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Topic: Reason for operator precedence
Replies: 15   Last Post: Mar 15, 2006 8:56 AM

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briggs@encompasserve.org

Posts: 404
Registered: 12/6/04
Re: Reason for operator precedence
Posted: Mar 14, 2006 8:46 AM
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In article <1142342196.542632.294210@i39g2000cwa.googlegroups.com>, matt271829-news@yahoo.co.uk writes:
>
> Tony wrote:

>> Hi all.
>>
>> Hope this isn't a silly question.
>>
>> I was wondering what the reason is for having multiple levels of operator
>> precedence?
>>
>> Phrased another way, why is it that we don't just evaluate everything from
>> left to right?
>>
>> Having multiple levels of precedence obviously adds complexity, so I assume
>> there must be some payback. However, I don't see what it is.
>>

>
> As far as addition/subtraction vs multiplication/division is concerned,
> one reason is to ensure that the distributive property of
> multiplication works sensibly. For example, we want 3*(4 + 6) = 3*4 +
> 3*6 = 3*(6 + 4) = 3*6 + 3*4.


Remember that what we're talking about here is merely a notational
convention. It has nothing whatsoever to do with the distributive
property of multiplication over addition.

You can express the distributive law for multiplication over division
using parentheses:

a*(b+c) = (a*b) + (b*c)

You can express the distributive law for multiplication over division
using a convention in which * binds tighter than +:

a*(b+c) = a*b + b*c

You can express the distributive law for multiplication over division
using a convention in which + binds tighter than *:

a*b+c = (a*b) + (b*c)

You can express the distributive law for multiplication over division
using Polish notation:

*a+bc = +*ab*bc

You can express the distributive law for multiplication over division
using Reverse Polish notation:

bc+a* = ab*bc*+




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