设机器数字长8位(含1位符号位)，若机器数BAH为原码，算术左移1位和算术右移1位分别得（ ）。
【答案】：C原码左、右移均补0，且符号位不变(注意与补码移位的区别)。BAH=(10111010)2，算术左移一位(11110100)2=F4H，算术右移一位得(100l1101)2=9DH。
绛旓細銆愮瓟妗堛戯細[A+B]琛=1锛10l111A+B=涓17锛64[A鈥擝]琛=1锛100011A涓B=43锛64銆俒A+B]琛=1锛10l111,A+B=涓17锛64,[A鈥擝]琛=1锛100011,A涓B=43锛64銆
绛旓細銆愮瓟妗堛戯細[A+B]琛=1锛01111A+B= 涓17锛32[A涓B]琛ワ細0锛11101婧㈠嚭銆俒A+B]琛=1锛01111,A+B=涓17锛32,[A涓B]琛ワ細0锛11101,婧㈠嚭銆
绛旓細璁炬満鍣ㄥ瓧闀涓 8 浣锛岀敤琛ョ爜杩愮畻瑙勫垯璁＄畻锛欰 + B銆侫 =锛11/64 = 锛0 . 00 1011锛 [ A ]琛 = 0000 1011銆侭 =锛15/32 = 锛0 . 01 1110锛 [ B ]琛 = 1110 0010銆俒 A + B]琛 = 1110 1101銆侫 + B = 锛0.01 0011銆
绛旓細棰樼洰鎴戝彲涓嶄細鍋氾紝浣嗘垜甯綘鎵惧埌浜嗚繖涓鐩殑绛旀锛屼緵浣犲弬鑰 璁炬満鍣ㄦ暟瀛楅暱涓8浣嶏紙鍚1浣嶇鍙蜂綅锛夛紝鐢ㄨˉ鐮佽繍绠楄鍒欒绠椾笅鍒楀悇棰樸傦紙1锛堿=9/64锛 B=-13/32锛 姹侫+B锛涳紙2锛堿=19/32锛孊=-17/128锛屾眰A-B锛涳紙3锛堿=-3/16锛孊=9/32锛 姹侫+B锛涳紙4锛堿=-87锛 B=53锛 姹侫-B锛涳紙5锛堿=...
绛旓細璁炬満鍣ㄦ暟闀8浣嶅惈1浣嶇鍙蜂綅,鍗佸叚杩涘埗鏁86H鍒嗗埆琛ㄧず涓哄師鐮併佽ˉ鐮佹椂锛屾墍瀵瑰簲鐨勫崄杩涘埗鏁板垎鍒负_-6_銆乢-122_銆
绛旓細绛1 (-13)鍘熺爜: (-13) = 1000 0000 + 1101 = 1000 1101 (-13)鍙嶇爜: (-13) = 1111 1111 - 1101 = 1111 0010 (-13)琛ョ爜: (-13) 鍙嶇爜鏈浣庝綅+1寰楄ˉ鐮 = 1111 0011 (-13)绉荤爜: (-13) 琛ョ爜绗﹀彿浣鍙嶅緱绉荤爜 = 0111 0011 姝ｆ暟鐨勫師鐮,鍙嶇爜,琛ョ爜鏄叾鑷韩 (64)鐨勫師鐮佸弽鐮佽ˉ鐮: ...
绛旓細鏈哄櫒瀛楅暱涓 8 浣銆1111 0011锛屽崄杩涘埗鏄 243銆傚鏋滄槸琛ョ爜锛屽叾鐪熷间负锛243锛256 = 锛13銆傚埆蹇樹簡閲囩撼銆
绛旓細[A]琛=10011111 [B]琛=00101001 [-B]琛=~[B]琛+1=11010111 [A-B]琛=[A]琛+[-B]琛 =10011111+11010111 =01110110 锛堜涪寮冭繘浣嶏紝婧㈠嚭浜嗭級
绛旓細棣栦綅鏄绗﹀彿浣锛0姝ｆ暟1璐熸暟 100=64+32+4=2^6+2^5+2^2 鍘熺爜锛01100100 姝ｆ暟鐨勫弽鐮佽ˉ鐮佷笌鍘熺爜鐩稿悓
绛旓細1銆璁炬満鍣ㄦ暟瀛楅暱涓8浣嶏紙鍚1浣嶇鍙蜂綅锛夛紝鐢ㄨˉ鐮佽繍绠楄鍒欒绠椾笅棰樸侫=-16 B=37 姹侫-B 鍜孉+B [A]鍘燂紳10010000B [B]鍘燂紳[B]琛ワ紳00100101B [A]琛ワ紳11110000B [锛岯]鍘燂紳10100101B [锛岯]琛ワ紳11011011B [A-B]琛=[A+(-B)]琛 锛漑A]琛+[锛岯]琛 锛11110000B+11011011B 锛11001011B...
