Consider:
echo 50 >> 4;
Output:
3
Why does it output 3?
The >> operator is called a binary right shift operator.
Shifting bits to the right 4 times is the same as dividing by two, four times in a row. The result, in this case would be 3.125. Since 50 is an int, bit shifting will return the floor of this, which is 3.
Put another way, 50 is 0b110010 in binary. Shifted 4 times we have 0b11, which is 3 in decimal.
For your convenience, one of the fastest ways to calculate the outputted value from a bitwise shift is to multiply or divide by 2.
For example echo 50 >> 4;
Given that this is a bitwise right, it literally means that the value will be decrease, then we can get the output by divide 50 for 2 and 4 times.
echo 50 >> 4; // 50/(2*2*2*2) ~ 3.
Given that (from) 48 -> (to) 63/16(2*2*2*2), the result will be more than 2 and less than 4. Then
echo 48 >> 4; // 48/(2*2*2*2) ~ 3.
echo 63 >> 4; // 63/(2*2*2*2) ~ 3.
However, when bitwise left, the result will be totally different as it multiplies by 2 with n times:
If echo 50 << 4; // 50*(2*2*2*2) ~ 800
If echo 51 << 4; // 51*(2*2*2*2) ~ 816
Live example: https://3v4l.org/1hbJe