thanks himanshu finally i got the reason!!
:)
On Sun, Jun 12, 2011 at 5:59 PM, himanshu kansal <
himanshukansal...@gmail.com> wrote:
> @anika:cz on gcc arguemnts r eval frm right to left and assgnment to a pre
> increment expression is delayed vry mch....
> so on eval frm right to left....
> frst a is incremented...(6) bt remember d new value is nt pushed on stack
> till nw....(cz assgnmnt is delayed)
> thn next value is 6...nd thn a is incremented.....here being a post
> increment op....assgnmnt is made 1st (2nd arg to fun is 6)and a is
> incrementd to 7....
> simalrly.....assgnmnt is made(1st arg is 7) and a is incremnted to 8.....
> nw d assgnmnt is made to the 3rd arg(d assgnmnt whch ws delayed till
> nw).....hence 3rd arg becomes 8.....
> so it prints 7 6 8......
>
> On Sun, Jun 12, 2011 at 5:33 PM, Anika Jain <anika.jai...@gmail.com>wrote:
>
>> 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
>>>> >
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>>
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>
>
>
> --
>
> Regards
> Himanshu Kansal
> Msc Comp. sc.
> (University of Delhi)
>
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