Question 1

The MUL instruction sets the Overflow flag if the upper half of the product is not equal to zero.

Accepted Answer

This statement is true. The MUL instruction multiplies two numbers and stores the result in a register. The upper half of the product is stored in a different register than the lower half, and if the upper half is not equal to zero, it means that the multiplication result was larger than what can be stored in the register and the Overflow flag is set.

Question 2

The binary value of AL after the following instructions have executed is 11101101.
mov al,01101011b
rol al,2

Accepted Answer

False

Question 3

The MUL instruction sets the Carry flag if the upper half of the product is not equal to zero.

Accepted Answer

This is true. The MUL instruction is used to multiply two unsigned integers and it sets the Carry flag if the upper half of the product is not equal to zero. If the upper half of the product is zero, the Carry flag is cleared.

Question 4

When the MUL BX instruction executes,the 32-bit product ends into the EAX register.

Accepted Answer

The MUL BX instruction multiplies the AX register by BX, and the result is stored in the DX:AX registers, not in the EAX register.

Question 5

The SAR instruction is suitable for signed integers.

Accepted Answer

SAR (Shift Arithmetic Right) is a signed shift instruction, which means it is designed to work with signed integers. It preserves the sign bit during the shift operation by shifting in copies of the sign bit rather than shifting in zeroes.

Question 6

The hexadecimal values of DX and AX after the following instructions have executed are DX=0005 and AX=0000.
mov dx,5000h
mov ax,20h
mul dx

Accepted Answer

When we multiply 20h with 5000h, the result in hexadecimal should be A0000h. However, since AX can only hold 16-bit values, the result is truncated to 0000h and stored in AX. The high-order 16 bits of the result (A0) are stored in DX. Therefore, the correct values of DX and AX after executing the above instructions should be DX=A000h and AX=0000h.

Question 7

RDX will equal 1 after the following instructions have executed.
mov rax,0FFFF0000FFFF0000h
mov rbx,2
mul rbx

Accepted Answer

The first instruction loads the 64-bit value 0FFFF0000FFFF0000h into the RAX register. The second instruction multiplies the value in RAX by 2, which results in the value 1FFFFFE000000000h. The value in RDX is equal to the upper 32 bits of the result, which is 1.

Question 8

The ROR instruction is only suitable for signed integers.

Accepted Answer

The ROR (rotate right) instruction is suitable for both signed and unsigned integers in most programming languages, including C and C++. However, in some specific cases, the programmer may need to take care of sign extension or carry flags when using the ROR instruction with signed integers.

Question 9

The binary value of AL after the following instructions have executed is 00001101.
mov al,01101011b
shr al,2

Accepted Answer

After executing `shr al,2`, the binary value of AL would be `00011010`. The `shr` instruction shifts the bits to the right by the specified count (in this case, 2), filling the leftmost bits with zeros.

Question 10

The SAR and SHR instructions perform the same operation.

Accepted Answer

The SAR (Shift Arithmetic Right) instruction performs a right arithmetic shift, which preserves the sign bit (most significant bit) of the operand, while the SHR (Shift Logical Right) instruction performs a right logical shift, which does not preserve the sign bit and shifts in zeros into the most significant bit.

Question 11

When the MUL BL instruction executes,the upper half of the product ends up in the AH register.

Accepted Answer

This is true. When the MUL BL instruction executes, the product of the multiplication is stored in the AX register, with the upper half in the AH register and the lower half in the AL register.

Question 12

The DIV instruction generates a divide overflow condition when the remainder is too large to fit into the destination operand.

Accepted Answer

The DIV instruction generates a divide overflow condition when the quotient is too large to fit into the destination operand, not the remainder.

Question 13

The destination operand of the IMUL instruction may be a memory operand.

Accepted Answer

In the IMUL instruction, the destination operand cannot be a memory operand; it must be a register.

Question 14

The binary value of AL after the following instructions have executed is 11100010.
mov al,11000101b
sar al,1

Accepted Answer

The instruction "mov al,11000101b" moves the binary value 11000101 into AL. 
The instruction "sar al,1" performs a signed arithmetic right shift of AL by 1 bit, which results in the binary value 11100010. Therefore, the answer is A – the binary value of AL is indeed 11100010.

Question 15

The binary value of AL after the following instructions have executed is 11111101.
mov al,11000101b
sar al,4

Accepted Answer

The instruction "sar al,4" performs a right shift operation on the binary value of AL by 4 bits, which means that the 4 least significant bits are dropped and 4 zeroes are added to the most significant bits. Therefore, the resulting binary value of AL should be 00001100, which is equivalent to 0x0C in hexadecimal or 12 in decimal, not 11111101.

Quiz 7: Integer Arithmetic Assessment

The MUL instruction sets the Overflow flag if the upper half of the product is not equal to zero.

The binary value of AL after the following instructions have executed is 11101101. mov al,01101011b rol al,2

The MUL instruction sets the Carry flag if the upper half of the product is not equal to zero.

When the MUL BX instruction executes,the 32-bit product ends into the EAX register.

The SAR instruction is suitable for signed integers.

The hexadecimal values of DX and AX after the following instructions have executed are DX=0005 and AX=0000. mov dx,5000h mov ax,20h mul dx

RDX will equal 1 after the following instructions have executed. mov rax,0FFFF0000FFFF0000h mov rbx,2 mul rbx

The ROR instruction is only suitable for signed integers.

The binary value of AL after the following instructions have executed is 00001101. mov al,01101011b shr al,2

The SAR and SHR instructions perform the same operation.

When the MUL BL instruction executes,the upper half of the product ends up in the AH register.

The DIV instruction generates a divide overflow condition when the remainder is too large to fit into the destination operand.

The destination operand of the IMUL instruction may be a memory operand.

The binary value of AL after the following instructions have executed is 11100010. mov al,11000101b sar al,1

The binary value of AL after the following instructions have executed is 11111101. mov al,11000101b sar al,4