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ST20-C1 数据表(PDF) 39 Page - STMicroelectronics

部件名 ST20-C1
功能描述  Instruction Set Reference Manual
PDF  205 Pages
Scroll/Zoom Zoom In 100%  Zoom Out
制造商  STMICROELECTRONICS [STMicroelectronics]
网页  http://www.st.com
标志 STMICROELECTRONICS - STMicroelectronics

ST20-C1 数据表(HTML) 39 Page - STMicroelectronics

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4 Using ST20-C1 instructions
®
ldc 0;
ldl Xlo; ldl Ylo; addc; stl Zlo;
ldl Xhi; ldl Yhi; addc; stl Zhi
The subscripts ‘lo’ and ‘hi’, used here and in subsequent text, specify the least and
most significant word respectively of the double word variable with which they are
associated.
Subtraction of two double length values,
Y from X giving Z, without overflow signalling
is compiled as
ldc 0;
ldl Xlo; ldl Ylo; subc; stl Zlo;
ldl Xhi; ldl Yhi; subc; stl Zhi
Overflow signalling for signed arithmetic may be added by performing an extra
addc or
subc to produce a final w ord which contains only a sign (0 for positive or -1 for
negative) unless an overflow has occurred. For example, the following code could be
used to perform double length signed addition with overflow signalling:
clear carry, overflow and underflow status bits
ld Xlo; ld Ylo; addc; st Zlo;
ld Xhi; ld Yhi; addc; st Zhi;
ldc 0; dup; addc;
dup; adc #7ffffff;
- overflows if and only if carry word > 0
rev; adc #8000001;
- underflows if and only if carry word < -1
Multiple length multiplication
The
umac instruction multiplies two single word unsigned operands in Areg and Breg,
and adds the single word carry operand in Creg to form a double length unsigned
result. The more significant (carr y) word of the result is left in Breg, the less significant
in Areg. No overflow can be signalled by this instruction.
Multiplication of a single length unsigned value
X by a double length unsigned value Y
(leaving the ‘carry’ in Areg) can be performed by:
ldc 0;
ldl X; ldl Ylo; umac; stl Zlo;
ldl X; ldl Yhi; umac; stl Zhi
Double length unsigned multiplication is more complex. The product of two unsigned
double length words
X and Y can be expressed as:
X* Y= (Xhi*2
32 +X
lo)*(Yhi*2
32 +Y
lo)
=(Xhi*Yhi)*2
64 +(X
hi*Ylo +Xlo*Yhi)*2
32 +(X
lo*Ylo)
This can be coded as follows:
ldc 0;
ldl Xlo; ldl Ylo; umac; stl Z0
ldl Xlo; ldl Yhi; umac; rev; stl Z2



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