第一种除法算法 Start: place dividend in remainder 对n位商和余数,需要n+1步. 1. Subtract the Divisor register from the 余数 商 除数 Remainder register. and place the result 0000 0111 0000 0010 0000 Lin the Remainder register. Remainder≥0 Test Remainder <o remainder 2a. shift the 2b Restore the original value by adding the Quotient register Divisor register to the Remainder register, to the left setting place the sum in the Remainder register. Also the new rightmost shift the Quotient register to the left, setting bit to 1 the new least significant bit to 0 3. Shift the divisor register right 1 bit n+1 No:<n+repetitions epetition Yes: n+l repetitions (n=4 here 北京大学计算机科学技术系 Done 计算机系统结构教研室
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第一种除法的启示 除数中12的位数总是为0 >64位加法器的12浪费!! =>1/2的除数浪费!! °是否可以用余数左移替代除数右移? 第一步不能在商中产生1(由于开始时余数的高32位为0,或者 被除数太大) =>变换次序到首先移位然后再减,可以减少一次迭代 北京大学计算机科学技术系 计算机系统结构教研室
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第二种除法算法 32位除数寄存器、32位ALU、64位余数寄存器、32位商寄存器 Divisor 32 bits Quotient 32-bit ALl 32 bits Shift Left Shift left Remainder Contre 64 bits Write 北京大学计算机科学技术系 计算机系统结构教研室
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第二种除法算法 Remainder Quotient Divisor 0000011100000010 Q:0000 D:0010 R:00000111 1: ShIR Q:0000 D:0010 R:00001110 2:R=RDQ:0000 D:0010 R:11101110 3b:+D,sQ,0Q:0000 D:0010 R:00001110 1: ShIR Q:0000 D:0010 R:00011100 2:R=R-DQ:0000 D:0010 R:11111100 3b:+D,s|Q,0Q:0000 D:0010 R:00011100 1: ShIR Q:0000 D:0010 R:00111000 2:R=RDQ:0000 D:0010 R:00011000 3a:SQ,1Q:0001 D:0010 R:00011000 1: ShIR Q:0000 D:0010 R:00110000 2:R=RDQ:0000 D:0010 R:00010000 3a:sQ,1Q:0011D:0010 R:00010000 北京大学计算机科学技术系 计算机系统结构教研室
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第二种除法算法 Start: Place dividend in remainder Remainder Quotient Divisor 1. Shift the remainder register left i bit 0000011100000010 2. Subtract the divisor register from the left half of the Remainder register, place the result in the left half of the remainder register Remainder≥0 Test Remainder <o Remained 3a. Shift the 3b. Restore the original value by adding the divisor Quotient registerregister to the left half of the Remainder register, to the left setting &place the sum in the left half of the Remainder the new rightmost register. Also shift the Quotient register to the left, bit to 1 setting the new least significant bit to O th HNo:<repetitions epetition Yes: n repetitions(n=4 here) 北京大学计算机科学技术系 Done 计算机系统结构教研室
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