Parse wordcodes in Python 3.6 #621
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Just a heads-up I've been tinkering with this in #536. I didn't get very far, but I did get the new dict opcode change done. (It's most other things that are still broken in my PR). |
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By all means please feel free to take anything that might be useful from my PR. I have very little time available to work on this. |
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Thanks cjrh, I have not touched the dict changes yet, so it works out. |
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Single wordcode is 2 bytes wide and consists of an opcode and its argument (or 0 if no argument is used)
Prior to 3.6 it was called CALL_FUNCTION_VAR_KW and operated in a slightly different manner
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@abonie Sorry for the delay! No credit necessary, just go for it 👍 |
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| this.push(dict) | ||
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| VirtualMachine.prototype.byte_STORE_MAP = function() { | ||
| switch (constants.BATAVIA_MAGIC) { | ||
| case constants.BATAVIA_MAGIC_35: |
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At column 5: (no-useless-return) Unnecessary return statement.
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| this.push(dict) | ||
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| VirtualMachine.prototype.byte_STORE_MAP = function() { | ||
| switch (constants.BATAVIA_MAGIC) { | ||
| case constants.BATAVIA_MAGIC_35: |
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At column 5: (no-useless-return) Unnecessary return statement.
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Python 3.6 uses 2 bytes for each instruction (instead of varying number of bytes as in previous versions). This PR will introduce changes to VM that check for Python version to parse bytecode appropriately.
Single wordcode is 2 bytes wide and consists of an opcode and its
argument (or 0 if no argument is used)
TODO opcodes:
EXTENDED_ARG
???