CSCI 310 Day 29

1. Administrivia
   1. Celebration of Student Research -- support your peers!
   2. office hours pushed back today: 4:30 - 6 pm.

2. Project 5 questions?

3. Intermediate representation trees, continued
   • Expressions
     • CONST(value)
     • NAME(n)
     • TEMP(t)
     • BINOP(op, left, right) (op one of PLUS, MINUS, MUL, DIV, AND, OR, XOR, LSHIFT, RSHIFT)
     • MEM(exp)
     • CALL(f, l) f is a label for a method, l is a list of expressions
     • ESEQ(s, exp)
   • Statements
     • MOVE(TEMP t, Expression e)
     • MOVE(MEM(e1), e2)
     • EXP(e)
     • JUMP(Expression e, labels) e is a name or address, labels is a list of valid labels/addresses that the expression can evaluate to. (Abbreviate as JUMP(e) if e is an obvious label or memory expression.)
     • CJUMP(op, e1, e2, t, f) t, f labels to jump to on true or false. op is one of EQ, NE, LT, GT, LE, GE.
     • SEQ(s1, s2) s1 evaluated, then s2
     • LABEL(n) (label this tree as n)

4. What we did last time: trees for
   • variable x
   • x + 5
   • y = x + 5
   • a[i] = x+ 5
   • a = new int [n]

5. Today
   • Continue a = new int[n]. Do we initialize to all 0's? how?
   • if (x < 5) a[i] = x + 5; else y = x + 5;
   • while (x < 5) { a[i] = x+5; x = x+1; i = i+1;}
   • x = new Foo();