我已经有基于野牛和Flex计算器程序,这需要从命令行参数的输入。
现在,我怎么可以重写程序,所以,我没有打电话给flex
但只建设过程中调用bison
和cc
做? (实现了类似于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
Makefile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("= %d\n> ", $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 { $$ = $2; }
;
%%
int main(int argc, char** argv)
{
// printf("> ");
if(argc > 1) {
if(argv[1]){
yy_scan_string(argv[1]);
}
}
yyparse();
return 0;
}
yyerror(char *s)
{
fprintf(stderr, "error: %s\n", s);
}
fb1-5.l:
/* recognize tokens for the calculator and print them out */
%{
# include "fb1-5.tab.h"
%}
%%
"+" { return ADD; }
"-" { return SUB; }
"*" { return MUL; }
"/" { return DIV; }
"|" { return ABS; }
"(" { return OP; }
")" { return CP; }
[0-9]+ { yylval = atoi(yytext); return NUMBER; }
"//".*
[ \t] { /* ignore white space */ }
. { yyerror("Mystery character %c\n", *yytext); }
%%
更新:
我试图按照建议的答复,请参见下面的修改后的代码。在main()
,为什么yyerror()
printf("argv[%d]: %s ", n, argv[n])
之前叫什么名字?不仅是由yyerror()
yyparse()
调用,不仅yyparse
在printf("argv[%d]: %s ", n, argv[n])
main()
main()
后调用。
$ ./fb1-5 2*4
2*4error: �
= 8
fb1-5.y
:
/* simplest version of calculator */
%{
# include <stdio.h>
FILE * fin;
int yylex (void);
void yyerror(char *s);
%}
/* declare tokens */
%token NUMBER
%token ADD SUB MUL DIV ABS
%token OP CP
%%
calclist: /* nothing */
| calclist exp { printf("= %d\n", $2); }
;
exp: factor
| exp ADD exp { $$ = $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: %s\n", s);
}
有一个基本的实现在examples section of the bison manual一个词法扫描。 (略碱性较弱的版本是手册后面。)
因为它是基于fscanf
,这意味着它可以在输入流这不会直接帮助你。大多数C库包含这让你把一个字符串作为FILE*
功能(参见,例如,POSIX标准fmemopen)。如果做不到这一点,你就必须更换基于字符串替代品GETC和scanf调用,这意味着你将需要跟踪缓冲器和输入指针的地方。因为第二个参数可以帮助你跟踪了多少字符串的使用数量strtoul
(或strtod
)将被证明是有用的。