Answer :
After executing the shift left instruction, the carry flag will hold the value 0, because the most significant bit shifted out of the 8-bit register is a 0. 1) 0
The question pertains to the behavior of the carry flag after performing a shift left operation on a binary number within an 8-bit register in assembly language. The shift left operation (SHL) moves each bit to the left by the number of positions specified.
In an 8-bit system, the most significant bit (MSB) is shifted out into the carry flag. The initial contents of the register are 01001101b and it is being shifted left by 1 (represented by 11b but since binary shift operations only use the least significant bits for the count, effectively the shift is by 1 bit, not 3 bits because binary 11 translates to 3 in decimal).
After the shift, the binary number becomes 10011010b, and the MSB 0 from the original value is shifted into the carry flag. Therefore, the carry flag will hold the value 0 after executing the code snippet.
1. The carry flag will hold the value 0 after the described shift operation.
The code mov dl, 01001101b sets the register dl to binary 01001101. Then, shl dl, 11b means we perform a logical left shift by 3 bits. The original binary value before the shift is 01001101, and after shifting by 3 bits, we get 01101000
In an 8-bit register:
- 1st bit shift: 10011011 (Carry flag: 0)
- 2nd bit shift: 00110110 (Carry flag: 0)
- 3rd bit shift: 01101100 (Carry flag: 0)
Since no bit shifted out of the most left bit during these shifts, the carry flag will be 0.
