@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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