1. Write A Program to move a block of data within the internal RAM

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1 UNIT 2: Example Programs. 1. Write A Program to move a block of data within the internal RAM Org 0h start1: mov r0,#40h ;r0 pointed to internal RAM 40h mov r1,#30h ;r1 pointing to internal RAM 030h mov r2,#5 ;r2 loaded with no. of elements in the array Start: ;data transfer djnz r2,start Sjmp Start1 ;decrement r2,if not equal to 0,continue with data ;transfer process. 2. WAP to move a block of data external RAM to internal RAM program for block data transfer from external RAM to internal RAM enter the elements from location 0500h(ext.RAM) Org 0h start1: mov dptr,#0500h //data pointer ppointed to external memory //0500h mov r1,#30h //r1 pointing to internal RAM 030h mov r2,#5 //r2 loaded with no. of elements in the array Start: movx A,@dptr inc dptr djnz r2,start Sjmp Start1 3. WAP to exchange data between internal RAM locations org 0000h mov r0,#30h ;r0 pointing to 030h(int. RAM)

2 mov r1,#40h ;r1 pointing to 040h(int. RAM) mov r2,#0ah ;r2 loaded with no. of elements to be exchanged start: is stored in temporary reg r3 mov r3,a mov is moved mov a,r3 ;data from r3 is moved ;increment data pointers djnz r2,start ;decrement counter r2,repeat the process if r2 is not zero nop 4. WAP to sort an array stored in the internal RAM org 0000h num equ 040h back1: mov r0,#50h ;store n elements(say n=5) from 50h mov a,r0 ;r0 and r1 are used as pointers mov r1,a mov r3,#04h ;load (n-1) to r3 (no. of passes) mov a,r3 mov r2,a ;load r3 to r2(no. of comparison in each pass) back: ;compare no. pointed to by r0 with no. pointed to by r1 mov num,@r1 cjne a,num,loop sjmp next loop: jc next ;if num at r0<no. at r1 continue with comparison process //jnc next for descing order mov r4,a ;else exchange the two numbers mov a,@r1 mov a,r4 next: djnz r2,back ;decrement no. of comparison djnz r3,back1 ;decrement no. of passes 5. WAP to find the largest element in ar array stored in the internal RAM.

3 org 0000h lar equ 040h ;location lar stores the largest element in the array start1: mov r0,#50h ;array location is 50h pointed to by r0 mov r2,#4h ;no. of elements in the array mov ;no. pointed to by r0 loaded to lar start: ;no. pointed to by r0+1 loaded to acc. cjne a,lar,big ;compare the two no.s sjmp next big: jc next ;jump to next if a<lar mov ;else acc loaded to lar next: djnz r2,start ;decrement count 6. WAP to add two 16 bit numbers // program to add two 16 bit numbers,result available in 40h,41h and 42h(40 lsb;41 msb;42 carry) start: mov r0,#20h ;r0 pointing to lsb of src1 mov r1,#30h ;r1 pointing to lsb of src2 ;add lsb of src1 and src2 add mov 40h,a ;result stored at 40h ;r0 pointing to msb of src1 ;r0 pointing to msb of src2 addc ;add msb of src1 and src2 with carry mov 41h,a ;result stored at 41h mov a,#0h addc a,#0h mov 42h,a ;carry stored at 42h 7. WAP to subtract two 16 bit numbers // program to subtract two 16 bit numbers,result available in 40h and 41h(40 lsb;41 msb) here: clr c ;clear carry bit

4 mov r0,#20h mov r1,#30h subb mov 40h,a subb mov 41h,a sjmp here ;r0 pointing to lsb of src1 ;r1 pointing to lsb of src2 ;sub lsb of src2 from lsb of src1 ;result stored at 40h ;r0 pointing to msb of src1 ;r0 pointing to msb of src2 ;sub msbs with borrow ;result stored at 41h 8. WAP to find the square of the number in the range 0h to ffh //Finds the square of the number in data RAM 20h and store the square in location30h and31h mov r0,#20h ;num in 20h is loaded into acc and b reg mov 0f0h,a mov 030h,a ;lower order product in 30h mov 031h,0f0h ;higher order product in 31h 9. Program to convert BCD to ASCII // The equivalent ASCII code will be found in Registers R2 and R6 10. WAP to find the cube of the number in the range 0h to fh //WAP to find the cube of the number in data RAM 20h and store the result in 30h mov r0,#20h ;num in 20h is loaded into acc and b reg mov 0f0h,a ;find the square of the no. mov 0f0h,@r0 ;multiply square of the no. by the no.to get the cube. mov 030h,a ;lower order product in 30h mov 031h,0f0h ;higher order product in 31h

5 11. WAP to convert BCD to ASCII start: mov r0,#20h ;r0 pointing to src location,loaded with a BCD no. ;no. moved to accumulator and added 30h to get equivalent ;ASCII add a,#30h mov 40h,a ;result stored at 40h 12. WAP to convert given decimal no. to equivalent ASCII start: mov r0,#30h ;r0 pointing to src locn. ;acc. loaded with decimal no. ;process of seperating the lower nibble and higher nibble of decimal no. anl a,#0fh mov r1,a anl a,#0f0h swap a mov r2,a ;mask higher nibble ;lower nibble stored at r1 ;mask lower nibble ;exchange higher and lower nibble positions ;higher nibble stored in r2 ;process of converting decimal to ASCII mov a,r1 add a,#30h mov 40h,a mov a,r2 add a,#30h mov 41h,a ;40h has ASCII value for lower nibble of decimal no. ;41h has ASCII value for higher nibble of decimal no.

6 13. WAP to convert given hex no. to equivalent decimal no. //Hex number has to be store at location x:5fffh and result to be stored in //next consecutive locations start: mov dptr,#5fffh movx mov 0f0h,#064h ;Load B reg with 100d or 64h div ab ;Hundreds inc dptr ;store in external ram mov a,0f0h ;remainder from b reg to acc mov 0f0h,#0ah ; Load B reg with 10d or 0ah div ab; inc dptr; ;store tens in external ram inc dptr mov a,0f0h ;store units in ext ram 14. WAP to convert given decimal no. to equivalent hexadecimal no. start: mov r0,#30h ;process of seperating the lower nibble and higher nibble of decimal no. anl a,#0fh mov r1,a swap a anl a,#0fh mov r2,a ;mask higher nibble ;lower nibble stored at r1 ;exchange higher and lower nibble position ;mask lower nibble ;higher nibble stored in r2

7 ;process of conversion ;dec no.=higher nibble*10d (or 0ah)+lower nibble*1 mov a,r1 ;lower nibble to acc mov 0f0h,#01h ;reg b=01 mov r3,a ;product in r3 mov a,r2 mov 0f0h,#0ah add a,r3 mov 40h,a ;higher nibble to acc ;reg b=10d ;product in acc. ;compute hex no. ;result at 40h

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