9 ms·

The entities and their destructors still have names that the compiler and linker understand. Programs just can't name them.

trealira 3y ago
I mean that it would complicate just the parser. For many compilers, as soon as the parser sees a left curly brace, it pushes a symbol table onto a stack, and when it sees the corresponding right curly brace, it pops the symbol table off the stack, and "forgets" any declarations that were made inside that scope. That is so things like this work as expected. { int x = 0; { int x = 1; printf("%d\n", x); // prints 1 } printf("%d\n", x); // prints 0 } But, in C++, using the auto keyword, declarations can escape their scope with auto. I'll change the C++ code that OP wrote. The C++ compiler has to correctly resolve cases like this, which means it can't just forget all the declarations within the scope of the function after the definition is done. #include <iostream> auto createVoldemortType(int value) { struct Voldemort { int value; }; return Voldemort{value}; } struct Voldemort { std::string value; }; int main() { auto voldemort = createVoldemortType(7); std::cout << voldemort.value << std::endl; // output: 7 }
- 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...

trealira 3y ago

I mean that it would complicate just the parser. For many compilers, as soon as the parser sees a left curly brace, it pushes a symbol table onto a stack, and when it sees the corresponding right curly brace, it pops the symbol table off the stack, and "forgets" any declarations that were made inside that scope. That is so things like this work as expected. { int x = 0; { int x = 1; printf("%d\n", x); // prints 1 } printf("%d\n", x); // prints 0 } But, in C++, using the auto keyword, declarations can escape their scope with auto. I'll change the C++ code that OP wrote. The C++ compiler has to correctly resolve cases like this, which means it can't just forget all the declarations within the scope of the function after the definition is done. #include <iostream> auto createVoldemortType(int value) { struct Voldemort { int value; }; return Voldemort{value}; } struct Voldemort { std::string value; }; int main() { auto voldemort = createVoldemortType(7); std::cout << voldemort.value << std::endl; // output: 7 }

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...