MID$(A$,5,2) = "AB" A$ 5 2 MID3$ "AB" AssignSubStrNotice the new code AssignSubStr and the lack of <ref> on A$. When the MID$ is processed by the Translator, the reference flags of its operands are cleared, hence no A$<ref>. However, the MID3$ needs to have its token reference flag set when the assign operator checks for a reference, since that MID3$ token will be on the done stack. Consider how this statement is processed at run-time.
At the A$, a reference string is pushed on the evaluation stack, that is, its pointer and length are copied to the stack. When the MID3$ is processed, it will pop the 2 and 5 off of the stack. The pointer on top of the stack will be changed to point to the fifth character in A$ and the length is set to 2 (assuming that A$ is at least 6 characters long).
When the AssignSubStr is processed, the "AB" string constant will be popped off of the stack. Two characters of this value are copied directly to the pointer that is on top of the stack (which is pointing to the fifth character in A$). A different AssignSubStr vs. AssignStr code is required because no allocation occurs, only a copy to an existing character array. An AssignSubStrTmp code is also required for the case that the value being assigned is a temporary string, which needs to be deleted after the characters are copied.
Since LEFT$ and RIGHT$ are also sub-string functions, there is no reason these functions can't also be used to assign sub-strings – and as it turns out, no extra code is required. Next, how sub-string assignments will be handled by the Translator...
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