【python】PLY词法分析实验「编译原理」

题

编写程序，能够把如下程序中的词法单元都识别出来

int asd = 0;
int bc = 10;
while ( asd < bc)
{
    
    
	if(bc - asd < 2)
		printf("they are close.");
	asd = asd + 1;
}

代码

import ply.lex as lex

# C语言保留字
reserved = {
    
    
    'if' : 'IF',
    'then' : 'THEN',
    'else' : 'ELSE',
    'while' : 'WHILE',
    'int' : 'INT',
    'printf' : 'PRINTF'
}
tokens = ['LPAREN', 'RPAREN', 'DIVIDE', 'TIMES', 'MINUS', 'PLUS', 'NUMBER', 'FENHAO', 'DENGHAO', 'XIAOYU', 'DAYU', 'ZUODAKUOHAO', 'YOUDAKUOHAO', 'YINHAO', 'DIANHAO', 'ID'] + list(reserved.values())

# ID规则
def t_ID(t):
    r'[a-zA-Z_][a-zA-Z_0-9]*'
    t.type = reserved.get(t.value,'ID') # Check for reserved words
    return t

# 规则
t_PLUS    = r'\+'
t_MINUS   = r'-'
t_TIMES   = r'\*'
t_DIVIDE  = r'/'
t_LPAREN  = r'\('
t_RPAREN  = r'\)'
t_FENHAO  = r';'
t_DENGHAO = r'='
t_XIAOYU = r'<'
t_DAYU = r'>'
t_ZUODAKUOHAO = r'{'
t_YOUDAKUOHAO = r'}'
t_YINHAO = r'"'
t_DIANHAO = r'\.'

# 规则
def t_NUMBER(t):
	# 识别数字
    r'\d+'
    t.value = int(t.value)    
    return t

# 对于空行的规则
def t_newline(t):
    r'\n+'
    t.lexer.lineno += len(t.value)

# 忽略空格
t_ignore  = ' \t'

# 输出错误的规则
def t_error(t):
    print ("Illegal character '%s'" % t.value[0])
    t.lexer.skip(1)

# Build the lexer
lexer = lex.lex()

# 测试数据
data = '''
int asd = 0;
int bc = 10;
while ( asd < bc)
{
	if(bc - asd < 2)
		printf("they are close.");
	asd = asd + 1;
}
'''

lexer.input(data)

while True:
	tok = lexer.token()
	if not tok: break 
	print (tok)

结果

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新手上路，有错请指正。