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<P><br>
MOV a,R0 <BR>
ADD b,R0 <BR>
MOV c,R1 <BR>
ADD d,R1 <BR>
MOV
R0,t<SPAN class=down><sub>1</sub></SPAN> <BR>
MOV e,R0 <BR>
SUB R1,R0 <BR>
MOV t<SPAN
class=down><sub>1</sub></SPAN>,R1 <BR>
SUB R0,R1 <BR>
MOV R1,t<SPAN class=down><sub>4</sub></SPAN>
<BR><br>
图8. 13代码序列</P>
<P>另一方面,假设我们重新组织语句序列如下,其中t<SPAN class=down><sub>1</sub></SPAN>的求值刚好在t<SPAN class=down>4</SPAN>的求值之前:<br>
t<SPAN class=down><sub>2</sub></SPAN>:=c+d <BR>
t<SPAN
class=down><sub>3</sub></SPAN>:=e-t<SPAN class=down><sub>2</sub></SPAN> <BR>
t<SPAN
class=down><sub>1</sub></SPAN>:=a+b <BR>
t<SPAN class=down><sub>4</sub></SPAN>:=t<SPAN
class=down><sub>1</sub></SPAN>-t<SPAN class=down><sub>3</sub><br>
</SPAN>那么,运用8.5节的算法,我们将得到图8.14的代码序列(仍然假定仅R0和R1可用)。以这种顺序完成计算,我们可以节省两条指令:MOV R0,t<SPAN
class=down><sub>1</sub></SPAN>和MOV t<SPAN
class=down><sub>1</sub></SPAN>,R<SPAN class=down>1</SPAN>。<br>
MOV c,R0 <BR>
ADD d,R0 <BR>
MOV e,R1 <BR>
SUB R0,R1 <BR>
MOV a,R0
<BR>
ADD b,R0 <BR>
SUB R1,R0 <BR>
MOV R0,t<SPAN class=down><sub>4</sub></SPAN>
<BR><br>
图8.14 重新改写的代码序列</P><br>
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