what does printf return??
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The number of characters printed.
On Sat, Sep 3, 2011 at 6:45 PM, priya ramesh love.for.programm...@gmail.com
wrote:
what does printf return??
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ok thank q :)
On Sat, Sep 3, 2011 at 6:51 PM, Nikhil Gupta nikhilgupta2...@gmail.comwrote:
The number of characters printed.
On Sat, Sep 3, 2011 at 6:45 PM, priya ramesh
love.for.programm...@gmail.com wrote:
what does printf return??
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What will be the output.
int i=5;
printf(%u,i);
What it will print:
i. 5
ii. Base address of the memory
iii. Physical address
iv. Logical address
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physical address i suppose...
On Fri, Aug 5, 2011 at 1:09 PM, anurag anurag19aggar...@gmail.com wrote:
What will be the output.
int i=5;
printf(%u,i);
What it will print:
i. 5
ii. Base address of the memory
iii. Physical address
iv. Logical address
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#includestdio.h
int main()
{
int i=1;
printf(%d%d%d%d,++i,++i,++i,i++);
return 0;
}
Explain how it will be evaluated ?? and the output obtained ..
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Values will be Pushed to the Stack from left to right and Evaluated to be
finally printed from the Stack.
On Fri, Jul 22, 2011 at 4:01 PM, rShetty rajeevr...@gmail.com wrote:
#includestdio.h
int main()
{
int i=1;
printf(%d%d%d%d,++i,++i,++i,i++);
return 0;
}
Explain how it will
Some more explanation taking an example will be helpful
take the example in the program.
Thanks
On Fri, Jul 22, 2011 at 4:03 PM, Prem Krishna Chettri hprem...@gmail.comwrote:
Values will be Pushed to the Stack from left to right and Evaluated to be
finally printed from the Stack.
On Fri,
Output:
5,4,3,1
*Explanation:*
Since the brackets acts as right precedence, the execution of the statement
is from right to left. The comma separates the individual.
For i++, it prints the current 'i' value and increments it by 1.
For ++i, it increments the value by 1 and prints the updated
undefined behaviour.
since value of i is changing more than once between two sequence points..
On Fri, Jul 22, 2011 at 5:42 PM, suresh srinivasan suree...@gmail.comwrote:
Output:
5,4,3,1
*Explanation:*
Since the brackets acts as right precedence, the execution of the statement
is from
associativity rule is compiler dependent ...thats why undefined...
On Fri, Jul 22, 2011 at 5:46 PM, Kamakshii Aggarwal
kamakshi...@gmail.com wrote:
undefined behaviour.
since value of i is changing more than once between two sequence points..
On Fri, Jul 22, 2011 at 5:42 PM, suresh srinivasan
There is NO universal way of doing this; every compiler does it differently,
in its own ways.
The problem here is that the above code changes the value of variable `i'
many times between two sequence points. So you cant be sure as to what value
`i' holds at what time!!!
For more insight and a
Its not the associativity which is undefined (Associativity has been defined
*clearly* by the C Standards for each and every operator). Its the order of
evaluation between 2 sequence
pointhttp://en.wikipedia.org/wiki/Sequence_points
which is undefined and hence compiler-dependent.
On gcc version
may be im wrongbut i read it somewhere about it...the arguments of
a function are evaluated in a associativity that is compiler
dependent...
for ex...
printf(%d %d %d ,fun1(),fun2(),fun3());
the order in which functions are evaluated are compiler dependent
On Fri, Jul 22, 2011 at 7:20
so many times such questions have come... behavior is entirely compiler
dependent
if you are preparing for interviews then believe me no one in the whole
world will ask such questions to you. there's no point discussing it
On Fri, Jul 22, 2011 at 8:15 PM, Gaurav Popli abeygau...@gmail.com
@Kamakhsi
In my ubuntu gcc this o/p is coming with warning of undefined %# :)
On Sat, Jul 16, 2011 at 5:43 AM, sukhmeet singh sukhmeet2...@gmail.comwrote:
@Anatony
the output will be compiler dependent
res1 is not defined .. as C don't allow to change the value of a variable
more than
and o/p is %#s Zi not %s Zi
On Sat, Jul 16, 2011 at 2:18 PM, sagar pareek sagarpar...@gmail.com wrote:
@Kamakhsi
In my ubuntu gcc this o/p is coming with warning of undefined %# :)
On Sat, Jul 16, 2011 at 5:43 AM, sukhmeet singh sukhmeet2...@gmail.comwrote:
@Anatony
the output will
tell me your output pls
On Sat, Jul 16, 2011 at 2:19 PM, sagar pareek sagarpar...@gmail.com wrote:
and o/p is %#s Zi not %s Zi
On Sat, Jul 16, 2011 at 2:18 PM, sagar pareek sagarpar...@gmail.comwrote:
@Kamakhsi
In my ubuntu gcc this o/p is coming with warning of undefined %# :)
@Antony
res1=++a + ++a + ++a;
Well it depends on the compiler but i know how gcc works :)
from left to right it will first do addition of first two 'a'
before addition it will increment the value of a by two cos of two pre
increments.
then resulting addition will then be added to the incremented
Hi,
See the below link for detailed explanation of special printf format
specifiers in C...
http://www.cplusplus.com/reference/clibrary/cstdio/printf/
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D Kranthi kumar
Computer Science Engg.
1st Mtech, IIT Madras.
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Whats wrong in this?
o/p is-- %.#s Zi
*%.#s *is printed as usual and *Zi *is printed coz of *%.2s *it will print
only 2 characters as written after decimal place. :)
It wil ignore 2nd str
On Fri, Jul 15, 2011 at 1:31 AM, rShetty rajeevr...@gmail.com wrote:
#includestdio.h
int main()
{
char
can any tell and explain the output of following code
#includestdio.h
main()
{ int a =5, b=5;
int res1=(++a)+(++a)+(++a);
int res2=(++b)+(++b)*10+(++b)*100;
printf(%d\n%d\n,res1,res2);
}
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@sagar:o/p is #s zi..can u explain y?
On Sat, Jul 16, 2011 at 1:45 AM, Antony Kotre antonyko...@gmail.com wrote:
can any tell and explain the output of following code
#includestdio.h
main()
{ int a =5, b=5;
int res1=(++a)+(++a)+(++a);
int
@Anatony
the output will be compiler dependent
res1 is not defined .. as C don't allow to change the value of a variable
more than once between a sequence point..
A sequence point occur while assigning a value , calling a function or
returning from it..
Hence both res1 and res2 would give arbitary
#includestdio.h
int main()
{
char str[] = Zingle Bell Zingle Bell;
printf(%.#s %.2s\n,str,str);
return 0;
}
I need the help to figure out the output of the program and please
expalin the behaviour of the printf statement here ?
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