C Programming - Arrays - Discussion

@Chris and @Sundar:

"printf("%d %d %d %d %d", a++, a--, ++a, --a, a); " evaluates the value passed in it from right to left(int turbo C).
That is
first "a" will get its value as 5
then --a will store 4
then ++a will store 5 (as "--a" had decremented a's value n ++a incremented it)
then "a--" will have 5 ( as a-- is post increment therefore a's value will be affected after evaluation of the expression "a--")
then "a++" will have value 4 (b'coz above "a--" has made its value 4)

Therefore output is: 45545
Hope u know preincrement increments the value first and then use it in the expression, whereas postincrement uses the previous value of the variable and then modifies it thet is increments by 1.

eg:
int x=5 ,y,z;
y= ++x; //first x will become 6 then y will get assigned 6
z= x++; //first x will get assigned in z that is z=6 then x gets incremented.

Good night :) :)
Hope u understood....

Any one can explin main difference between a[] & *ptr[] with an example.

*ptr (10) means.

It is a pointer type aaray whch have 10 integers in it.

@Chris:
Turbo C uses __Cdecl__ type of Argument Passing mechanism in which rightmost arg is passed 1st and leftmost at atlast..
In printf(), as u might know, the leftmost format specifier is associated with leftmost value,but due to __Cdecl__, the evaluation of values is done from rightmost position..
So, 1st %d -> a++
2nd %d -> a--
3rd %d -> ++a
4th %d -> --a
5th %d -> a
This is binding is done,but as rightmost is evaluated 1st,
5th %d has value of a(5)
4th %d has --a(4) as preincrement has higher priority
3rd %d has ++a(5)
2nd %d has a--(5) as post increment is done after assignment
1st %d has a++(4) same reason as above..
So the output is as given..
Hope u have understood..
gd:)

I have got that ++a is pre-increment so it will increment and then assign the value.

Similarly a++ means it assigns and then increment so its value cannot be printed by this only it shows the value what we assigned..... clear per-increment post-increment is the concept used there...!!!!!

*ptr[10] means 'ptr is an array of pointer type 10 elements'.

(*ptr)[10] means there is an array of 10 elements with no array varaible but 'ptr is pointer type variable' that has base address of that array.

int (*ptr)[10];
here "ptr" is name of array,[10] is index of that array.
but '*'-it denotes the pointer varable is created of specified name. name-(ptr).

Here ptr is the pointer for the memory location occupied by array having 10 elements.

int (*arr)[10] /* It refers that arr is a pointer to an array of 10 integers*/
and,
int *arr[10] refers to an array of pointers which can hold the starting address of 10 different array of integer data type....

int (*ptr)[10];

Start reading what ever is in brackets 1st ,(ptr is a pointer)
Then go towards left till you hit ; (to an array of 10)
Then go backwards and read what ever is left out, (integers).