1. Code Generation
My suggestion: Pick three registers to be used throughout
Assuming stmt of form dest = s1 op s2
Generate code by:
Load s1 into r5
Load s2 into r6
r7 = r5 op r6
Store r7 into dest

2. Array Declaration Algorithm
Dimension Node {
int min;
int max;
int size; }

3. Declaration Algorithm (2)
Doubly linked list of dimension nodes
Pass 1 – while parsing
Build linked list from left to right
Insert min, max
Size = size of an element (e.g. 4 for int)
Append node to end of list
min = max = size = 1

4. Declaration Algorithm (3) Pass 2
Traverse list from tail to head
For each node, n, going “right” to “left”
Factor = n.max – n.min + 1
For each node, m, right to left starting with n
m.size = m.size * factor
For each node, n, going right to left
max = N->left->max; min = N->left->min
Save size of first node as size of entire array
Delete first element of list
Set tail->size = size of an element (e.g. 4 for int)

5. Array Declaration (Row Major)
int weight[ ][1..12][1..31]; list of “dimension” nodes int min, max, size size of element of this dimension
1488
124 4

6. Array Offset (Row Major)
Traverse list summing (max-min) * size int weight[ ][1..12][1..31]; x = weight [2002][5][31] ( ) * (5-1) * (31-1) * 4
1488
124 4

7. Array Offset (Row Major)
Traverse list summing (max-min) * size int weight[ ][1..12][1..31]; x = weight [i][j][k] (i ) * (j-1) * (k-1) * 4
1488
124 4

8. Your Turn Assume “Show” A’s dimension list
int A[10][20][30];
Row major order
“Show” A’s dimension list
Show hypothetical 3-addr code for
X = A[2][3][4] ;
A[3][4][5] = 9

9. My “Assembly” code X = A[2][3][4]; r1 = 4800; # 2 * 2400 r2 = 360;
r3 = r1 + r2;
r4 = 16;
r5 = r3 + r4;
r6 = r5 + 64; #offset of A
iptr1 = &r6;
r7 = *iptr1;

10. Your Turn 2 Assume “Show” A’s dimension list
int A[10][20][30];
Column major order
“Show” A’s dimension list
Show hypothetical 3-addr code for
X = A[2][3][4] ;
A[3][4][5] = 9

11. Control Constructs Can be cumbersome, but not difficult
“Write” control construct in 3-addr pseudo code using labels and gotos.
Map that “control construct” to grammar rule action(s).

12. Semantic Hooks
Selection_statement : IF ‘(‘ comma_expr ‘)’ stmt | IF ‘(‘ comma_expr ‘)’ stmt ELSE stmt ; 1 shift/reduce error

13. Add actions (1)
Selection_statement : IF ‘(‘ comma_expr ‘)’ {printf(“start IF \n”);} stmt {printf(“IF Body \n”);}

14. Add actions (2)
| IF ‘(‘ comma_expr ‘)’ {printf(“start IF \n”);} stmt {printf(“Then Body \n”);} ELSE stmt {printf(“ELSE Body \n”);} ; 31 reduce/reduce errors !

15. Solution (1)
Selection_statement : if_start | if_start ELSE stmt {printf(“ELSE body”);} }

16. Solution (2)
if_start : IF ‘(‘ comma_expr ‘)’ {printf(“start IF \n”);} stmt {printf(“Then Body \n”);} ; 1 shift-reduce