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
<[email protected]>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 <[email protected]>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 <[email protected]>
>> 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
>> >
>> > --
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