Re: clear the concept of 2D array and pointers

Sumant wrote:
> Your concern is right!
> The question you are asking here is quite subtle.
> Apart from the address of elements you have to pay attention at the
> type of the element.
> See,
> when you have an "int s[10]"
> s gives you a value which is an address of first element of the
> array. (This is called "decay")
> The TYPE of this value is " int * ". Which means a pointer to an
> integer.
> &s[0] also gives you a value which is an address of the first element
> of the array but there is no decay here. The TYPE of this value is "
> int * ". Which means a pointer to an integer.
> Real interesting part starts here:
> &s is different thing all together as far as the TYPE is concerned. The
> value given by &s is same as s or &s[0] but the TYPE of &s is " int
> (*)[10] ". Note the difference. It says it is a pointer to an array of
> 10 integers. This is definitely not the same as pointer to an integer.
> The concept of "decay of array name to the pointer to the first
> element" in C really blurs this distinction!
>
> Now apply this concept to 2D arrays.
> int s[5][2]
> You understand s[0][0] and &s[0][0].
> Now s[0] is an array of integers. so s[0] DECAYS to "int *" A pointer
> to the first element of s[0] which is infact &s[0][0] by value. But
> &s[0] is not the same (though the address is same) The TYPE of &s[0] is
> " int (*)[2] " which means a pointer to an array of 2 integers. Now
> imagine what s is : It is, after decay, int (*)[2]. and &s is int
> (*)[5][2]. Note the difference. First is a pointer to an array of 2
> integers. Second one is a pointer to a 2 dimentional array of integers.
> Also note that decay happens only once for the last dimention of the
> array.
> So for s[5][2][3]
> s means int (*)[5][2]
> and
> &s means int (*)[5][2][3]
> I know this is getting very involved! Play with a c++ compiler instead
> of C compiler which spits out " type errors". such as int *p = &s; C
> compiler won't. Last note: CONCENTRATE ON TYPE NOT THE VALUE. Hope that
> helps.

hi frnd,
Very very thanks for my help. Ur concept is very useful
for me.I have found my answer.
Now i have a new problem.
let i=2;
so, j=i++ + ++i;
its anw would be ;j=6 and i=4 after excecution.
But now problem starts from here:
if i=2;
j= ++i + ++i;
then o/p after execution willbe:
j= 8 or unpredicted .
i cant explation howthis process has been done. pls help me.
take care .bye