Change bold code snippets to inline code blocks
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12-04.md
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12-04.md
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@ -76,9 +76,9 @@ in this case—one cycle before using a register to address memory
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produces no penalty; loading 2 cycles ahead is the only case that
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normally incurs a penalty. However, think of Rule \#4 as meaning that
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loading a byte register disrupts the memory addressing pipeline as it
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starts up. Viewed that way, we can see that **MOV BX,OFFSET MemVar**
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starts up. Viewed that way, we can see that `MOV BX,OFFSET MemVar`
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interrupts the addressing pipeline, forcing it to start again, and then,
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presumably, **MOV CL,AL** interrupts the pipeline again because the
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presumably, `MOV CL,AL` interrupts the pipeline again because the
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pipeline is now on its first cycle: the one that loading a byte register
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can affect.
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@ -150,7 +150,7 @@ Note that Listings 12.1 and 12.2 each repeat the timing of the code
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under test a second time, to make sure that the instructions are in the
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cache on the second pass, the one for which results are displayed. Also
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note that the code is less than 8K in size, so that it can all fit in
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the 486's 8K internal cache. If I double the **REPT** value in Listing
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the 486's 8K internal cache. If I double the `REPT` value in Listing
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12.2 to 2,000, making the test code larger than 8K, the execution time
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more than doubles to 224 µs, or 3.7 cycles per repetition; the extra
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seven-tenths of a cycle comes from fetching non-cached instruction
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