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Joined 2026-09-21 Message
Incoming links: Literal value
Abstract syntax tree 2026-10-13
An abstract syntax tree is a rooted tree recording the meaningful syntactic pieces of a program and how they fit together. A program identifier such as
x, a literal value such as 2, and a node representing addition have different roles. Their arrangement answers structural questions: which program expression supplies an operand, which name is a variable assignment target, and which construct encloses another. This makes an abstract syntax tree useful to a compiler, an interpreter, and the type system that checks a program.The word "abstract" describes what is omitted. Lexical analysis first recognizes lexical tokens, such as
x, =, 2, +, 3, *, 4. A concrete parse tree can retain grammar derivation structure, including punctuation and intermediate context-free grammar categories such as expression, term and factor. An abstract syntax tree usually keeps the operation and its operands without all those grammar scaffolds. Parentheses can determine its shape without surviving as separate nodes. A parser may construct the abstract syntax tree directly; it need not first allocate a concrete parse tree.An abstract syntax tree records programming language syntax, but does not by itself supply all programming language semantics. In
a+b, the names do not reveal their current values, and + might perform addition of integers or concatenate strings. The programming language rules, name bindings, and operand types determine the meaning. In Python, successfully calling ast.parse also does not guarantee that later compilation will accept the result: parsing return 42 succeeds, but compiling that statement outside a function fails. [ref]ast — Abstract syntax trees, Python 3.14 documentation. ast.parse: parsing performs no scoping checks; the return 42 example
An abstract syntax tree is not machine code. Its nodes describe language constructs. A compiler can check and transform this structure before translating it into executable instructions; an interpreter can evaluate an appropriate tree directly. The little evaluator below demonstrates this distinction without pretending to implement an entire programming language.
Abstract syntax tree precedence example 2026-10-13
Consider Python programs
x=2+3*4 and x=(2+3)*4. Operator precedence places multiplication inside the right operand of addition in the first abstract syntax tree. The parentheses instead place addition inside the left operand of multiplication in the second. Their right-hand sides evaluate to and . No rule that simply reads all operators left to right would recover both structures. [ref]https://docs.python.org/3.14/reference/expressions.html#operator-precedence. Python language reference, section 6.17, multiplicative operators bind more tightly than additive operators
Two abstract syntax trees for variable assignment: operator precedence gives
. Details. x=2+3*4 the value , while parentheses give x=(2+3)*4 the value These are schematic views of Python trees:
Module is omitted, and the Add/Mult and Store markers are written inside their owning nodes. In the actual schema, BinOp has left, op and right fields; Assign has a list of targets and a value. Name(id='x', ctx=Store()) marks the name as a variable assignment target. Reading x in a program expression instead uses Load. A variable assignment is a statement, not an equation asserting that two values are equal. [ref]A restricted recursive evaluator can follow those fields. It accepts only integer literal values and binary addition, subtraction and multiplication; the surrounding variable assignment remains a separate action. Using
type(value) is int excludes Boolean constants, even though Python treats bool as a subclass of int.import ast
OPS = {ast.Add: lambda a, b: a + b,
ast.Sub: lambda a, b: a - b,
ast.Mult: lambda a, b: a * b}
def evaluate(node):
if isinstance(node, ast.Constant) and type(node.value) is int:
return node.value
if isinstance(node, ast.BinOp) and type(node.op) in OPS:
return OPS[type(node.op)](evaluate(node.left), evaluate(node.right))
raise ValueError("outside the integer-expression language")
for source in ("x=2+3*4", "x=(2+3)*4"):
tree = ast.parse(source, optimize=0)
statement = tree.body[0]
print(statement.targets[0].id, evaluate(statement.value))
# x 14
# x 20The children in a rooted tree have ordered roles:
left and right cannot be interchanged. For subtraction, 8-3 gives while 3-8 gives . Operator associativity also matters: 8-3-1 parses as (8-3)-1, giving , whereas 8-(3-1) gives . Here the syntax groups operators to the left; this does not assert the mathematical associativity of subtraction. Operator precedence determines grouping, rather than a general permission to reorder execution. For example, Python evaluates the calls in f()+g()*h() in the order f, g, h, although the multiplication result must be ready before the outer addition. [ref]https://docs.python.org/3.14/reference/expressions.html#evaluation-order. Python language reference, sections 6.16 and 6.17: left-to-right evaluation and left grouping of additive operators
