python中的小语言

Question

我正在编写甚至可能在python中称为语言的东西.我目前有几家运营商:+,-,*,^,fac,@,!!.fac计算阶乘,@返回变量的值,!!设置变量.代码如下.我将如何编写一种用这种简单语言定义函数的方法？

编辑:我更新了代码!

import sys, shlex, readline, os, string
List, assign, call, add, sub, div, Pow, mul, mod, fac, duf, read,\
kill, clr, STO, RET, fib, curs = {}, "set", "get", "+", "-", "/", "^", "*",\
"%", "fact", "func", "read", "kill", "clear", ">", "@", "fib", "vars"
def fact(num):
    if num == 1: return 1
    else: return num*fact(num-1)
def Simp(op, num2, num1):
    global List
    try: num1, num2 = float(num1), float(num2)
    except:
        try: num1 = float(num1)
        except:
            try: num2 = float(num2)
            except: pass
    if op == mul: return num1*num2
    elif op == div: return num1/num2
    elif op == sub: return num1-num2
    elif op == add: return num1+num2
    elif op == Pow: return num1**num2
    elif op == assign: List[num1] = num2; return "ok"
    elif op == call: return List[num1]
    elif op == fac: return fact(num1)
    elif op == duf: return "%s %s %s"%(duf, num1, num2)
    elif op == mod: return num1%num2
    elif op == kill: del List[num1]; return "ok"
    elif op == clr: os.system("clear")
    elif op == STO: List[num2] = num1; return "ok"
    elif op == RET: return List[num1]
    elif op == curs: return List
    elif op == read: List[num1] = Eval(raw_input("%s "%num1)); return "ok"
def Eval(expr):
    ops = "%s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s"%(mul, add, sub, div, Pow, assign, call, fac, duf, mod, read, kill, clr, STO, RET, curs)
    stack, expr, ops = [], shlex.split(string.lower(expr)), ops.split()
    for i in expr:
        if i[0] != ';':
            if i not in ops: stack.append(i)
            elif i in ops: stack.append(Simp(i, stack.pop(), stack.pop()))
        else: stack.append("ok")
    return stack[0]
def shell():
    try:
        x = ""
        while x != "quit":
            x = raw_input("star>   ")
            try: l = Eval(x)
            except KeyError: l = "does not exist"
            except: l = "parse error!"
            if l != None: print "   =>",l,"\n"
    except (EOFError, KeyboardInterrupt): print
if len(sys.argv) > 1:
    x = open(sys.argv[1], 'r'); l = x.readlines(); x.close()
    for i in l:
        if i[0] != ";":
            i = ' '.join(i.split())
            x = Eval(i)
            if x != None: print i,"\n   =>",x,"\n"
        else: pass
    shell()
else: shell()

Answer 1

您的程序非常困惑,需要先修复它才能修改它以支持定义函数.我将分几步完成这项工作,当我完成它们时,我会将它们添加到答案中.这个答案将会很长.

此外,您显然还没有决定您的语言定义应该是什么.您决定使您的语言定义遵循您的实现技术,这有点破碎,并导致很多痛苦.

首先,你的Simp功能定义真的被打破了.它要求所有内容都从堆栈中取出两个值,然后重新放入一个值.这已破了.阶乘函数不能以这种方式工作,Fibonacci函数也不起作用,所以你被迫拥有一个从未使用过的"虚拟"参数.此外,分配给全局列表或字典的元素之类的东西没有理由将值推送到堆栈,因此您只需要按"确定"即可.这是破碎的,需要修复.

