can anybody explain that in following code y output is coming to be: 7 6 8
void call(int a,int b,int c)
{
printf("%d %d %d",a,b,c);
}
int main()
{
int a=5;
call(a++,a++,++a);
return 0;
}
On Sat, Jun 11, 2011 at 8:21 PM, PRAMENDRA RATHi rathi
<prathi...@gmail.com>wrote:
> IN second program:
> in function value are always push in the stack from right.
> so first value is --i that will make i=1 and value 1 will be passed to
> function
> and
> after that i++ that's means i will be passed.
> so 1 will be passed and after passing value. i will changed to 2.
>
> if u want to know why reverse order than can go through:
> http://cs.nyu.edu/courses/fall03/V22.0201-003/c_param.html<http://www.google.com/url?sa=D&q=http://cs.nyu.edu/courses/fall03/V22.0201-003/c_param.html>
> -----------------------------------------
> PRAMENDRA RATHI
> NIT ALLAHABAD
>
>
>
>
> On Sat, Jun 11, 2011 at 7:28 PM, Vishal Thanki <vishaltha...@gmail.com>wrote:
>
>> In 1st program, 2nd printf requires one more argument. And basically
>> %a is used for printing a double value in hex. see "man 3 printf".
>>
>> On Sat, Jun 11, 2011 at 5:29 PM, nicks <crazy.logic.k...@gmail.com>
>> wrote:
>> > Hello friends..plz help me in understanding the following C Output
>> >
>> > first one is --
>> >
>> > #include<stdio.h>
>> > #include<conio.h>
>> > main()
>> > {
>> > int a=5;
>> > printf("a=%d\n",a);
>> > printf("%a=%d",a);
>> > getch();
>> > }
>> > OUTPUT -
>> > a=5
>> > 0x1.2ff380p-1021=4199082
>> >
>> >
>> > and the other one is --
>> >
>> > #include<stdio.H>
>> > # include <conio.h>
>> > int i=2;
>> > main()
>> > {
>> > void add();
>> > add(i++,--i);
>> > printf("\ni=%d \n",i);system("pause");
>> > }
>> > void add(int a ,int b)
>> > {
>> > printf("\na=%d b=%d",a,b);
>> > }
>> >
>> > OUTPUT -
>> > a=1 b=1
>> > i=2
>> >
>> > --
>> > 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.
>> >
>>
>> --
>> 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.
>>
>>
> --
> 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.
>
--
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.
