Write program in Assembly language which accepts two decimal digits from
its user, prints a new line on the display
screen, and if the first digit is larger than the second, displays the
average of the two digits (to the
nearest whole number) and otherwise displays the square root of the
product of the two digits (to
the nearest whole number). Then the program should return to DOS control.
Thus, if the user types
93
then the program displays
6
and if the user types
39
then the program displays
5
Hint: Think of easier way for square root calculation
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Light Poster
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| 16Aug2008,20:31 | #31 |
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Light Poster
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| 5Dec2008,11:07 | #32 |
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i have just checke this site. I found it very useful.
great job,buddy!!!! |
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Go4Expert Founder
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| 5Dec2008,13:18 | #33 |
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Newbie Member
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| 5Dec2008,16:47 | #34 |
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difference between addition of 8-bit numbers and 16-bit numbers
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Newbie Member
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| 8Dec2008,12:23 | #35 |
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salam
sir i need a exponential series program in assembly language. i m looking forward for ur replay because i need it urgently i have to submitted on firday. |
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Newbie Member
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| 22Sep2009,13:14 | #36 |
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Now i am learning assembly language programming.Would you mind providing me the details about this programming that how to write this?
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Newbie Member
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| 6Oct2009,21:53 | #37 |
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When we declare a variable like
var dw 92F3h its OK, but if we write it like var dw B2F3, it gives an overflow error. Now to overcome this, we need to give a 0 before it. var dw 0B2F3h. Now this works. My question is why in certain cases we have to give the 0 ? |
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Newbie Member
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| 31Oct2009,20:14 | #38 |
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39.A block starts at 0C20H and ends where 4 consecutive 00H are followed by FFH. Write a program to estimate the size of the block and store the count in BCD in COUNT.
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Newbie Member
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| 31Oct2009,20:19 | #39 |
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Write a program to fill 6410 locations with nos. as shown below: 1st 4 locations with 0FH
2nd 4 locations with 0EH 3rd 4 locations with 0DH last 4 locations with 00H and repeat the same till all locations are fully filled. |
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Newbie Member
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| 31Oct2009,20:23 | #40 |
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58. Two blocks of bytes exist in 0C20H and 0C40H. When the interrupt 6. 5 key is pressed, for each pressing the larger of 1st bytes of each block is to be written in location 0C60H. Each block consists of 10 bytes and hence after 10 interrupts a new block shall exist in OC60H with larger bytes of the two blocks. Disable the interrupt after the new block formation.
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