这是修复此问题的版本.请注意,我已重命名Simp为builtin_op更准确地反映其目的:

import sys, shlex, readline, os, string
List, assign, call, add, sub, div, Pow, mul, mod, fac, duf, read,\
kill, clr, STO, RET, fib, curs = {}, "set", "get", "+", "-", "/", "^", "*",\
"%", "fact", "func", "read", "kill", "clear", ">", "@", "fib", "vars"
def fact(num):
    if num == 1: return 1
    else: return num*fact(num-1)
def builtin_op(op, stack):
    global List
    if op == mul: stack.append(float(stack.pop())*float(stack.pop()))
    elif op == div: stack.append(float(stack.pop())/float(stack.pop()))
    elif op == sub: stack.append(float(stack.pop())-float(stack.pop()))
    elif op == add: stack.append(float(stack.pop())+float(stack.pop()))
    elif op == Pow: stack.append(float(stack.pop())**float(stack.pop()))
    elif op == assign: val = List[stack.pop()] = stack.pop(); stack.append(val)
    elif op == call: stack.append(List[stack.pop()])
    elif op == fac: stack.append(fact(stack.pop()))
    elif op == duf: stack.append("%s %s %s" % (duf, stack.pop(), stack.pop()))
    elif op == mod: stack.append(float(stack.pop())%float(stack.pop()))
    elif op == kill: del List[stack.pop()]
    elif op == clr: os.system("clear")
    elif op == STO: val = List[stack.pop()] = stack.pop(); stack.append(val)
    elif op == RET: stack.append(List[stack.pop()])
    elif op == curs: stack.append(List)
    elif op == read: prompt = stack.pop(); List[prompt] = Eval(raw_input("%s "%prompt)); stack.append(List[prompt])
def Eval(expr):
    ops = "%s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s"%(mul, add, sub, div, Pow, assign, call, fac, duf, mod, read, kill, clr, STO, RET, curs)
    stack, expr, ops = [], shlex.split(string.lower(expr)), ops.split()
    for i in expr:
        if i[0] != ';':
            if i not in ops: stack.append(i)
            elif i in ops: builtin_op(i, stack)
        else: stack.append("ok")
    return stack[0]
def shell():
    try:
        x = ""
        while x != "quit":
            x = raw_input("star>   ")
            try: l = Eval(x)
            except KeyError: l = "does not exist"
            except: l = "parse error!"
            if l != None: print "   =>",l,"\n"
    except (EOFError, KeyboardInterrupt): print
if len(sys.argv) > 1:
    x = open(sys.argv[1], 'r'); l = x.readlines(); x.close()
    for i in l:
        if i[0] != ";":
            i = ' '.join(i.split())
            x = Eval(i)
            if x != None: print i,"\n   =>",x,"\n"
        else: pass
    shell()
else: shell()

这里仍有许多问题没有修复,我将不会修复任何未来的版本.例如,堆栈上的值可能无法解释为浮点数.这将导致异常,并且在从堆栈读取另一个值之前可能抛出此异常.这意味着如果堆栈上存在错误的"类型",则在"解析错误"之后堆栈可能处于模糊状态.通常,您希望在语言中避免这种情况.

定义函数是一个有趣的问题.用您的语言,即时评估.您没有延迟评估的机制,直到稍后.但是,您正在使用该shlex模块进行解析.它有一种方式可以说一组完整的字符(包括空格等)是一个实体的一部分.这为我们提供了一种快速简便的实现功能的方法.你可以这样做:

star>   "3 +" add3 func

创建你的功能,并:

star>   2 add3 get

打电话给它.我用过,get因为这是你call在程序中分配的内容.

唯一的问题是该函数需要访问堆栈的当前状态才能工作.您可以轻松地将函数的字符串反馈到其中Eval,但Eval每次调用时都会创建一个全新的堆栈.为了实现功能,需要修复它.所以我stack在Eval函数中添加了一个默认参数.如果此参数保留为其默认值,Eval则仍将创建一个新堆栈,就像之前一样.但是如果传入现有堆栈,Eval则会使用它.

