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Join Date: Mar 2006
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Newbie Poster
what is the difference between statements below?
a kind of optimization problem.
X = Y*0; X = 0;
a kind of optimization problem.
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Join Date: Jun 2005
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I think you can see the difference right there for yourself. This has nothing to do with optimization, unless you enable the -fretarded flag on your compiler.
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I've never really heard of optimization on a high level programming language. In assembly perhaps.
So I would expect that X = 0 is more efficent that X = Y * 0;
The issue lies with memory access and compution. X = 0 accesses memory once where as the X = Y * 0 access memory twice and performs a calculation. If you look at the assembly commands to carry out these operations, you'll see the actual costs involved.
So I would expect that X = 0 is more efficent that X = Y * 0;
The issue lies with memory access and compution. X = 0 accesses memory once where as the X = Y * 0 access memory twice and performs a calculation. If you look at the assembly commands to carry out these operations, you'll see the actual costs involved.
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Join Date: Feb 2007
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You don't optimize wrong code.. You correct it..
Why would anyone do x= y * 0; if they want to set x to 0 ?!
Why would anyone do x= y * 0; if they want to set x to 0 ?!
Last edited by thekashyap : Mar 28th, 2007 at 2:44 pm. Reason: it -> if
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yeah as anything timesed by 0 = 0
i suppose x = 0 would be faster than x = y * 0 as its less steps?
i suppose x = 0 would be faster than x = y * 0 as its less steps?
Last edited by jbennet : Mar 28th, 2007 at 4:39 pm.
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Join Date: Jun 2005
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You might have x = y * CONSTANT where #define CONSTANT 0 is located someplace. That's how you could get that in your code.
Whether x = y * 0 is compiled to anything different than x = 0 is compiled to depends on the compiler.
Whether x = y * 0 is compiled to anything different than x = 0 is compiled to depends on the compiler.
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