In Assembly code: We have seen that multiplication by a power of two is easy with binary numerals: To multiply a number by 2^p corresponds to p left shifts. This works both for signed and unsigned numbers as long as we are careful with the range. Below is an attempt on a subroutine to multiply by a positive power of two. As noted in the contract, the assumption is that p is a strictly positive integer (i.e., 1, 2, 3 etc.). What happens when this subroutine is called with R5 = 3 and R6 = 11? Give a short but precise answer at the register level of the MCU
In Assembly code: We have seen that multiplication by a power of two is easy with binary numerals: To multiply a number by 2^p corresponds to p left shifts. This works both for signed and unsigned numbers as long as we are careful with the range. Below is an attempt on a subroutine to multiply by a positive power of two. As noted in the contract, the assumption is that p is a strictly positive integer (i.e., 1, 2, 3 etc.). What happens when this subroutine is called with R5 = 3 and R6 = 11? Give a short but precise answer at the register level of the MCU
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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In Assembly code: We have seen that multiplication by a power of two is easy with binary numerals: To
multiply a number by 2^p corresponds to p left shifts. This works both for signed and
unsigned numbers as long as we are careful with the range.
Below is an attempt on a subroutine to multiply by a positive power of two. As noted in
the contract, the assumption is that p is a strictly positive integer (i.e., 1, 2, 3 etc.).
multiply a number by 2^p corresponds to p left shifts. This works both for signed and
unsigned numbers as long as we are careful with the range.
Below is an attempt on a subroutine to multiply by a positive power of two. As noted in
the contract, the assumption is that p is a strictly positive integer (i.e., 1, 2, 3 etc.).
What happens when this subroutine is called with R5 = 3 and R6 = 11? Give a short
but precise answer at the register level of the MCU
Transcribed Image Text:; Subroutine: times_power_of_two
; input: R5 16-bit strictly positive number p --
;
R6 16-bit number x
;
; output: R6 16-bit number y = x*2^p
;
;
y is obtained by p left-shifts
times_power_of_2:
push
rla.w R6
dec.w R5
jne
R5
pop
ret
times_power_of_2
R5
returned unchanged
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