这是修改后的代码:

import sys, shlex, readline, os, string
List, assign, call, add, sub, div, Pow, mul, mod, fac, duf, read,\
kill, clr, STO, RET, fib, curs = {}, "set", "get", "+", "-", "/", "^", "*",\
"%", "fact", "func", "read", "kill", "clear", ">", "@", "fib", "vars"
funcdict = {}
def fact(num):
    if num == 1: return 1
    else: return num*fact(num-1)
def builtin_op(op, stack):
    global List
    global funcdict
    if op == mul: stack.append(float(stack.pop())*float(stack.pop()))
    elif op == div: stack.append(float(stack.pop())/float(stack.pop()))
    elif op == sub: stack.append(float(stack.pop())-float(stack.pop()))
    elif op == add: stack.append(float(stack.pop())+float(stack.pop()))
    elif op == Pow: stack.append(float(stack.pop())**float(stack.pop()))
    elif op == assign: val = List[stack.pop()] = stack.pop(); stack.append(val)
    elif op == call: Eval(funcdict[stack.pop()], stack)
    elif op == fac: stack.append(fact(stack.pop()))
    elif op == duf: name = stack.pop(); funcdict[name] = stack.pop(); stack.append(name)
    elif op == mod: stack.append(float(stack.pop())%float(stack.pop()))
    elif op == kill: del List[stack.pop()]
    elif op == clr: os.system("clear")
    elif op == STO: val = List[stack.pop()] = stack.pop(); stack.append(val)
    elif op == RET: stack.append(List[stack.pop()])
    elif op == curs: stack.append(List)
    elif op == read: prompt = stack.pop(); List[prompt] = Eval(raw_input("%s "%prompt)); stack.append(List[prompt])
def Eval(expr, stack=None):
    ops = "%s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s"%(mul, add, sub, div, Pow, assign, call, fac, duf, mod, read, kill, clr, STO, RET, curs)
    if stack is None:
        stack = []
    expr, ops = shlex.split(string.lower(expr)), ops.split()
    for i in expr:
        if i[0] != ';':
            if i not in ops: stack.append(i)
            elif i in ops: builtin_op(i, stack)
        else: stack.append("ok")
    return stack[0]
def shell():
    try:
        x = ""
        while x != "quit":
            x = raw_input("star>   ")
            try: l = Eval(x)
            except KeyError: l = "does not exist"
            except: l = "parse error!"
            if l != None: print "   =>",l,"\n"
    except (EOFError, KeyboardInterrupt): print
if len(sys.argv) > 1:
    x = open(sys.argv[1], 'r'); l = x.readlines(); x.close()
    for i in l:
        if i[0] != ";":
            i = ' '.join(i.split())
            x = Eval(i)
            if x != None: print i,"\n   =>",x,"\n"
        else: pass
    shell()
else: shell()

在基于堆栈的语言中,两个非常有用的内置运算符是dup和swap.dup采用顶部堆栈元素并复制它.swap交换前两个堆栈元素.

如果你有,dup你可以实现这样的square功能:

star>   "dup *" square func

以下是您的计划dup并swap实施:

import sys, shlex, readline, os, string
List, assign, call, add, sub, div, Pow, mul, mod, fac, duf, read,\
kill, clr, STO, RET, fib, curs, dup, swap = {}, "set", "get", "+", "-", "/", "^", "*",\
"%", "fact", "func", "read", "kill", "clear", ">", "@", "fib", "vars", "dup", "swap"
funcdict = {}
def fact(num):
    if num == 1: return 1
    else: return num*fact(num-1)
def builtin_op(op, stack):
    global List
    global funcdict
    if op == mul: stack.append(float(stack.pop())*float(stack.pop()))
    elif op == div: stack.append(float(stack.pop())/float(stack.pop()))
    elif op == sub: stack.append(float(stack.pop())-float(stack.pop()))
    elif op == add: stack.append(float(stack.pop())+float(stack.pop()))
    elif op == Pow: stack.append(float(stack.pop())**float(stack.pop()))
    elif op == assign: val = List[stack.pop()] = stack.pop(); stack.append(val)
    elif op == call: Eval(funcdict[stack.pop()], stack)
    elif op == fac: stack.append(fact(stack.pop()))
    elif op == duf: name = stack.pop(); funcdict[name] = stack.pop(); stack.append(name)
    elif op == mod: stack.append(float(stack.pop())%float(stack.pop()))
    elif op == kill: del List[stack.pop()]
    elif op == clr: os.system("clear")
    elif op == STO: val = List[stack.pop()] = stack.pop(); stack.append(val)
    elif op == RET: stack.append(List[stack.pop()])
    elif op == curs: stack.append(List)
    elif op == dup: val = stack.pop(); stack.append(val); stack.append(val)
    elif op == swap: val1 = stack.pop(); val2 = stack.pop(); stack.append(val1); stack.append(val2)
    elif op == read: prompt = stack.pop(); List[prompt] = Eval(raw_input("%s "%prompt)); stack.append(List[prompt])
def Eval(expr, stack=None):
    ops = "%s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s %s"%(mul, add, sub, div, Pow, assign, call, fac, duf, mod, read, kill, clr, STO, RET, curs, dup, swap)
    if stack is None:
        stack = []
    expr, ops = shlex.split(string.lower(expr)), ops.split()
    for i in expr:
        if i[0] != ';':
            if i not in ops: stack.append(i)
            elif i in ops: builtin_op(i, stack)
        else: stack.append("ok")
    return stack[0]
def shell():
    try:
        x = ""
        while x != "quit":
            x = raw_input("star>   ")
            try: l = Eval(x)
            except KeyError: l = "does not exist"
            except: l = "parse error!"
            if l != None: print "   =>",l,"\n"
    except (EOFError, KeyboardInterrupt): print
if len(sys.argv) > 1:
    x = open(sys.argv[1], 'r'); l = x.readlines(); x.close()
    for i in l:
        if i[0] != ";":
            i = ' '.join(i.split())
            x = Eval(i)
            if x != None: print i,"\n   =>",x,"\n"
        else: pass
    shell()
else: shell()

