我已经有一个基于bison和flex的计算器程序,它从命令行参数中获取输入。
现在如何重写程序,这样我就不必调用flex
,而只需在构建过程中调用bison
和cc
?(实现类似于 https://unix.stackexchange.com/questions/499190/where-is-the-official-documentation-debian-package-iproute-doc#comment919875_499225 的东西(。
$ ./fb1-5 '1+3'
= 4
制作文件:
fb1-5: fb1-5.l fb1-5.y
bison -d fb1-5.y
flex fb1-5.l
cc -o $@ fb1-5.tab.c lex.yy.c -lfl
FB1-5.Y
/* simplest version of calculator */
%{
# include <stdio.h>
%}
/* declare tokens */
%token NUMBER
%token ADD SUB MUL DIV ABS
%token OP CP
%%
calclist: /* nothing */
| calclist exp { printf("= %dn> ", $2); }
;
exp: factor
| exp ADD exp { $$ = $1 + $3; }
| exp SUB factor { $$ = $1 - $3; }
| exp ABS factor { $$ = $1 | $3; }
;
factor: term
| factor MUL term { $$ = $1 * $3; }
| factor DIV term { $$ = $1 / $3; }
;
term: NUMBER
| ABS term { $$ = $2 >= 0? $2 : - $2; }
| OP exp CP { $$ = $2; }
;
%%
int main(int argc, char** argv)
{
// printf("> ");
if(argc > 1) {
if(argv[1]){
yy_scan_string(argv[1]);
}
}
yyparse();
return 0;
}
yyerror(char *s)
{
fprintf(stderr, "error: %sn", s);
}
FB1-5.L:
/* recognize tokens for the calculator and print them out */
%{
# include "fb1-5.tab.h"
%}
%%
"+" { return ADD; }
"-" { return SUB; }
"*" { return MUL; }
"/" { return DIV; }
"|" { return ABS; }
"(" { return OP; }
")" { return CP; }
[0-9]+ { yylval = atoi(yytext); return NUMBER; }
"//".*
[ t] { /* ignore white space */ }
. { yyerror("Mystery character %cn", *yytext); }
%%
更新:
我尝试遵循回复中的建议,请参阅下面的修改代码。 在main()
中,为什么在printf("argv[%d]: %s ", n, argv[n])
之前调用yyerror()
?不是yyerror()
只被yyparse()
调用,也不是yyparse
只在main()
main()
printf("argv[%d]: %s ", n, argv[n])
之后调用。
$ ./fb1-5 2*4
2*4error: �
= 8
fb1-5.y
:
/* simplest version of calculator */
%{
# include <stdio.h>
FILE * fin;
int yylex (void);
void yyerror(char *s);
%}
/* declare tokens */
%token NUMBER
%token ADD SUB MUL DIV ABS
%token OP CP
%%
calclist: /* nothing */
| calclist exp { printf("= %dn", $2); }
;
exp: factor
| exp ADD exp { $$ = $1 + $3; }
| exp SUB factor { $$ = $1 - $3; }
| exp ABS factor { $$ = $1 | $3; }
;
factor: term
| factor MUL term { $$ = $1 * $3; }
| factor DIV term { $$ = $1 / $3; }
;
term: NUMBER
| ABS term { $$ = $2 >= 0? $2 : - $2; }
| OP exp CP { $$ = $2; }
;
%%
/* The lexical analyzer returns a double floating point
number on the stack and the token NUM, or the numeric code
of the character read if not a number. It skips all blanks
and tabs, and returns 0 for end-of-input. */
#include <ctype.h>
#include <string.h>
int yylex (void)
{
char c;
/* Skip white space. */
while ((c = getc(fin)) == ' ' || c == 't'){
continue;
}
// printf("%s", &c);
/* Process numbers. */
if (c == '.' || isdigit (c))
{
ungetc(c, fin);
fscanf (fin, "%d", &yylval);
return NUMBER;
}
/* Process addition. */
if (c == '+')
{
return ADD;
}
/* Process sub. */
if (c == '-')
{
return SUB;
}
/* Process mult. */
if (c == '*')
{
return MUL;
}
/* Process division. */
if (c == '/')
{
return DIV;
}
/* Process absolute. */
if (c == '|')
{
return ABS;
}
/* Process left paren. */
if (c == '(')
{
return OP;
}
/* Process right paren. */
if (c == ')')
{
return CP;
}
/* Return a single char. */
yyerror(&c);
return c;
}
int main(int argc, char** argv)
{
// evaluate each command line arg as an arithmetic expression
int n=1;
while (n < argc) {
if(argv[n]){
// yy_scan_string(argv[n]);
// fin = stdin;
fin = fmemopen(argv[n], strlen (argv[n]), "r");
printf("%s ",argv[n]);
fflush(stdout);
yyparse();
}
n++;
}
return 0;
}
void yyerror(char *s)
{
fprintf(stderr, "error: %sn", s);
}
野牛手册的示例部分中有一个词法扫描器的基本实现。(稍不那么基本的版本在手册的后面。
这不会直接帮助您,因为它基于 fscanf
,这意味着它适用于输入流。大多数 C 库都包含允许您将字符串视为FILE*
的函数(例如,请参阅 Posix 标准 fmemopen(。如果做不到这一点,你将不得不用基于字符串的替代方案替换getc和scanf调用,这意味着你需要在某处跟踪缓冲区和输入指针。 strtoul
(或strtod
(将证明是有用的,因为第二个参数可以帮助您跟踪数字使用了多少字符串。