Transcript Arithmetic and Logical instruction and CPU Flag

CPU Flags and Boolean
Instructions
Sahar Mosleh
California State University San Marcos
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1
The CPU flags
• Each instruction affects the CPU flags.
• The zero flag is set when the result of an operation equal zero.
• The carry flag is set when an instruction generates a result that is
too large (or too small) for the destination operand.
• The sign flag is a copy of the high bit of the destination operand
indicating that it is negative if set and positive if clear.
• The overflow flag is set when an instruction generates an invalid
signed result.
• The parity flag is set when an instruction generates an even
number of 1 bits in the low byte of the destination operand
Sahar Mosleh
California State University San Marcos
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Zero and Sign Flags:
• The zero flag is set when the destination operand of an arithmetic
instruction is assigned a value of zero.
• example:
mov cx,1
sub cx,1
; cx = 0, ZF = 1
mov ax, 0FFFFh
Inc ax
; ax = 0, ZF = 1
Inc ax
; ax = 1, ZF = 0
• The sign flag is set when the result of an arithmetic operation is
negative
Sahar Mosleh
California State University San Marcos
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• Example:
mov cx,0
sub cx,1
add cx,2
; CX = -1,
; CX= 1,
SF=1
SF = 0
Carry Flag:
• Carry flag is significant only when the CPU performs unsigned
arithmetic.
• The result of an unsigned addition operation is too large (or too small)
for the destination operand, the carry flag is set.
• Example:
mov al, 0FFh
add al,1
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; al = 00,
CF= 1
California State University San Marcos
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• The following diagram shows what happens at the bit level
1
Carry:
1
1
1 1 1 1 1
1 1 1 1 1 1 1 1
+
0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0
• On the other hand, if we add 1 to 00FFh in ax, the sum easily
fits into 16 bits and the carry flag is cleared
Sahar Mosleh
California State University San Marcos
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mov ax, 00FFh
add ax,1
;CF = 0,
ax = 0100h
• If we add 1 to FFFFh in the ax register, a carry is generated out
of the high bit of ax.
mov ax, 0FFFFh
add ax1
; CF = 1,
ax = 0000h
• If we subtract large unsigned integer from a smaller one, the
carry flag is set and the value in al is invalid.
mov al, 1
sub al, 2
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; CF = 1
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The Overflow flag:
• The overflow flag is relevant only when performing signed
arithmetic.
• It is set when an arithmetic operation generates a signed result
that can not fit in the destination operand.
• Example:
mov al, +127
add al, 1
; OF = 1
• Similarly,
mov al, -128
sub al, 1
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; OF = 1
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• Overflow has occurred if:
• Two positive operands were added and their sum is negative
• Two negative operands were added and their sum is positive
• Overflow never occurs, when the sign of two addition operands
are different.
• Example:
• When adding the binary values 10000000 and 11111110, there
is no carry from bit 6 to bit 7 but there is a carry from bit 7 into
the carry flag
• Overflow has occurred as displayed in the following
7 6 5
CF = 1
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1 0 0 0 0 0 0 0
+ 1 1 1 1 1 1 1 0
0 1 1 1 1 1 1 0
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Neg Instruction and flag
• This can produce an invalid result if the destination operand can not
be stored correctly
• Example, if we move -128 to al and try to negate it, the value + 128
can not stored in al.
• This causes the overflow flag to be set, and an invalid value to be
moved to al
mov al, -128
; al = 10000000b
neg al
; al = 10000000b,
OF=1
• On the other hand 1f +127 is negated, the result is valid and the
overflow flag is clear.
mov al, +127
; al = 01111111b
neg al
; al = 1000001b,
OF = 0
Sahar Mosleh
California State University San Marcos
Page
9
Boolean and comparison instruction
• We are going to begin the study of the conditional processing by
working at the binary level, using the four basic operations from
the boolean algebra AND , OR, XOR, and NOT.
• These operations are used in the design of the computer
hardware and software.
• The IA-32 instruction set contains the AND, OR, XOR, NOT,
TEST, and BTop instruction which directly implement boolean
operations between bytes, words and doublewords.