最后,这是我在Python中的版本(我认为无论如何)比你编写的Python更清晰:

import shlex, functools, sys, StringIO

def bin_numeric_op(func):
    @functools.wraps(func)
    def execute(self):
        n2, n1 = self._stack.pop(), self._stack.pop()
        n1 = float(n1)
        n2 = float(n2)
        self._stack.append(func(n1, n2))
    return execute

def relational_op(func):
    @functools.wraps(func)
    def execute(self):
        n2, n1 = self._stack.pop(), self._stack.pop()
        self._stack.append(bool(func(n1, n2)))
    return execute

def bin_bool_op(func):
    @functools.wraps(func)
    def execute(self):
        n2, n1 = self._stack.pop(), self._stack.pop()
        n1 = bool(n1)
        n2 = bool(n2)
        self._stack.append(bool(func(n1, n2)))
    return execute

class Interpreter(object):
    def __init__(self):
        self._stack = []
        self._vars = {}
        self._squarestack = []

    def processToken(self, token):
        if token == '[':
            self._squarestack.append(len(self._stack))
        # Currently inside square brackets, don't execute
        elif len(self._squarestack) > 0:
            if token == ']':
                startlist = self._squarestack.pop()
                lst = self._stack[startlist:]
                self._stack[startlist:] = [tuple(lst)]
            else:
                self._stack.append(token)
        # Not current inside list and close square token, something's wrong.
        elif token == ']':
            raise ValueError("Unmatched ']'")
        elif token in self.builtin_ops:
            self.builtin_ops[token](self)
        else:
            self._stack.append(token)
    def get_stack(self):
        return self._stack
    def get_vars(self):
        return self._vars
    @bin_numeric_op
    def add(n1, n2):
        return n1 + n2
    @bin_numeric_op
    def mul(n1, n2):
        return n1 * n2
    @bin_numeric_op
    def div(n1, n2):
        return n1 / n2
    @bin_numeric_op
    def sub(n1, n2):
        return n1 - n2
    @bin_numeric_op
    def mod(n1, n2):
        return n1 % n2
    @bin_numeric_op
    def Pow(n1, n2):
        return n1**n2
    @relational_op
    def less(v1, v2):
        return v1 < v2
    @relational_op
    def lesseq(v1, v2):
        return v1 <= v2
    @relational_op
    def greater(v1, v2):
        return v1 > v2
    @relational_op
    def greatereq(v1, v2):
        return v1 > v2
    @relational_op
    def isequal(v1, v2):
        return v1 == v2
    @relational_op
    def isnotequal(v1, v2):
        return v1 != v2
    @bin_bool_op
    def bool_and(v1, v2):
        return v1 and v2
    @bin_bool_op
    def bool_or(v1, v2):
        return v1 or v2
    def bool_not(self):
        stack = self._stack
        v1 = stack.pop()
        stack.append(not v1)
    def if_func(self):
        stack = self._stack
        pred = stack.pop()
        code = stack.pop()
        if pred:
            self.run(code)
    def ifelse_func(self):
        stack = self._stack
        pred = stack.pop()
        nocode = stack.pop()
