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During semantic analysis a parser usually attaches symbol info (of some kind) to the already existing abstract syntax tree, or creates a new tree entirely. When

by danhau 3y ago

During semantic analysis a parser usually attaches symbol info (of some kind) to the already existing abstract syntax tree, or creates a new tree entirely. Whenever it needs to know about a type, it just walks the tree to the node with the type definition. That way there’s really never any data that‘s forgotten. At least that’s how I think the parsers work I‘m familiar with.

trealira 3y ago
Yeah, it depends on the compiler. I've read a book about a BLISS compiler [1] that does this, but still uses a stack like I described [2]. It implements a hash table that used linked list nodes for collision. A new declaration adds a new name to the table, and it attaches the node to uses of the name in expressions of the syntax tree. When a scope is exited, the declarations from that scope are removed from the symbol table, but because they're still attached to the syntax tree, they can't just be freed. They're added to a linked list of "purged" nodes, so that the information they contain can be used later during code generation, and then freed. One-pass compilers don't have this problem; they really can just free the memory for reuse, because after they exit a scope, they've already generated the assembly or machine code from the high-level language. However, I don't know what LLVM or GCC, or any other remotely modern compiler, does. I haven't read the code much. [1]: https://en.wikipedia.org/wiki/The_Design_of_an_Optimizing_Compiler https://en.wikipedia.org/wiki/The_Design_of_an_Optimizing_Co... [2]: Actually, it intertwines the stack and the symbol table in a complicated way, so there's only one hash table, and multiple stacks within it. It's explained by a diagram they include on page 13. You can find a PDF of it here: https://kilthub.cmu.edu/articles/journal_contribution/The_design_of_an_optimizing_compiler/6610535 https://kilthub.cmu.edu/articles/journal_contribution/The_de...

by danhau 3y ago

trealira 3y ago

Yeah, it depends on the compiler. I've read a book about a BLISS compiler [1] that does this, but still uses a stack like I described [2]. It implements a hash table that used linked list nodes for collision. A new declaration adds a new name to the table, and it attaches the node to uses of the name in expressions of the syntax tree. When a scope is exited, the declarations from that scope are removed from the symbol table, but because they're still attached to the syntax tree, they can't just be freed. They're added to a linked list of "purged" nodes, so that the information they contain can be used later during code generation, and then freed. One-pass compilers don't have this problem; they really can just free the memory for reuse, because after they exit a scope, they've already generated the assembly or machine code from the high-level language. However, I don't know what LLVM or GCC, or any other remotely modern compiler, does. I haven't read the code much. [1]: https://en.wikipedia.org/wiki/The_Design_of_an_Optimizing_Compiler https://en.wikipedia.org/wiki/The_Design_of_an_Optimizing_Co... [2]: Actually, it intertwines the stack and the symbol table in a complicated way, so there's only one hash table, and multiple stacks within it. It's explained by a diagram they include on page 13. You can find a PDF of it here: https://kilthub.cmu.edu/articles/journal_contribution/The_design_of_an_optimizing_compiler/6610535 https://kilthub.cmu.edu/articles/journal_contribution/The_de...