I have a small query in c, I am using the bitwise left shift on number 69 which is 01000101 in binary
01000101 << 8 and I get answer as 100010100000000
Shouldn't it be all 8 zeros i.e. 00000000 as we shift all the 8 bits to left and then pad with zeros.
It is because of the literal (default data type) for a number (int) is, in most of nowadays CPU, greater than 8-bit (typically 32-bit) and thus when you apply
69 << 8 //note 69 is int It is actually applied like this
00000000 00000000 00000000 01000101 << 8 Thus you get the result
00000000 00000000 01000101 00000000 If you use, say, unsigned char specifically, then it won't happen:
unsigned char a = 69 << 8; //resulting in 0 This is because though 69 << 8 itself will still result in
01000101 00000000 But the above value will be casted to 8-bit unsigned char, resulting in:
00000000 Bit shift operators act on entire objects, not individual bytes. If the object storing 69 is wider than 1 byte (int is typically 4 bytes for example), then the bits that are shifted outside of the first (lowest/rightmost) byte overflow and are "pushed into" the second byte. For example:
00000000 00000000 00000000 01000101 //The number 69, stored in a 32 bit object 00000000 00000000 01010000 00000000 //shifted left by 8 If you had stored the number in a 1-byte variable, such as a char, the result would indeed have been zero.
01000101 //The number 69, stored in an 8 bit object (01000101) 00000000 //shifted left by 8 ^^^^^^^^ these bits have been shifted outside the size of the object. The same thing would happen if you shifted an int by 32.
00000000 00000000 00000000 01000101 //The number 69, stored in a 32 bit int 00000000 00000000 01010000 00000000 //shifted left by 8 00000000 01010000 00000000 00000000 //shifted left by 16 01010000 00000000 00000000 00000000 //shifted left by 24 00000000 00000000 00000000 00000000 //shifted left by 32, overflow 
(a+b) <<c Or a literal (like in the example): 1234 << x
intprecision