[CS61C FA20] Lecture 09.1 - RISC-V Decisions II: Logical Instructions — Transcript

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1. 0:00[Music]
2. 0:09welcome back to risk 5 assembly we
3. 0:12have learned 15 assembly instructions so
4. 0:14far
5. 0:15and we are just about to add a dozen
6. 0:21we are going to look into logical
7. 0:24instructions in
8. 0:25this five assembly they should be fairly
9. 0:27familiar to us because we have seen in
10. 0:30high-level languages like c or java
11. 0:33bitwise ands and ors and xors
12. 0:36have their direct counterparts in the
13. 0:39assembly language
14. 0:40and so do shifts left or right
15. 0:44they're often used for isolating a byte
16. 0:47or a nibble
17. 0:48out of a word or by packing bytes or
18. 0:51nibbles
19. 0:52into a word for more efficient
20. 0:56storing of data let's take a look at a
21. 0:58few examples here
22. 1:00logical instructions always have two
23. 1:03variants
24. 1:04in risk five they have a register based
25. 1:06version
26. 1:07and an immediate version and that should
27. 1:10come
28. 1:10to no surprise to you because both of
29. 1:13them
30. 1:13are frequently needed and frequently
31. 1:15used and they're
32. 1:17relatively inexpensive to implement in
33. 1:20hardware
34. 1:21what we'll see is that they share the
35. 1:23same data path
36. 1:24as arithmetic instructions
37. 1:28so whatever we needed to implement ads
38. 1:31and subs
39. 1:31we'll be using for implementing
40. 1:35ors ants xors and shifts
41. 1:39so in this case let's take all an
42. 1:41example
43. 1:42um register based and bitwise and
44. 1:46so we'll take the contents of a register
45. 1:48of x6
46. 1:50and it with the contents of a register
47. 1:52x7 and store the result
48. 1:54in the destination register x5 and its
49. 1:56immediate counterpart will take the
50. 1:58value from the register x6
51. 2:01and it with an immediate binary value
52. 2:04that represents a three and stored
53. 2:06result in the
54. 2:08register x5
55. 2:11this kind of operation is used for
56. 2:13masking as i've said
57. 2:15um the previous immediate instruction
58. 2:17would isolate
59. 2:19the most uh the
60. 2:22the least significant bits on the right
61. 2:25and zero out everything else
62. 2:27um in this case and immediate
63. 2:30with an ff a hexadecimal will isolate
64. 2:34the least significant byte
65. 2:36if we end immediate with ff
66. 2:39in the most significant byte position we
67. 2:41would isolate
68. 2:42the most significant byte keep in mind
69. 2:46that the instruction itself is still
70. 2:48lean
71. 2:49in risk five and we don't have
72. 2:51unnecessary instructions the most
73. 2:52notable one that is missing there is no
74. 2:54not in this client
75. 2:56why because we don't need it if we
76. 2:58simply xor
77. 3:01our content of a register a value in the
78. 3:03register with all one's
79. 3:05binary um then we invert every single
80. 3:08one of those bits
81. 3:10remember this is done always for
82. 3:11simplicity unnecessary things are not
83. 3:13going to be there
84. 3:15let's take a look at logical shifting so
85. 3:19shift left logical comes in two forms
86. 3:22sll that operates on the register
87. 3:24contents
88. 3:25and the immediate that works with an
89. 3:28immediate value
90. 3:29so in this case ss
91. 3:33x11 x12 2 will take the value
92. 3:37in the register x12 shift it to
93. 3:40the left by two bit positions
94. 3:44and fill the two least significant bit
95. 3:47positions with zeros
96. 3:48so if we have a hex value of two in the
97. 3:53register beforehand
98. 3:55after shifting it by two to the left
99. 3:59we are going to end up with eight we
100. 4:01have essentially
101. 4:02moved this one zero
102. 4:05to the left by two bit positions and
103. 4:08inserted
104. 4:09zeros in the least significant bit
105. 4:11positions
106. 4:13so a quick question for you what
107. 4:15arithmetic
108. 4:16operation we have just executed
109. 4:19think about that for a sec we have
110. 4:22multiplied it
111. 4:23by four so two times four equals to
112. 4:26eight
113. 4:26and you can think of that this extends
114. 4:29easily to
115. 4:30multiplication with any power
116. 4:33of two so we can easily multiply with
117. 4:35two to the end
118. 4:37um but here's a quick question for
119. 4:41thinking about this how do we do
120. 4:42multiplication by
121. 4:4412 well we can
122. 4:48shift to the right by three which would
123. 4:51multiply by eight then
124. 4:52we multiply it by four and add
125. 4:56those two results we ended up with
126. 4:58multiplying
127. 5:00by 12. this is by the way commonly done
128. 5:02in dsp
129. 5:04now shift rate logical does the opposite
130. 5:08and shifts the the word
131. 5:12to the right by
132. 5:15a given amount and fills the most
133. 5:18significant bit positions with zeros
134. 5:21so that's a logical shift when we are
135. 5:24working inside numbers
136. 5:26we generally perform the arithmetic
137. 5:28shifting arithmetic shifting
138. 5:30does the same shift write arithmetic
139. 5:33that exist in register in in immediate
140. 5:36forms moves
141. 5:38and bits to the right and inserts the
142. 5:40sine
143. 5:41bit into the empty bits it basically
144. 5:43replicates that
145. 5:45topmost bit into
146. 5:48the empty positions that we have emptied
147. 5:50by shifting to the right
148. 5:52okay so in this example if x10 contained
149. 5:58a decimal value of minus 25
150. 6:01see notice its topmost bit is negative
151. 6:05so it's a negative
152. 6:06number if we execute shift right
153. 6:08arithmetic immediate
154. 6:10by four bit positions the result
155. 6:14is going to be this we are going to be
156. 6:16we
157. 6:17will shift everything to the right by
158. 6:19four bits
159. 6:21and fill the top most ones all with once
160. 6:25what's the result of this it's
161. 6:28kind of close to a division we took 25
162. 6:32and divided by that by 16 and we ended
163. 6:34up with
164. 6:35one point something minus one point
165. 6:37something
166. 6:40uh there is a slight issue here by
167. 6:43convention c arithmetic requires us
168. 6:47to round towards zero always so in this
169. 6:51case
170. 6:51the result should have been rounded
171. 6:54towards
172. 6:55minus one not towards minus two
173. 6:58which is the result that we got but
174. 7:01you will have to fix that to implement
175. 7:03through division
176. 7:05that's doable but requires a bit more
177. 7:07work
178. 7:08a few more instructions actually
179. 7:13that's it for the logical instructions
180. 7:16we'll continue with the rest of this 5
181. 7:18assembly
182. 7:19in just a bit see you there