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ST9 数据表(PDF) 37 Page - STMicroelectronics

部件名 ST9
功能描述  USER GUIDE
PDF  146 Pages
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制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

ST9 数据表(HTML) 37 Page - STMicroelectronics

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ST9 USER GUIDE
This “eight bits signed by eight bits unsigned” multiplication with a “sixteen bits signed” result
takes a maximum of 36 cycles.
There are two divide instructions.
The div instruction divides a word by a byte, and returns the quotient and the remainder as the
low and the high bytes of the destination respectively. For example:
ldw
rr0,#31184
; rr0=#31184
ld
r2,#201
; r2=#201
div
rr0,r2
; rr0=1D9Bh, 1Dh=#29 and 9Bh=#155
This puts the value 155 in r1 (the quotient) and the value 29 (the remainder) in r0, and r2 still
contains 201.
If the divider is greater than the dividend, nothing is done. If the divisor is zero, a trap is trig-
gered that acts like an interrupt request, and uses the vector at locations 2 and 3 in program
memory. It is up to you to write the appropriate code to handle this trap. Finally, the numbers
to be divided should be such that the quotient be less than 256, that is, can be stored in a
single byte. Otherwise, the results are undefined.
The usable result is only the data stored in r1 which is 155 (for the previous example), the re-
mainder must be divided by the divisor (201) to give more precision (16-bits precision).
ldw
rr0,#31184
; rr0=#31184
ld
r2,#201
; r2=#201
div
rr0,r2
; rr0=1D9Bh, 1Dh=#29 the remainder and
; 9Bh=#155 the quotient
ld
r4,r1
; r4=9Bh=#155
clr
r1
; rr0=1D00h
div
rr0,r2
; rr0=0BC24h, 0BCh=#188 the remainder and
; 24h=#36 the quotient
ld
r0,r4
; rr0=09B24h, 09B24h means in fix-point
; with the point in the 16-bits middle
; 09B24h=155.140625 instead of
; #31184/#201=155.1442786
In the best case, the number of cycles required to divide a word by a byte with 16-bits preci-
sion is 80 cycles. This program has to manage overflow and divide by zero functions in order
to be able to be used.
The divws instruction performs one of the sixteen partial divides required to divide a double
word by a word, so you need to write a subroutine to perform the division completely. An ex-
ample subroutine is given in the ST9 Programming Manual.



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