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why this printf not working...

have answer?
it goes into infinite loop, look at the for loop condition. i<=5 ,
which actually(array[5]) outer to the array. And in storage in stack
this will overwrite the i=0 ,( array[5] is the same as i's mem location.)
and it goes to infinite loop ,
cheers..

pretty good q to understand the stack...if following printf() prints same addr(for a[5] & i) then u will go in infinite loop..

int
 main()
{
int i=0;
int array[5];
printf("addr a[5]:%p i:%p\n", &array[5], &i);
 
for(i=0;i<=5;i++)
   array[i]=0;

printf("why this printf not working?");
 return 0;
}

Its working.... I tried in Dev-C++

we have array of size 5
'0' will be assigned to
array[0]....1....2...3.....array[4]
in case of array[5] it will accept assignment but will not throw any error...
array[5]=0 is like nothing.... depends on C++ compiler...!

I still didn't understand why the address of i is the same as the address of array[5]? Even if you switch the order of the variable declarations from

int i=0;
int array[5];

to

int array[5];
int i=0;

, it seems to have the same behavior.
This is strictly compiler dependent I feel. This is the behavior under gcc.
Anybody?

Technically this code should work.
It worked at my end.

i am surprised which version of gcc you are using.

It may or may not work. We are accessing 6th element of array , where as it is declared to be array of 5 element.

working in my borland c compiler

its working on dev c++ compiler.!!

go to in infinite loop

go to in infinite loop

gcc -o2 <file.c> -o <exeName>

Friends try this solution:
............................
void main()
{
int i=0;
int j =0;
int array[5];
for(i=0;i<=5;i++)
array[i]=0;
printf("how this printf is working fine?");
}
.........................................
Above program'll clarify your doubts.

i tried in visual studio.. addresses are different and it is printed as usual..

Dear rockstar...
Visual studio is giving wrong results...How did you checked my dear friend ?

array[5] allocates space for 5 intergers.
when for loop executes it exceeds array boundaries.

Address of variable i - -4201868
Address of array element 0 - -4201896
Address of variable i - -4201868
Address of array element 1 - -4201892
Address of variable i - -4201868
Address of array element 2 - -4201888
Address of variable i - -4201868
Address of array element 3 - -4201884
Address of variable i - -4201868
Address of array element 4 - -4201880
Address of variable i - -4201868
Address of array element 5 - -4201876

As you can see compiler kept some spacing between array allocation and int allocation, if you run the loop 2 more times, it will overwrite address of i and then it becomes infinite loop.