Cout << "0123456789ABCDEF"[((n >> i*4) & 0xF)]; Any one plz can explain the meaning of this line,

It prints the hexadecimal value of n, where i is the digit number.

Each hexadecimal digit represents 4 bits, so you start by looping i from 0 to b/4, where b is the number of bits in n (8 for a char, 16 for a short int, 32 for a long int etc). The part of the code that is "n >> i * 4" shifts the bits in n right by the required amount while ".. & 0x0F" masks off any higher bits. What you are left with is exactly 4 bits containing the digit you are printing i.e. a number from 0 to 15. This is then used to index a string to send the appropriate char to cout.

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