Assignment 3 Solution Background

1. Assignment 3 Solution Background

2. Constant Expression : Infix to postfix 2 + 3 * 4 ( 2 + (3 * 4 ) ) 2 3 4 * + • Evaluating postfix expression using stack | 2 | | 2 | 3 | 4 | | 2 | 12 | |14|

3. Evaluating postfix expression using stack | 2 | | 2 | 3 | 4 | | 2 | 12 | |14| • Compiling constant expression for a stack machine Push 2 Push 3 Push 4 Mul Add

4. Generalizing to expressions with variables i + j * k Push i Push j Push k Mul Add • Conversion to abstract syntax tree + i * j k

5. Generalizing to expressions with variables i + j * k Push i Push j Push k Mul Add • Byte code generation for static f(int i,j,k) iload_0 iload_1 iload_2 imul iadd

6. Right associative “+” iload_0 iload_1 iload_2 iadd iadd Left associative “+” iload_0 iload_1 iadd iload_2 iadd Byte code for i + j + k for static f(int i,j,k)

7. Introducing numeric types with real variables a + b Push a Push b Add • Byte code generation for static f(double a,b) dload_0 dload_2 dadd

8. Mixing int and double variables (requiring coercion code) for static f(double a,int i, j) i + j * a iload_2 i2d iload_3 i2d dload_0 dmul dadd

9. Translation algorithm essence trans (e1 * e2) = trans(e1) [type coercion code?] trans(e2) [type coercion code?] trans(*) • Map grammar rules to control structures • E.g., alternatives, while-loop, etc