8051 program to find the smallest number

Let range be largest minus smallest. Output range. For more details visit www. A Discussion has been started about this question. To view it and take part, please click on the middle button shown below. Finding something on the music program Spotify. Log in. Math and Arithmetic. Study now. See answer 1. Best Answer. Clr psw. Study guides.

Algebra 20 cards. A polynomial of degree zero is a constant term. The grouping method of factoring can still be used when only some of the terms share a common factor A True B False. The sum or difference of p and q is the of the x-term in the trinomial. A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials. Q: Finding largest smallest number for microcontroller program? Write your answer Related questions. Program for microcontroller ?

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Write a program that will find the smallest largest and average values in a collection of N numbers Get the value of N before scanning each value in the collection of N numbers? Differences between 89s52 microcontroller and 89c51 microcontroller? In 89S51 what s means 89c51 what c means? What is a token in the context of a C program? What is the smallest addressable memory unit in C? In java What is a smallest individual unit in a program? Mov r0, 0FFh ; save the content of 50h in r0. Mov 0FFh, a ; move a to 50h.

Mov a, r0 ; get content of 50h in a. Movx dptr, a ; move it to h. Statement 2: — store the higher nibble of r7 in to both nibbles of r6. Solution: — first we shall get the upper nibble of r7 in r6. Then we swap nibbles of r7 and make OR operation with r6 so the upper and lower nibbles are duplicated. Mov a, r7 ; get the content in acc. Anl a, 0F0h ; mask lower bit. Mov r6, a ; send it to r6. Swap a ; xchange upper and lower nibbles of acc.

Orl a, r6 ; OR operation. Mov r6, a ; finally load content in r6. Statement 3: — treat r6-r7 and r4-r5 as two 16 bit registers.

Perform subtraction between them. Store the result in 20h lower byte and 21h higher byte. Solution: — first we shall clear the carry. Then subtract the lower bytes afterward then subtract higher bytes. Clr c ; clear carry. Mov a, r4 ; get first lower byte. Subb a, r6 ; subtract it with other. Mov 20h, a ; store the result. Mov a, r5 ; get the first higher byte. Subb a, r7 ; subtract from other. Mov 21h, a ; store the higher byte.

Statement 4: — divide the content of r0 by r1. Store the result in r2 answer and r3 reminder. Then restore the original content of r0. Solution:- after getting answer to restore original content we have to multiply answer with divider and then add reminder in that. Mov a, r0 ; get the content of r0 and r1. Mov b, r1 ; in register A and B. Div ab ; divide A by B.

Mov r2, a ; store result in r2. Mov r3, b ; and reminder in r3. Mov b, r1 ; again get content of r1 in B. Mul ab ; multiply it by answer. Add a, r3 ; add reminder in new answer. Mov r0, a ; finally restore the content of r0. Statement 5: — transfer the block of data from 20h to 30h to external location h to h. Solution: — here we have to transfer 10 data bytes from internal to external RAM.

So first, we need one counter. Then we need two pointers one for source second for destination. Mov r7, 0Ah ; initialize counter by 10d. Mov r0, 20h ; get initial source location. Mov dptr, h ; get initial destination location. Nxt: Mov a, r0 ; get first content in acc. Movx dptr, a ; move it to external location. Inc r0 ; increment source location. Inc dptr ; increase destination location. Djnz r7, nxt ; decrease r7. Statement 6: — find out how many equal bytes between two memory blocks 10h to 20h and 20h to 30h.

Solution: — here we shall compare each byte one by one from both blocks. Increase the count every time when equal bytes are found. Mov r0, 10h ; get initial location of block1. Mov r1, 20h ; get initial location of block2. Mov r6, 00h ; equal byte counter. Starts from zero. Nxt: Mov a, r0 ; get content of block 1 in acc. Mov a, r1 ; get content of block 2 in acc. Cjne a, b, nomatch ; compare both if equal.

Inc r6 ; increment the counter. Nomatch: inc r0 ; otherwise go for second number. Inc r1. Statement 7: — given block of h to h. Find out how many bytes from this block are greater then the number in r2 and less then number in r3. Store the count in r4. Solution: — in this program, we shall take each byte one by one from given block. Now here two limits are given higher limit in r3 and lower limit in r2.