compiler-constructiongrammar
A grammar is not LL(1) even though it has no left recursion or common prefixes
Supposedly, following grammar is not LL(1). However I cannot find a reason for that. I think it is perfectly LL(1) because it has no left recursion and there is no common prefix issue. Can someone clarify?
Solution
LL(1) means that a parser can be constructed that uses a single token look-ahead, and in stmt, both assignment and subr_call start with ID so a parser cannot distinguish between the two cases just by looking at the first token (ID).
- Compiled vs. Interpreted Languages
- How to add keyword to a C++ compiler
- How does C-- compare to LLVM?
- Why won't javac accept `x = x+++++y`?
- Representing an Abstract Syntax Tree in C
- How do I generate sentences from a formal grammar?
- Why do lexers usually define a var as not being able to start with a number?
- Which bitset implementation should I use for maximum performance?
- Branch prediction in compiler level
- Do any languages / compilers utilize the x86 ENTER instruction with a nonzero nesting level?
- Compiling NASM on Mac OSX
- Calculating administrative normal form
- Parsing extended lambda calculus using recursive descent
- How to handle forward references with type inference
- Do debugging information reveal code in C++ / MSVC?
- From Object Code To executable
- What is the order of the scala compiler phases?
- Where is LLVM's interface for describing the instructions of an ISA?
- Translation limits of an identifier in C
- Register based VM - byte order - push/load
- What was ALGOL compiler originally programmed in?
- Unpack / pack operator
- Issue with Bison Parser for Handling for Loops (Syntax Error)
- vba-style compiler/gui for python
- Writing a compiler in its own language
- How does the code generated by JIT compiler link with the program?
- Convert EBNF to C# for compiler
- What is the exact compiler optimization applied here expect for tail recursion elimination?
- Compiling vs Transpiling
- Why is (void) 0 a no operation in C and C++?