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使用Python编写一个Lisp语言的解释器

作者:skywalk8163

这篇文章主要为大家详细介绍了如何使用Python编写一个简单的Lisp语言的解释器,文中的示例代码讲解详细,感兴趣的小伙伴可以跟随小编一起学习一下

一般的源代码程序经过编译器解析生成解析树。Lisp的奇特之处就在于,你可以完全卸除程序,控制这种解析树,进行任意的存取操作,也就是可以用程序生成程序。

Python号称最接近Lisp的语言,但它终究不是。但是因为几乎所有语言都是图灵完备的,所以即使Python无法实现Lisp的某个功能,也可以通过在Python中写一个Lisp解释器来实现那个功能。很奇妙是不是? 

我们来写一个简单的基于Scheme语法的Lisp解析器吧:

先导入库

################ lis.py: Scheme Interpreter in Python 3.10
## (c) Peter Norvig, 2010-18; See http://norvig.com/lispy.html
## Type hints and minor additions by Luciano Ramalho
 
import math
import operator as op
from collections import ChainMap
from itertools import chain
from typing import Any, NoReturn
from typing import Union, List, MutableMapping, Optional, Iterator
 
Symbol = str
Atom = Union[float, int, Symbol]
Expression = Union[Atom, List]
 
Environment = MutableMapping[Symbol, object]
 
print(Atom, Expression)
print(Environment)

创建Parse解析

def parse(program: str) -> Expression:
    "Read a Scheme expression from a string."
    return read_from_tokens(tokenize(program))
 
def tokenize(s: str) -> List[str]:
    "Convert a string into a list of tokens."
    return s.replace('(', ' ( ').replace(')', ' ) ').split()
 
def read_from_tokens(tokens: List[str]) -> Expression:
    "Read an expression from a sequence of tokens."
    if len(tokens) == 0:
        raise SyntaxError('unexpected EOF while reading')
    token = tokens.pop(0)
    if '(' == token:
        exp = []
        while tokens[0] != ')':
            exp.append(read_from_tokens(tokens))
        tokens.pop(0)  # discard ')'
        return exp
    elif ')' == token:
        raise SyntaxError('unexpected )')
    else:
        return parse_atom(token)
 
def parse_atom(token: str) -> Atom:
    "Numbers become numbers; every other token is a symbol."
    try:
        return int(token)
    except ValueError:
        try:
            return float(token)
        except ValueError:
            return Symbol(token)

创建环境

def standard_env() -> Environment:
    "An environment with some Scheme standard procedures."
    env: Environment = {}
    env.update(vars(math))   # sin, cos, sqrt, pi, ...
    env.update(
        {
            '+': op.add,
            '-': op.sub,
            '*': op.mul,
            '/': op.truediv, # 小数除
            'quotient': op.floordiv, # 商 地板除法 整数除
            '>': op.gt,
            '<': op.lt,
            '>=': op.ge,
            '<=': op.le,
            '=': op.eq,
            'abs': abs,
            'append': lambda *args: list(chain(*args)),          
            'apply': lambda proc, args: proc(*args),
            'begin': lambda *x: x[-1],
            '起': lambda *x: x[-1],
            'car': lambda x: x[0],
            'cdr': lambda x: x[1:],
            'cons': lambda x, y: [x] + y,
            'eq?': op.is_,
            'equal?': op.eq,
            'filter': lambda *args: list(filter(*args)),
            'length': len,
            'list': lambda *x: list(x),
            'list?': lambda x: isinstance(x, list),
            'map': lambda *args: list(map(*args)),
            'max': max,
            'min': min,
            'not': op.not_,
            'null?': lambda x: x == [],
            'number?': lambda x: isinstance(x, (int, float)),
            'procedure?': callable,
            'round': round,
            'symbol?': lambda x: isinstance(x, Symbol),
            'display': lambda x: print(lispstr(x), end=''),
            '显': lambda x: print(lispstr(x), end=''),
            'newline': lambda: print(),
        }
    )
    return env

执行函数

def evaluate(x: Expression, env: Environment) -> Any:
    "Evaluate an expression in an environment."
    if isinstance(x, str):                       # variable reference
        return env[x]
    elif not isinstance(x, list):                # constant literal
        return x
    elif x[0] == 'define':                       # (define var exp)
        _, var, exp = x
        env[var] = evaluate(exp, env)
    elif x[0] == 'lambda':                       # (lambda (var...) body)
        _, parms, body = x
        return Procedure(parms, body, env)
    elif x[0] == 'quote':                        # (quote exp)
        _, exp = x
        return exp
    elif x[0] == 'if':                           # (if test consequence alternative)
        _, test, consequence, alternative = x
        if evaluate(test, env):
            return evaluate(consequence, env)
        else:
            return evaluate(alternative, env)
    elif x[0] == '设':                       # (define var exp)
        _, var, exp = x
        env[var] = evaluate(exp, env)
    elif x[0] == '函':                       # (lambda (var...) body)
        _, parms, body = x
        return Procedure(parms, body, env)
    elif x[0] == '引':                        # (quote exp)
        _, exp = x
        return exp
    elif x[0] == '若':                           # (if test consequence alternative)
        _, test, consequence, alternative = x
        if evaluate(test, env):
            return evaluate(consequence, env)
        else:
            return evaluate(alternative, env)
    else:                                        # (proc arg...)
        proc_exp, *args = x
        proc = evaluate(proc_exp, env)
        arg_values = [evaluate(exp, env) for exp in args]
        return proc(*arg_values)

交互执行函数

def run_lines(source: str, env: Optional[Environment] = None) -> Iterator[Any]:
    global_env: Environment = ChainMap({}, standard_env())
    if env is not None:
        global_env.update(env)
    tokens = tokenize(source)
    while tokens:
        exp = read_from_tokens(tokens)
        yield evaluate(exp, global_env)
 
 
def run(source: str, env: Optional[Environment] = None) -> Any:
    # 实际上,这个函数只是简单地迭代了run_lines的所有结果,并没有对其进行任何操作。
    # 最后,返回run_lines的最后一个结果。
    for result in run_lines(source, env):
        pass
    return result

运行测试

percent = """
(define a 126)
(define b (* 6 50))
(* (/ a b) 100)
"""
run(percent)

输出:42

当然我们也可以用中文关键字:

percent = """
(设 a 126)
(设 b (* 6 50))
(* (/ a b) 100)
"""
run(percent)

这样看起来是不是更亲切一些了呢?

以上代码节选自:https://github.com/fluentpython/lispy

附:

scheme学习资料:The Scheme Programming Language, 4th Edition

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