convert left shifted LSB bits to 1

Question

I need to convert LSB 0 to 1 in left shift:

x=5;
int num = 0x02;//0b00000010
shiftVal = num << x;// 0b01000000

shiftVal should convert to 0b01011111

Answer 1

Simply set them to 1 using the or operator. To get all one bits just do (1 << x) - 1

hiftVal = (num << x) | ((1 << x) - 1);

See also

• Understanding the arithmetic in `(1 << j) - 1`
• Creating a mask with N least significant bits set
• Why does (x-1) toggle all the bits from the rightmost set bit of x?