In c++ what happens by the statement s>>a . Where s and a both are integer variables.

Question

but when the a is printed after this statement the output is 0(only when we don't enter a value in a before this… else the output of a is value of a) I couldn't get the reason behind this 0 as the output . As output should be the garbage value in a.

Answer 1

Just for clarification:

n >> m

its equal to

n/(2^m)

so 4>>2 == 4 /(2^2) = 1
512 >> 8 == 512 /(2^8)= 2

Answer 2

Arithmetic (signed) right shift operator >> and logical/arithmetic left shift operator <<, are responsible for manipulating data in bit level. Consider the following

int a = -32;
unsigned b = 32;
unsigned shift = 3;

Now, the binary representation of a and b values in 8bit words would be as

a = 1110 0000 = -32
b = 0010 0000 = 32

Shifting the number to right causes the number gets back filled with the sign bit value (MSB) for signed types (arithmetic shifting) and filling with 0 for unsigned types (logical shifting).

a >>= shift;

Yields 1111 1100 which is -4. But

b >>= shift;

Yields 0000 0100 which is 4.

Note: Left shift is always gets back filled with zero no matter which integral type you would use as data type.