C Programming - Arrays - Discussion

* means its represent as a pointer variable and here within a array bracket they give us 10 integers.

So that the ans is B.

Can you explain me the concept?

Yes p is pointer of 10 integers.

*ptr is a pointer and[10] is array declarations.

The brackets are the key to this problem. Always read C declarations from right to left (backwards) and start will stuff inside brackets.

[] = array * = pointer

int *ptr [10] backwards is [10] ptr * int

so this is Array of 10 pointers to int

int (*ptr)[10] bracket first and backwards is * ptr [10] int

so this is pointer to an array of 10 integers

#include<stdio.h>
int main()
{
int a=5;
printf("%d %d %d %d %d", a++, a--, ++a, --a, a);
return 0;
}

// Ans : 4 5 5 4 5

Can any one explain the above code please ?

@Chris

The given output '4 5 5 4 5' is matches in Turbo C - 16 OS (DOS).


But, I tried in GCC, the details are as below:

Warnings:=

prog.c: In function 'main':
prog.c:5: warning: operation on 'a' may be undefined
prog.c:5: warning: operation on 'a' may be undefined
prog.c:5: warning: operation on 'a' may be undefined
prog.c:5: warning: operation on 'a' may be undefined

Output:=

4 5 5 5 5

Please can anyone explain with full details here. Thanks in advance.

@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.