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pointless arguing whether the bug is that Byte2Str uses CX, or that Word2Str assumes that no one else is using CX To make things work again, you would have to stash the value somewhere other than in CX Pushing it onto the stack is your best bet if you run out of registers (You might hit on the idea of stashing it in an unused segment register such as ES-but I warn against it! Later on, if you try to use these utility routines in a program that makes use of ES, you'll be in a position to mess over your memory addressing royally, and once you move to protected mode you can't play with the segment registers at all Let segment registers hold segments Use the stack instead) Virtually everything that Word2Str does involves getting the converted digits into the proper positions in the target string A word requires four hexadecimal digits altogether In a string representation, the high byte occupies the left two digits, and the low byte occupies the right two digits Since strings are indexed from the left to the right, it makes a certain sense to convert the left end of the string first This is the reason for the XCHG instruction It swaps the high and low bytes of AX, so that the first time Byte2Str is called, the high byte is actually in AL instead of AH (Remember that Byte2Str converts the value passed in AL) Byte2Str does the conversion and stores the two converted digits in the first two bytes of the string at DS:SI For the second call to Byte2Str, AH and AL are not exchanged Therefore, the low byte will be the one converted Notice the following instruction: ADD SI,2 This is not heavy-duty math, but it's a good example of how to add a literal constant to a register in assembly language The idea is to pass the address of the second two bytes of the string to Byte2Str as though they were actually the start of the string This means that when Byte2Str converts the low byte of AX, it stores the two equivalent digits into the second two bytes of the string For example, if the high byte was 0C7H, the digits C and 7 would be stored in the first two bytes of the string, counting from the left Then, if the low byte were 042H, the digits 4 and 2 would be stored at the third and fourth bytes of the string, respectively The whole string would look like this when the conversion was complete: C742 As I've said numerous times before: Understand memory addressing and you've got the greater part of assembly language in your hip pocket Most of the trick of Byte2Str and Word2Str lies in the different ways they address memory If you study them, study the machinery behind the lookup table and target string addressing The logic nd shift instructions are pretty obvious and easy to figure out by comparison. asp.net upc-a reader .NET UPC-A Barcode Reader for C#, VB.NET, ASP.NET Applications
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As in Duf eld and Whitt [14], the conditional mean can be thought of as a deterministic uid approximation; for example, see Chen and Mandelbaum [8]. Since the withinlevel variation process has mean zero, the within-level variation process has no effect on this conditional mean. Hence, the conditional mean of the aggregate bandwidth process is just the sum of the conditional means of the component level processes. Unlike the more elementary M =G=I model considered in Duf eld and Whitt [14], however, the conditional mean here is not available in closed form. In order to rapidly compute the time-dependent conditional mean aggregate demand, we exploit numerical inversion of Laplace transforms. It follows quite directly from the classical theory of semi-Markov process that explicit expressions can be given for the Laplace transform of the conditional mean. More recently, it has been shown that numerical inversion can be an effective algorithm; see Abate et al. [1]. 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