Compilation arithmetic example 2026-10-14
Consider the program expression
x*(2+3)-4 in a deliberately small programming language. Its only program identifier is x, supplied as an integer input. Its literal values are nonnegative decimal numerals, interpreted as integers. Whitespace is ignored between lexical tokens. Its context-free grammar is:expression ::= term (("+" | "-") term)*
term ::= factor ("*" factor)*
factor ::= numeral | "x" | "(" expression ")"
numeral ::= digit digit*
digit ::= "0" | "1" | "2" | "3" | "4"
| "5" | "6" | "7" | "8" | "9"Here
* outside quotation marks means zero or more repetitions. Repeated operators are grouped to the left, specifying operator associativity; term gives multiplication higher operator precedence than addition and subtraction. Parentheses override that grouping. There is no unary minus, variable assignment, function call, or side effect in this language. Evaluation obtains each left operand before its right operand and uses exact integer arithmetic, with no fixed-width overflow.Lexical analysis recognizes
x, *, (, 2, +, 3, ), -, 4. Parsing constructs the concrete abstract syntax tree Sub(Mul(Name(x), Add(Int(2), Int(3))), Int(4)). Parentheses have done their work by making Add a child of Mul; they need not survive as nodes. Constant folding changes this to Sub(Mul(Name(x), Int(5)), Int(4)). It cannot replace Name(x) by a literal value because the input is supplied at execution. The resulting function of the input is .A compiler for our stack machine emits
LOAD x; PUSH 5; MUL; PUSH 4; SUB. LOAD x pushes the current input; PUSH n pushes the integer literal value n. Each binary instruction removes the top item as its right operand, then the next item as its left operand, and pushes the result. The instruction sequence is an intermediate representation that an interpreter can execute; no machine code has been emitted. A compact binary encoding could make it bytecode, but these printed instruction names are only a readable notation.Why does this compilation preserve the result? A structural induction supplies a short proof outline. Translating a literal value or
x appends its value to any initial stack. For a binary abstract syntax tree node, first translating the left child and then the right child appends the two operand values; the matching instruction replaces them by exactly the original operation's result. Previously present stack items remain intact. Constant folding also preserves the result because in the specified integer arithmetic. Together these facts explain agreement for every integer input, rather than merely agreement for one trial. Constant folding 2026-10-13
Constant folding replaces a program expression whose value is known during translation by a literal value. For
x=2+3*4, first replace the multiplication subtree by Constant(value=12), then replace the addition subtree by Constant(value=14). The variable assignment and its Store target survive: the result still assigns to x. This illustrates why a bottom-up abstract syntax tree traversal is convenient: replacing one child can make its parent eligible for the next replacement.
Here is a deliberately small Python transformer. Calling
generic_visit first transforms the children; returning a node replaces the visited node. The location helpers retain or supply the metadata expected for compilation. [ref]ast — Abstract syntax trees, Python 3.14 documentation. NodeTransformer, generic_visit, copy_location and fix_missing_locations
class FoldIntegers(ast.NodeTransformer):
def visit_BinOp(self, node):
self.generic_visit(node)
if (isinstance(node.left, ast.Constant)
and isinstance(node.right, ast.Constant)
and type(node.left.value) is int
and type(node.right.value) is int
and type(node.op) in OPS):
value = OPS[type(node.op)](node.left.value, node.right.value)
return ast.copy_location(ast.Constant(value=value), node)
return node
tree = ast.parse("x=2+3*4", optimize=0)
folded = ast.fix_missing_locations(FoldIntegers().visit(tree))
namespace = {}
exec(compile(folded, "<folded example>", "exec"), namespace)
assert namespace["x"] == 14
assert ast.unparse(folded) == "x = 14"This transformer folds literal integer operands only. It leaves
a+3*4 as a+12, without inventing a value for the program identifier a. It leaves calls intact: replacing 0*f() by 0 could discard a side effect or a failure. Nor does it reassociate floating-point arithmetic, where changing parentheses can change rounding. For the supported integer arithmetic, a short structural induction explains correctness: a retained literal value has the same value; at each folded binary node, the already-correct child values are combined by exactly the original operation in the original left/right order. This claim concerns the supported fragment's values, not unrestricted equivalence of arbitrary Python programs. Lexical token 2026-10-14