4 Sep - Compiler - Intermediate Code and Code Optimization
Duration: 1 hr 50 min
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This lecture provides a comprehensive overview of Intermediate Code Generation in compiler design. The instructor begins by defining intermediate code as an internal representation generated after parsing and before final machine code production, acting as a bridge between high-level source code and low-level assembly. Key reasons for its use include portability, allowing the same intermediate code to be translated to different target machines (e.g., Java bytecodes), and optimization, which is easier to perform on this abstract representation. The lecture details two main categories of intermediate code: Tree Representation (Syntax Trees and Directed Acyclic Graphs) and Linear Representation (Three-Address Code and Postfix). Specific focus is given to constructing DAGs to identify common sub-expressions and converting complex expressions into Three-Address Code using temporary variables. The concept of Single Static Assignment (SSA) form is introduced, where each variable is assigned exactly once. The session concludes with practical examples of converting control structures like for-loops and array indexing into three-address code sequences.
Chapters
0:00 – 2:00 00:00-02:00
The video begins with a title card displaying the name 'Sanchit Jain' against a black background. This serves as an introduction to the instructor or the session before the main content begins. There is no visible lecture material yet, just the branding for the educational content.
2:00 – 5:00 02:00-05:00
The lecture starts with the handwritten heading 'Intermediate Code'. The definition is provided: Intermediate Code is an internal representation of a program generated by a compiler after the source code is parsed and before the final machine code is produced. It is described as a 'Bridge between high level Source Code & low level assembly/Machine Code'. This establishes the fundamental role of intermediate code in the compilation process.
5:00 – 10:00 05:00-10:00
The instructor explains why intermediate code is used, listing two main reasons. First is Portability: the compiler translates source to intermediate code, and then intermediate code to the target machine. An example given is Java bytecodes. Second is Optimization: optimizations are easier to perform on intermediate code than on raw source or machine code. This highlights the strategic advantage of using an intermediate layer.
10:00 – 15:00 10:00-15:00
The lecture discusses the cons of using Intermediate Representation (IR). These include needing an extra compilation step, consuming memory because IR must be stored and processed, not being directly executable, and the complexity of design. The conclusion states that intermediate code is used because it provides portability and optimization opportunities, with the main strength being simplifying compiler design by acting as a bridge, despite the trade-off of extra steps and complexity.
15:00 – 20:00 15:00-20:00
The instructor revisits IR, noting it is closer to the target machine than the source language, making it easier to generate target code. Unlike machine language, intermediate code is more or less machine-independent, which makes it easier to retarget the compiler. It allows a lot of optimizations to be performed in a machine-independent way. Generally, intermediate code generation can be implemented via syntax-directed translation. In the analysis-synthesis model, the front-end analyzes source code and creates an intermediate representation, from which the backend generates target code.
20:00 – 25:00 20:00-25:00
The choice of intermediate representation design varies from compiler to compiler. The lecture categorizes intermediate representations into two types: Linear representation and Tree representation. Linear representation includes Three-address code and Postfix. Tree representation includes Syntax tree and Directed Acyclic Graph (DAG). This classification sets the stage for detailed explanations of each type.
25:00 – 30:00 25:00-30:00
The focus shifts to Tree Representation. A Syntax tree, also known as an Abstract Syntax Tree (AST), is a tree representation of the syntactic structure of source code. An example expression `a * a * (b-c) + (b-c) * d` is shown. A syntax tree is drawn for this expression, illustrating the hierarchical structure of operations. The instructor points out a 'Common Sub-expression' in the tree, which is the term `(b-c)` appearing twice.
30:00 – 35:00 30:00-35:00
The concept of a Directed Acyclic Graph (DAG) is introduced as a special kind of abstract syntax tree. A DAG for an expression identifies the common sub-expressions of the expression. Specifically, it handles 'Sub-expressions that occur more than once' by merging them into a single node. The instructor notes that a DAG is an intermediate code as well as a code optimization technique.
35:00 – 40:00 35:00-40:00
The method for constructing a DAG is explained. Interior nodes always represent the operators, while exterior nodes (leaf nodes) always represent the names, identifiers, or constants. The rule is to always check if there exists any node with the same operator and operands. If a node already exists, it is reused instead of creating a new one. This process effectively eliminates redundant computations in the graph structure.
40:00 – 45:00 40:00-45:00
An example of DAG construction is shown for the expression `a * a * (b-c) + (b-c) * d`. The syntax tree is drawn first, showing the duplication of the `(b-c)` subtree. Then, the DAG is drawn below it. In the DAG, the node for `(b-c)` is shared by both multiplication operations, demonstrating how common sub-expressions are identified and merged to optimize the representation.
45:00 – 50:00 45:00-50:00
More examples of DAG construction are provided. The expression `(a+b) * (a+b+c)` is analyzed. The syntax tree shows separate subtrees for `(a+b)`, but the DAG merges them. Another example `a+a+a+a` is shown, where the DAG structure collapses the repeated additions into a single chain of operations, highlighting the efficiency of DAGs in representing repeated computations.
50:00 – 55:00 50:00-55:00
The lecture moves to Linear Representation, specifically Three-Address Code. In three-address code, there is at most one operator on the right side of an instruction, meaning no built-up arithmetic expressions are permitted. An example `x + y * z` is broken down into `t1 = y * z` and `t2 = x + t1`. This linearizes the tree structure into a sequence of simple instructions.
