我正在尝试学习解析表达式。我发现递归下降解析似乎很容易做到这一点。从维基百科,我在C中找到了一个例子。所以,我开始阅读和编辑这段代码,以了解它是如何工作的。我根据维基百科页面上的描述编写了缺失的例程,但它不能像我预期的那样在任何表达式中起作用。例如:1+2*3+1;
返回
错误:声明:语法错误
有人可以解释我错过了什么吗?
当前的C代码:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef enum {ident, number, lparen, rparen,
times, // symbol *
slash, // symbol \ not added yet
plus, // symbol +
minus, //symbol -
eql, //symbol ==
neq, // !=
lss, // <
leq,// <=
gtr, // >
geq, // >=
callsym, // not added yet
beginsym, // not added yet
semicolon, // ;
endsym,
ifsym, whilesym, becomes, thensym, dosym, constsym,
comma, //:
varsym, procsym, period, oddsym,
not, // !
eq // =
} Symbol;
Symbol sym;
int peek;
void getsym(void);
void error(const char*);
void expression(void);
void program(void);
void nexttok(void);
#define is_num(c) ((c) >= '0' && (c) <= '9')
#define is_letter(c) ((c) <= 'a' && (c) <= 'z' || (c) >= 'A' && (c) <= 'Z')
int main(void)
{
program();
return 0;
}
void nexttok(void)
{
peek = getchar();
}
void getsym(void)
{
for(;;) {
nexttok();
if(peek == ' ' || peek == '\t') continue;
else if(peek == EOF) break;
else break;
}
//static char buf[256];
switch(peek) {
case '+': sym = plus; return;
case '-': sym = minus; return;
case '*': sym = times; return;
case ';': sym = semicolon; return;
case ':': sym = comma; return;
case '=':
nexttok();
if(peek == '=')
sym = eql;
else
sym = eq;
return;
case '!':
nexttok();
if(peek == '=')
sym = neq;
else
sym = not;
return;
case '<':
nexttok();
if(peek == '=')
sym = leq;
else
sym = lss;
return;
case '>':
nexttok();
if(peek == '=')
sym = geq;
else
sym = gtr;
return;
}
if(is_num(peek)) {
sym = number;
return;
}
}
int accept(Symbol s) {
if (sym == s) {
getsym();
return 1;
}
return 0;
}
int expect(Symbol s) {
if (accept(s))
return 1;
error("expect: unexpected symbol");
return 0;
}
void factor(void) {
if (accept(ident)) {
;
} else if (accept(number)) {
;
} else if (accept(lparen)) {
expression();
expect(rparen);
} else {
error("factor: syntax error");
getsym();
}
}
void term(void) {
factor();
while (sym == times || sym == slash) {
getsym();
factor();
}
}
void expression(void) {
if (sym == plus || sym == minus)
getsym();
term();
while (sym == plus || sym == minus) {
getsym();
term();
}
}
void condition(void) {
if (accept(oddsym)) {
expression();
} else {
expression();
if (sym == eql || sym == neq || sym == lss || sym == leq || sym == gtr || sym == geq) {
getsym();
expression();
} else {
error("condition: invalid operator");
getsym();
}
}
}
void statement(void) {
if (accept(ident)) {
expect(becomes);
expression();
} else if (accept(callsym)) {
expect(ident);
} else if (accept(beginsym)) {
do {
statement();
} while (accept(semicolon));
expect(endsym);
} else if (accept(ifsym)) {
condition();
expect(thensym);
statement();
} else if (accept(whilesym)) {
condition();
expect(dosym);
statement();
} else {
error("statement: syntax error");
getsym();
}
}
void block(void) {
if (accept(constsym)) {
do {
expect(ident);
expect(eql);
expect(number);
} while (accept(comma));
expect(semicolon);
}
if (accept(varsym)) {
do {
expect(ident);
} while (accept(comma));
expect(semicolon);
}
while (accept(procsym)) {
expect(ident);
expect(semicolon);
block();
expect(semicolon);
}
statement();
}
void program(void) {
getsym();
block();
expect(period);
}
void error(const char *msg)
{
fprintf(stderr,"error: %s\n",msg);
exit(1);
}
答案 0 :(得分:10)
statement
从不调用expression
,因此所有表达式都将是语法错误。要解决此问题,您需要更改statement
,以便expression
如果sym
是启动表达式的有效符号,则会调用accept
。 (expression
是不可接受的,因为它会使用符号而{{1}}不会看到它。)