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There's an unusual behavior with the printing of *p .Following will be the output of *p :
Enter the pointer -------> Output in printf (*p)
1 -----------1
3 ----------- 3
5 ----------- 5
9 ----------- 9
10 ----------- 16
11 ----------- 17
14 ----------- 20
20 ----------- 32

scanf("%p",p); will take the input in hex , if we change it to scanf("%d",p); the output will be as expected, number entered will be in pure decimal and hence no conversion , it is the format specifier which is actually doing this.

The behavior is undefined.The explanation goes as follows:
If you want to change the pointer p to point to another memory location, you must use &p in scanf.
However, if you write p, then you must supply the format specifier as %d, because by p, you are actually passing the memory location where the pointer p is pointing[i.e. i]. So, whatever you input that gets stored in i indirectly.
Note that in printf, if the format specifier is changed then the programmer is responsible for all kinds of side effects.

Hey *p does not print 10 because of the scanf statement. You are taking a user input for p and then printing the value in *p that is why

The output is as desired in Dev C++

here throug you change the pointer of p which initially represent the 10 but after exection of scanf it has changed