Example
| . Assume each
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100: ACTION-l
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|
action |
120: MOV 140, 364
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| block takes 20 |
132: GOTO 200
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| bytes of space
|
140: ACTION-2
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| .Start address |
160: HALT
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| of code for c |
:
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| and p is |
200: ACTION-3
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| 100 and 200
|
220: GOTO *364
|
| . The activation |
:
|
| records |
300:
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| arestatically |
304:
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allocated starting |
:
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| at addresses |
364:
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| 300 and 364.
|
368:
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This example corresponds to the code shown in slide 57. Statically we say that the code for c starts at 100 and that for p starts at 200. At some point, c calls p. Using the strategy discussed earlier, and assuming that callee.staticarea is at the memory location 364, we get the code as given. Here we assume that a call to 'action' corresponds to a single machine instruction which takes 20 bytes.
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