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fix calculation of 8086 available memory range
Thanks to @rodggerbr
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@ -164,7 +164,7 @@ just as explained above. We have only 16-bit general purpose registers, which ha
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'0x10ffef'
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```
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where `0x10ffef` is equal to `1MB + 64KB - 16b`. An [8086](https://en.wikipedia.org/wiki/Intel_8086) processor (which was the first processor with real mode), in contrast, has a 20-bit address line. Since `2^20 = 1048576` is 1MB, this means that the actual available memory is 1MB.
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where `0x10ffef` is equal to `(1MB + 64KB - 16B) - 1`. An [8086](https://en.wikipedia.org/wiki/Intel_8086) processor (which was the first processor with real mode), in contrast, has a 20-bit address line. Since `2^20 = 1048576` is 1MB and `2^20 - 1` is the maximum address that could be used, this means that the actual available memory is 1MB.
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In general, real mode's memory map is as follows:
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