Code Generation Issues and Target - Compiler Construction - Lecture Slides, Slides of Compiler Construction

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Compiler

Construction

Lecture 42

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Code Generation

 The code generation problem

is the task of mapping intermediate code to machine code.

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Issues:

 Input language: intermediate

code (optimized or not)

 Target: machine code,

assembly language

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Issues:

 Memory management: mapping names to data objects in the run-time system.

 Instruction Selection

 Instruction Scheduling

 Register allocation

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Issues:

 Memory management: mapping names to data objects in the run-time system.

 Instruction Selection

 Instruction Scheduling

 Register allocation

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Issues:

 Memory management: mapping names to data objects in the run-time system.

 Instruction Selection

 Instruction Scheduling

 Register allocation

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Target: Assembly Language

Operations:

• MOV source, destination
• ADD source, destination
• SUB source, destination (dest = dest – source)
• GOTO address
• CJ conditional jump

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Addressing Modes

MODE FORM ADDRESS ADDED COST absolute M M 1 register R R 0 indexed c(R) c + contents(R) 1 indirect register *R contents(R) 0 indirect indexed *c(R) contents(c+contents(R)) 1 literal #c c 1 stack SP SP 0 indexed stack c(SP) c + contents(SP) 1

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Simple Code Generation

 Define a target code sequence for each intermediate code statement type.

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Simple Code Generation

Intermediate becomes… a = b MOV b,a

a = b[c] MOV addr(b),R ADD c, R MOV *R0,a

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 Consider the C statement: a[i] = b[c[j]]; t1 := c[j] MOV addr(c), R ADD j, R MOV *R0, t t2 := b[t1] MOV addr(b), R ADD t1, R MOV *R0, t a[i] := t2 MOV address(a), R ADD i, R MOV t2, *R  The cost of this code is 18 and we are forced to allocate space for two temporaries.

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Problems

 Local decisions do not produce good code.  Do not take temporary variables into account

• t1 := c[j] MOV addr(c), R a[i] = b[c[j]];
  • ADD j, R
  • MOV *R0, t
• t2 := b[t1] MOV addr(b), R
  • ADD t1, R
  • MOV *R0, t
• a[i] := t2 MOV address(a), R
  • ADD i, R
  • MOV t2, *R
• The cost of this code is - MOV addr(c), R - ADD j, R - MOV addr(b), R - ADD *R0, R - MOV addr(a), R - ADD i, R - MOV *R1, *R

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Can optimize further …

MOV addr(c), R ADD j, R MOV addr(a), R ADD i, R MOV *addr(b)(R0), *R

The cost of this code is 10.