HI, I'm seeking for a source for learning c-aps and i don get any valuable
ones. Can anyone plz suggest me or share the valuable sources u know. Thank
u in anticipation
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radhika .. :)
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Hi all, as i find it hard to find a site that has a collection of non
obvious or fairly hard c aps sites anywhere, i seek help here. plz post if
u find any source proving to be useful.
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Vicky
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This depends on how the compiler evaluates the actual parameters and
push to the stack when printf is called.
this is totally depeding on the compiler architecture [ basically
implemetion of activation record in intermediate code genaration
phase, read intermidiate language and virtual machine @ co
thanks ppl :)
On Mon, Jul 18, 2011 at 3:59 PM, schrodinger <6fae1ce6347...@gmail.com>wrote:
> AFAIK, official answer is "due to undefined behavior".
>
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AFAIK, official answer is "due to undefined behavior".
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@keyan
i examined many times in ubuntu that it delays the o/p of pre-increment
:)
but in dev C++ o/p is as given by atul
On Mon, Jul 18, 2011 at 3:15 PM, keyan karthi wrote:
> @boobal
> can u pls elaborate on "sequence points"
> @atul
> i compiled this in ubuntu
>
>
> On Mon, Jul 18, 2011 at 3:13
@boobal
can u pls elaborate on "sequence points"
@atul
i compiled this in ubuntu
On Mon, Jul 18, 2011 at 3:13 PM, Boobal Subramaniyam wrote:
> sequence points...
>
>
> On Mon, Jul 18, 2011 at 3:11 PM, Amol Sharma wrote:
>
>> it's not defined in the standard.unpredictable behaviour !!
>> -
sequence points...
On Mon, Jul 18, 2011 at 3:11 PM, Amol Sharma wrote:
> it's not defined in the standard.unpredictable behaviour !!
> --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
> MNNIT Allahabad
>
>
>
>
> On Mon, Jul 18, 2011 at 2:36 PM, keyan karthi
>
it's not defined in the standard.unpredictable behaviour !!
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Amol Sharma
Third Year Student
Computer Science and Engineering
MNNIT Allahabad
On Mon, Jul 18, 2011 at 2:36 PM, keyan karthi wrote:
> #include
> using namespace std;
>
> int main()
> {
> int p=1;
> printf("%d %d %d %d",++p,p
In dev C++ output is coming 5 3 2 2
On Mon, Jul 18, 2011 at 2:36 PM, keyan karthi wrote:
> #include
> using namespace std;
>
> int main()
> {
> int p=1;
> printf("%d %d %d %d",++p,p++,p++,++p);
> }
>
> output : 5 3 2 5
>
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#include
using namespace std;
int main()
{
int p=1;
printf("%d %d %d %d",++p,p++,p++,++p);
}
output : 5 3 2 5
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On Fri, May 20, 2011 at 3:19 PM, siva viknesh wrote:
>
> main()
> {
> int i = 257;
> int *iPtr = &i;
> printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
> }
> Answer:
> 1 1
>
i = 10001
first case *((char *)iPtr) cast to char 8 bits, discard first bit
0001 ==> 1
second *((char *)iPtr+1
hi
#include
main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("\n%d %d %d",a,*f1,*f2);
}
for this code in code blocks IDE got 8 8 8 as op
in http://ideone.com/ok850 got 12 12 12
in 175 c aps pdf it has been given as 16 16 16 as
ou
main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
..can anybody explain how??
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the printf function returns 1 when it prints a string..
for (1;1&&i--;1)
{
printf("bat");
}
note that i=2 has been cunningly initialized before. and the decrement of i
has been done in middle expression of the for loop
hence, actually its like
for(;i--;)
{
}
how many times will this execute? Twice
#include
int main()
{
int i=2;
for(printf("cat ");printf("rat ")&&i--;printf("mat "))
{
printf("bat ");
}
}
ouput : cat rat bat mat rat bat mat rat
can anybody plz explain how we get this output??
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the evalution will be done from right to left in printf
statement..so ++x will be evaluted and then x++.may be the out
put is 6 and 6 i think
On 13 December 2010 22:39, siva viknesh wrote:
> #include
> int main()
> {
> int x = 5;
> printf("%d %d", x++, ++x);
>
> return 0;
>
> }
>
#include
int main()
{
int x = 5;
printf("%d %d", x++, ++x);
return 0;
}
for this output is "6 7" ... how evaluation proceeds??
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