP
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8149198343
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Re: [algogeeks] please explain the output
Hi Arvind,
These are preprocessor specific operators. Check out
http://msdn.microsoft.com/en-us/library/wy090hkc(v=vs.80).aspx
-Vandana
On Tue, Apr 5, 2011 at 12:45 PM, Arvind akk5...@gmail.com wrote:
#includestdio.h
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
int main()
{
printf(%s,g(f(1,2)));
printf(\t%s,h(f(1,2)));
return 0;
}
i have run this program in gcc compiler and getting : f(1,2) 12 as
output.
can anyone explain the reason for getting this output?
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Re: [algogeeks] please explain the output
Few Important things about macros, before I explain the output
1. Macros are replaced in passes.
2. Macros are not recursive.
regarding the output remember the rule for expansion
A parameter in the replacement list, *UNLESS* preceded by a # or ##
preprocessing token or followed by a ## preprocessing token, is replaced by
the corresponding argument after all macros contained therein have been
expanded.
In other words, macros are replaced inside out unless # or ## exists
printf(%s,g(f(1,2))); is replaced as #f(1,2) --- f(1,2) according to
the replacement rule.
printf(\t%s,h(f(1,2))); As this does not replace with # or ## directly,
inside out expansion leads to h(1,2) -- g('1,2) -- 1,2
for the first pass
printf(%s, f(1,2)); --- g(a) #a
printf(\t%s, h(1,2));
second pass
printf(%s, f(1,2)); not processed(exhausted)
printf(\t%s, g(1,2)); -- h(a) g(a)
Third pass
printf(%s, f(1,2)); not processed(exhausted)
printf(\t%s, 1,2); -- g(a) #a
Hope this answers your question.
-
Azhar.
On Tue, Apr 5, 2011 at 3:22 PM, Vandana Bachani vandana@gmail.comwrote:
Hi Arvind,
These are preprocessor specific operators. Check out
http://msdn.microsoft.com/en-us/library/wy090hkc(v=vs.80).aspx
-Vandana
On Tue, Apr 5, 2011 at 12:45 PM, Arvind akk5...@gmail.com wrote:
#includestdio.h
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
int main()
{
printf(%s,g(f(1,2)));
printf(\t%s,h(f(1,2)));
return 0;
}
i have run this program in gcc compiler and getting : f(1,2) 12 as
output.
can anyone explain the reason for getting this output?
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Re: [algogeeks] please explain the output
*The prototype of printf is*
*int printf(const char ***format**, ...);*
*
*
*Thus it takes a string and then variable number of arguments.*
*
*
*Every argument passed after (char *format)string is to resolve the%
inside the string (which is passed as the first argument)*
*
*
*Thus anuj will be printed as normal as printf(anuj);*
*Since anuj which is char *format for the printf function does not need to
resolve % because there aren't any.*
*
*
It will be good to read the printf implementation in Dennis Ritchie to get
a better understanding
On Fri, Nov 5, 2010 at 7:58 PM, ANUJ KUMAR kumar.anuj...@gmail.com wrote:
#includestdio.h
int main()
{
printf(anuj,kumar);
return 0;
}
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