#### Transcript S.E.Extc Sub : MPP 32 bit Muliplication PPT

Prof.V.P.Chitragar,KGCE,Karjat 32 bit multiplication Explanation: First number: 12345678H Second number: 11111111H As one memory location can store 1 byte of data, the above numbers will be requiring 4 memory locations ,each i.e. 78 will be stored at SI, 56 will be stored at (SI+1), 34 will be stored at (SI+2) and 12 will be stored at (SI+3). 1 Prof.V.P.Chitragar,KGCE,Karjat With LEA SI,A we are loading the starting address of number 12345678H in SI, i.e. for this 32 bit number the least significant byte is 78 and the address for 78 is the starting address . E.g. Address (SI) Data 2000 78 (SI+1) 2001 56 (SI+2) 2002 34 (SI+3) 2003 12 Lower word of A Higher word of A 2 Prof.V.P.Chitragar,KGCE,Karjat MOV AX,[SI] This instruction will do the following: Address Data (SI) 2000 (SI+1) 2001 (SI+2) 2002 (SI+2) 2003 78 56 34 12 AL AH 78 56 1st 2 words multiplication MUL B[BX] The second number is pointed by register BX ,i.e. 3 Prof.V.P.Chitragar,KGCE,Karjat BX points to the starting address of second number B . Address Data (BX) 3000 11 Lower word of B (BX+1) 3001 11 (BX+2) 3002 11 Higher word of B (BX+3) 3003 11 When MUL B[BX] is executed, the contents of AX are multiplied by the contents of [BX] and [BX+1], i.e. contents of 3000 and 3001 ,i.e. 5678H × 1111. 4 Prof.V.P.Chitragar,KGCE,Karjat The result of the multiplication is stored in i.e. DX AX 05C3 B5F8 As AX is used every time for arithmetic operations, we store the contents of AX to the memory location reserved for storing the result of multiplication. This is done by the following instruction: MOV C[DI],AX 5 Prof.V.P.Chitragar,KGCE,Karjat DI acts as a pointer for storing the result of multiplication , e.g. Address (DI) (DI+1) (DI+2) (DI+3) (DI+4) (DI+5) (DI+6) (DI+7) Data 4000 4001 4002 4003 4004 4005 4006 4007 F8 B5 6 Prof.V.P.Chitragar,KGCE,Karjat The contents of DX act as carry to be added to the next multiplication result. DX also is used every time for storing the higher word of the result of multiplication of two 16 bit numbers. Hence we store the contents of DX into CX with MOV CX,DX 2nd 2 words multiplication Now we multiply the higher word of A with lower word of B, i.e. 1234H × 1111H. 7 Prof.V.P.Chitragar,KGCE,Karjat But first, we move the higher word of A stored at (SI+2) and (SI+3) in AL and AH respectively, by the instruction MOV AX,[SI+2] i.e. AL 34 AH 12 So the 2nd 2 words multiplication will be executed by the instruction MUL B[BX] and the result will be stored in DX AX 0136 A974 8 Prof.V.P.Chitragar,KGCE,Karjat Now the present contents of AX have to be added with the previous contents of DX, which is stored in CX i.e. A974 + 05C3 = AF37 This is done with the help of the instruction ADD CX,AX The result of addition are copied to the memory location C[DI+2] and C[DI+3] with the help of the instruction MOV C[DI+2],CX The higher word of multiplication i.e. 0136H, which is stored in DX, is copied in CX with MOV CX,DX 9 Prof.V.P.Chitragar,KGCE,Karjat 3rd 2 words multiplication The next multiplication is between lower word of A and higher word of B i.e. 5678H × 1111H This is done by the instructions: MOV AX,[SI] MUL B[BX+2] As we know the result of multiplication is stored in DX:AX, the contents of AX are added to C[DI+2] and C[DI+3] with ADD C[DI+2],AX i.e. C[DI+2]=AF37+B5F8 = 1 652F 10 The higher word of multiplication result , which is stored in DX is added with contents of CX with ADC CX,DX i.e. CX=0136+05C3+1 =6FA 4th 2 words multiplication The higher word of A ,i.e. 1234H is copied to AX with MOV AX,[SI+2] This word is multiplied with the upper word of B, i.e. 1111H with MUL B[BX+2] The contents of AX are added to contents of CX with ADD CX,AX i.e. CX= 6FA+A974 =B06E Prof.V.P.Chitragar,KGCE,Karjat 11 Prof.V.P.Chitragar,KGCE,Karjat The added result stored in CX is copied to the memory location C[DI+4] and C[DI+5] with MOV C[DI+4],CX Any carry generated from previous addition is added to the present contents of DX with ADC DX,0000H This result in DX is copied in memory location with MOV C[DI+6],DX For knowing the starting address of the result C we use the instruction LEA SI,C 12