Question Bank (2 marks) · Q9
Q.Write the summary of unsigned multipliers.
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MUL AB multiplies the unsigned bytes in A and B, leaving the low byte of the product in A and the high byte in B.
The 8051 performs unsigned multiplication with the single instruction MUL AB (there is no comma between A and B). Both operands must first be placed in the accumulator A and register B. On execution the two 8-bit numbers are multiplied to give a 16-bit result: the lower byte is stored in A and the upper byte in B. For example, 35H × 45H gives 0E49H, so B = 0EH and A = 49H.
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