        yescode = stack.pop()
        code = yescode if pred else nocode
        self.run(code)
    def store(self):
        stack = self._stack
        value = stack.pop()
        varname = stack.pop()
        self._vars[varname] = value
    def fetch(self):
        stack = self._stack
        varname = stack.pop()
        stack.append(self._vars[varname])
    def remove(self):
        varname = self._stack.pop()
        del self._vars[varname]
    # The default argument is because this is used internally as well.
    def run(self, code=None):
        if code is None:
            code = self._stack.pop()
        for tok in code:
            self.processToken(tok)
    def dup(self):
        self._stack.append(self._stack[-1])
    def swap(self):
        self._stack[-2:] = self._stack[-1:-3:-1]
    def pop(self):
        self._stack.pop()
    def showstack(self):
        print"%r" % (self._stack,)
    def showvars(self):
        print "%r" % (self._vars,)
    builtin_ops = {
        '+': add,
        '*': mul,
        '/': div,
        '-': sub,
        '%': mod,
        '^': Pow,
        '<': less,
        '<=': lesseq,
        '>': greater,
        '>=': greatereq,
        '==': isequal,
        '!=': isnotequal,
        '&&': bool_and,
        '||': bool_or,
        'not': bool_not,
        'if': if_func,
        'ifelse': ifelse_func,
        '!': store,
        '@': fetch,
        'del': remove,
        'call': run,
        'dup': dup,
        'swap': swap,
        'pop': pop,
        'stack': showstack,
        'vars': showvars
        }

def shell(interp):
    try:
        while True:
            x = raw_input("star>   ")
            msg = None
            try:
                interp.run(shlex.split(x))
            except KeyError:
                msg = "does not exist"
            except:
                sys.excepthook(*sys.exc_info())
                msg = "parse error!"
            if msg != None:
                print "   =>",msg,"\n"
            else:
                print "   => %r\n" % (interp.get_stack(),)
    except (EOFError, KeyboardInterrupt):
        print

interp = Interpreter()
if len(sys.argv) > 1:
    lex = shlex.shlex(open(sys.argv[1], 'r'), sys.argv[1])
    tok = shlex.get_token()
    while tok is not None:
        interp.processToken(tok)
        tok = lex.get_token()
shell(interp)

这个新版本支持if和ifelse声明.通过this和函数调用,可以在语言中实现fib和fact功能.我将在后面添加如何定义这些内容.

以下是定义fib函数的方法:

star>   fib [ dup [ pop 1 0 + ] swap [ dup 1 - fib @ call swap 2 - fib @ call + ] swap 0 + 2 0 + < ifelse ] !
   => []

star>   15 fib @ call
   => [987.0]

在0 + 2 0 +之前的顺序<很给力的比较是数值比较.

还要注意如何[与]单个字符引用运营商.它们导致它们之间的所有内容都不会执行,而是作为单个项目列表存储在堆栈中.这是定义函数的关键.函数是可以与call运算符一起执行的一系列标记.它们也可以用于"匿名块",它们是lambda表达式和标准Python块之间的交叉.它们在fib函数中用于ifelse语句的两个可能路径.

解析器非常简单.而shlex充足的足够强大,因为这个简单的语言词法分析器.其他项目将从列表中提取单个项目.创建仅包含先前列表的一部分的新列表."弄明白"堆栈中的单个令牌.实现while原语.对整数进行操作的数值运算符(在实际的Forth中,默认情况下,数值运算操作整数,您需要指定类似于+.获得浮点版本的东西).以及允许字符串操作的符号标记的一些操作.也许"拆分"和"连接"操作将令牌转换为字符的单个令牌列表或将列表连接成单个令牌就足够了.