Sahar Mosleh
California State University San Marcos
Page 10
Operation
Description
AND
Boolean AND operation between a source operand and
the destination operand
OR
Boolean OR operation between a source operand and the
destination operand
XOR
Boolean XOR operation between a source operand and
the destination operand
NOT
Boolean NOT operation on a destination operand
TEST
Implies boolean AND operation between a source and
destination operand, setting the CPU flags appropriately
BT, BTC, BTR, BTS
Copy bit n from the source operand to the carry flag and
complement/reset/set the same bit in the destination
operand
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California State University San Marcos
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• The AND instruction performs a boolean (bitwise) AND
operation between each pair of matching bits in 2 operands and
place the result in the destination operand
AND destination, source
• The following operand combination are permitted
•AND reg, reg
•AND reg, mem
•AND reg, imm
•AND mem, reg
•AND mem, imm
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California State University San Marcos
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• The operand can be 8, 16, or 32 bits, and they must be the same
size.
• The following truth table labels the input bits x and y . The third
column shows the value of expression x AND y
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X
Y
X AND Y
0
0
0
0
1
0
1
0
0
1
1
1
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• The AND instruction is often used to clear selected bits and
preserve others.
• In the following example, the upper 4 bits are cleared and the
lower 4 bits are unchanged
00111011
AND 00001111
-----------00001011
mov
and
a1, 00111011b
a1, 00001111b
• The lower 4 bits might contain useful information while we don’t
care about the upper 4 bits
• This technique is bit extraction because the lower 4 bits are
pulled from AL
• The AND instruction always clears the overflow and carry flag. It
modifies the sign, zero, parity flag according to the value of the
destination operand
Sahar Mosleh
California State University San Marcos
Page 14
• Example:
• Converting characters to upper case:
• The AND instruction provides an easy way to translate a letter
from a lower case to upper case.
• If we compare the ASCII code for A and a, it becomes clear that
only bit 5 is different
001100001= 61h (‘a’)
001000001= 41h (‘A’)
• The rest of the alphabetic characters have the same relationship.
• If we AND any character with 11011111 binary, all bits are
unchanged except for the bit 5 which is clear.
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California State University San Marcos
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• The OR instruction performs a boolean (bitwise) OR operation
between each pair of matching bits in 2 operands and place the
result in the destination operand.
OR
destination, source
• The following operand combination are permitted.
OR reg, reg
OR reg, mem
OR reg, imm
OR mem, reg
OR mem, imm
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California State University San Marcos
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• The operand can be 8, 16, or 32 bits, and they must be the same
size.
• The following truth table labels the input bits x and y . The third
column shows the value of expression x OR y
Sahar Mosleh
X
Y
X OR Y
0
0
0
0
1
1
1
0
1
1
1
1
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• The OR instruction is often used to set selected bits and preserve
others.
• In the following example, 3Bh is ORed with 0Fh. The lower 4
bits of the result are set and the high 4 bits are unchanged
OR
00111011
00001111
-----------00111111
• The OR instruction can be used to convert a byte containing an
integer between 0 and 9 into an ASCII digit.
• To do this, you must set bits 4 and 5 if for example AL=05h, you
can OR it with 30h to convert it to ASCII code for the digit 5
(35h)
Sahar Mosleh
California State University San Marcos
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• Example:
OR
00000101
00110000
-----------00110101
05h
30h
35h, ‘5’
• The assembly language instruction to do this are as follows:
mov d1, 5
or d1, 30h
; binary value
; convert to ASCII
• Flags: The OR instruction always clears the carry and overflow
flags. It modifies the sign, zero, and parity flag according to the
value of the destination operand
Sahar Mosleh
California State University San Marcos
Page 19
• The XOR instruction performs a boolean (bitwise) XOR
operation between each pair of matching bits in 2 operands and
place the result in the destination operand.
XOR destination, source
• The following operand combination are permitted.
XOR reg, reg
XOR reg, mem
XOR reg, imm
XOR mem, reg
XOR mem, imm
Sahar Mosleh
California State University San Marcos
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• The operand can be 8, 16, or 32 bits, and they must be the same
size.
• The following truth table labels the input bits x and y . The third
column shows the value of expression x XOR y
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X
Y
X+Y
0
0
0
0
1
1
1
0
1
1
1
0
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• Special Quality for XOR is that it reverses itself when applied
twice to the same operand.
X
Y
0
0
1
1
0
1
0
1
X + Y (X + Y) + Y
0
1
1
0
0
0
1
1
• This reversible property of XOR makes it ideal tool for a simple
form of data encryption
• Flag: The XOR instruction always clears the overflow and carry
flags. It modifies the sign, zero, and parity flags according to the
value of the destination operand
Sahar Mosleh
California State University San Marcos
Page 22