55:00 – 60:00 55:00-60:00
Details about Three-Address Code are elaborated. The temporary names `t1` and `t2` are compiler-generated. Three-address code is a linearized representation of a syntax tree or a DAG in which explicit names correspond to the interior nodes of the graph. This ensures that the complex structure of the expression is flattened into a sequence of assignments that a machine can execute.
60:00 – 65:00 60:00-65:00
A detailed example of converting `a * a * (b-c) + (b-c) * d` to Three-Address Code is shown. The sequence is: `t1 = b-c`, `t2 = t1 * d`, `t3 = a * a`, `t4 = t3 * t1`, `t5 = t2 + t4`. This demonstrates how the DAG or syntax tree is traversed to generate a linear sequence of instructions, reusing the temporary variable `t1` for the common sub-expression `(b-c)`.
65:00 – 70:00 65:00-70:00
A note is added about the structure of Three-Address Code. Each instruction typically has at most three addresses: two source operands (the values to be used) and one destination operand (where the result is stored). The assignment `t = a + b` is treated as a single instruction performing one operation (addition) and storing the result. The equals sign is just notation, not an extra operation.
70:00 – 75:00 70:00-75:00
The concept of Single Static Assignment (SSA) form is introduced as a property of three-address code. In SSA, each variable is assigned exactly once. No value ever gets a new value, meaning no reassignment occurs. Every time a new value is computed, it gets a new version of the variable. This ensures that every variable name corresponds to exactly one definition in the program.
75:00 – 80:00 75:00-80:00
An example of SSA form is shown. The code `p = a + b`, `q = p - c`, `p = q * d`, `p = e - p`, `q = p + q` is converted. The variables are renamed to `p1, q1, p2, q2, p3, q3` etc., and the original variables `a, b` are renamed to `a0, b0`. This illustrates how SSA prevents reassignment by creating new versions for each computation.
80:00 – 85:00 80:00-85:00
A multiple-choice question is presented regarding SSA form. The question asks to identify the correct SSA form for a given intermediate program. The options show different variable renaming schemes. The instructor analyzes the options, noting that in one option, `p1` is assigned more than once, which is false for SSA. This reinforces the rule that each variable must be assigned exactly once.
85:00 – 90:00 85:00-90:00
Another question is discussed involving a code segment: `x = u - t`, `y = x * v`, `x = y + w`, `y = t - z`, `y = x * y`. The question asks for the minimum number of total variables required to convert this to static single assignment form. The instructor begins to write down the SSA versions, such as `x1 = u0 - t0`, `y1 = x1 * v0`, etc., to count the total variables needed.
90:00 – 95:00 90:00-95:00
A question asks for the least number of temporary variables required to create a three-address code in static single assignment form for the expression `q + r/3 + s - t * 5 + u * v/w`. The instructor breaks down the expression into steps: `t1 = r/3`, `t2 = t * 5`, `t3 = u * v`, `t4 = t3 / w`, `t5 = q + t1`, `t6 = t5 + s`, `t7 = t6 - t2`, `t8 = t7 + t4`. This demonstrates the process of counting temporaries for a complex expression.
95:00 – 100:00 95:00-100:00
The lecture lists common three-address code instruction forms. 1. Assignment instructions of the form `x = y op z`. 2. Assignments of the form `x = op y` (unary operation). 3. Copy instructions of the form `x = y`. 4. Unconditional jump `goto L`. 5. Conditional jumps of the form `if x goto L` and `if x goto L else`. These forms cover the basic building blocks of intermediate code.
100:00 – 105:00 100:00-105:00
An example of converting a `for` loop to three-address code is shown. The loop `for (i=0; i<=20; i++) x = y + z;` is converted into a sequence of instructions. It includes initialization `i=0`, a conditional jump `if (i<=20) goto 4`, an unconditional jump `goto 9`, the body `t1 = y + z`, `x = t1`, the increment `t2 = i + 1`, `i = t2`, and the loop back `goto 2`. This illustrates how control flow is managed in intermediate code.
105:00 – 109:47 105:00-109:47
The final topic covers array indexing in three-address code. For a C program accessing `X[i][j][k]`, the intermediate code is generated to calculate the memory address. The steps are: `t0 = i * 1024`, `t1 = j * 32`, `t2 = k * 4`, `t3 = t1 + t0`, `t4 = t3 + t2`, `t5 = X[t4]`. This shows how multi-dimensional array access is flattened into a series of multiplications and additions to compute the final offset.
The lecture systematically builds an understanding of Intermediate Code Generation, starting from its definition as a bridge between source and machine code. It emphasizes the benefits of portability and optimization while acknowledging the overhead of extra compilation steps. The core of the lesson distinguishes between Tree Representation (Syntax Trees and DAGs) and Linear Representation (Three-Address Code and Postfix). Detailed examples illustrate how DAGs optimize common sub-expressions and how complex expressions are linearized into Three-Address Code using temporary variables. The concept of Single Static Assignment (SSA) is introduced to manage variable reassignments, ensuring each variable is defined only once. Finally, practical applications like converting control structures (for-loops) and multi-dimensional array accesses into three-address code sequences are demonstrated, providing a complete picture of how high-level constructs are translated into executable intermediate instructions.