[TUHS] Having trouble with V6 source code

[Jose R Valverde]

No. 

On PDP 11 assembler the left and right shifts were a single < and >.

You made a typo when transcribing the source statement: you typed

> 0636: mov $USIZE-1/<8|6, (r1)+

but the actual code from m40.s reads

> 0636:         mov     $usize-1\<8|6,(r1)+

However, since < and > where also used to delimit strings, there was a
need to escape them so as to distinguish their usage as shifts from the
string delimiters. The escaping was achieved by \ which is what you see
in the code.

See the assembler section of vol 2b of the Unix V7 manuals for details at
	http://plan9.bell-labs.com/7thEdMan/v7vol2b.pdf
or
	http://web.cuzuco.com/%7Ecuzuco/v7/v7vol2b.pdf
namely:

6.1 Expression operators
	The operators are:
	(blank) when there is no operand between operands, the effect is 
		exactly the same as if a ‘‘+’’ had appeared.
	+ addition
	– subtraction
	* multiplication
	\/ division (note that plain ‘‘ / ’’ starts a comment)
	& bitwise and
	| bitwise or
	\> logical right shift
	\< logical left shift
	...   ...   ...

				j

On Mon, 2 Oct 2006 09:44:58 +0300
jigsaw <jigsaw at gmail.com> wrote:
> hi all,
> 
> I just started to read the source code of V6 with Lion's book.
> 
> But before I went far I was stopped by m40.s
> 
> 0636: mov $USIZE-1/<8|6, (r1)+
> 
> What does the slash "/" stand for?
> 
> I guess this line should be
> 
> mov $USIZE-1<<8|6, (r1)+
> 
> Is "/<" the same as "<<"?
> 
> I checked in Unix PDP11 Assemble Refrence Manual but didn't find a clue.
> 
> Is it the right place to ask such question?
> 
> Thanks in advance
> 
> Regards,
> 
> Qinglai


	
	
		
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