How to shift left one specific bit?

Question

I want to shift left only one bit in a specific place leaving its position 0 , so I do not want to shift the whole variable with << operator, here is an example: say the variable has the value 1100 1010 and I want to shift the fourth bit then the result should be 1101 0010 .

Answer 1

Steps to get there.

1. Pull out bit value from the original number.
2. Left shift the bit value by one.
3. Merge the bit-shifted value back to the original number.

// Assuming C++14 or later to be able to use the binary literal integers
int a = 0b11001010;  
int t = a & 0b00001000;  // Pull out the 4-th bit.
t <<= 1;                 // Left shift the 4-th bit.
a = a & 0b11100111;      // Clear the 4-th and the 5-th bit
a |= t;                  // Merge the left-shifted 4-th bit.

Answer 2

For C++, I'd just use a std::bitset . Since you set the bit of pos + 1 to the value of the bit at pos , and then set the bit at pos to 0 this translate into bitset code that is quite easy to read. That would give you a function like

unsigned char shift_bit_bitset(unsigned char val, unsigned pos)
{
    std::bitset<8> new_val(val);
    new_val[pos + 1] = new_val[pos];
    new_val[pos] = 0;
    return new_val.to_ulong();
}

Answer 3

Maybe not the shortest/cleanest way, but this'll do it:

unsigned shift_bit = 4;
unsigned char val = 0xCA; // 1100 1010

unsigned char bit_val = val & (1 << shift_bit - 1); // Get current bit value
val = val & ~(1 << shift_bit - 1);                  // Clear initial bit location
val = bit_val ?                                     // Update next bit to 0 or 1
        val | (1 << shift_bit) : 
        val & ~(1 << shift_bit);

See it work with the test cases specified in your question and comments here: ideone

Answer 4

A simpler way is

(x & 0b11101111) + (x & 0b00001000)

that is, clear the bit that will be shifted into and add the bit to be shifted, which will overflow to the left if it is 1.