我该如何重写程序，这样我就不必调用`flex`而是仅调用`bison`和`cc`了? [英] How can I rewrite the programs, so that I don't have to call `flex` but only call `bison` and `cc`?
问题描述
我已经有一个基于bison和flex的计算器程序，该程序从命令行参数中获取输入.
I already have a calculator program based on bison and flex which takes input from command line arguments.
现在我该如何重写程序，这样我就不必在构建过程中调用flex而只调用bison和cc了? (实现与 .stackexchange.com/questions/499190/where-is-the-of-official-documentation-debian-package-iproute-doc#comment919875_499225 ).
Now how can I rewrite the programs, so that I don't have to call flex but only call bison and cc during building process? (Achieve something similar to https://unix.stackexchange.com/questions/499190/where-is-the-official-documentation-debian-package-iproute-doc#comment919875_499225).
$ ./fb1-5 '1+3'
= 4
Makefile:
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 { $$ = $2; }
;
%%
int main(int argc, char** argv)
{
// printf("> ");
if(argc > 1) {
if(argv[1]){
yy_scan_string(argv[1]);
}
}
yyparse();
return 0;
}
yyerror(char *s)
{
fprintf(stderr, "error: %s\n", s);
}
fb1-5.l:
/* recognize tokens for the calculator and print them out */
%{
# include "fb1-5.tab.h"
%}
%%
"+" { return ADD; }
"-" { return SUB; }
"*" { return MUL; }
"/" { return DIV; }
"|" { return ABS; }
"(" { return OP; }
")" { return CP; }
[0-9]+ { yylval = atoi(yytext); return NUMBER; }
"//".*
[ \t] { /* ignore white space */ }
. { yyerror("Mystery character %c\n", *yytext); }
%%
更新:
我试图按照回复中的建议进行操作，请参阅下面的修改后的代码.在main()中，为什么在printf("argv[%d]: %s ", n, argv[n])之前调用yyerror()?不仅yyerror()仅由yyparse()调用，而且yyparse仅在main()中main()中的printf("argv[%d]: %s ", n, argv[n])之后被调用.
I tried to follow the advice in the reply, see the modified code below. in main(), why is yyerror() called before printf("argv[%d]: %s ", n, argv[n])? Isn't yyerror() called only by yyparse(), and isn't yyparse only called after printf("argv[%d]: %s ", n, argv[n]) in main() in main().
$ ./fb1-5 2*4
2*4error: �
= 8
fb1-5.y:
/* simplest version of calculator */
%{
# include <stdio.h>
FILE * fin;
int yylex (void);
void yyerror(char *s);
%}
/* declare tokens */
%token NUMBER
%token ADD SUB MUL DIV ABS
%token OP CP
%%
calclist: /* nothing */
| calclist exp { printf("= %d\n", $2); }
;
exp: factor
| exp ADD exp { $$ = $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);
}
推荐答案
There is a basic implementation of a lexical scanner in the examples section of the bison manual. (Slightly less basic versions are later in the manual.)
这不会直接为您提供帮助，因为它基于fscanf，这意味着它适用于输入流.大多数C库包含使您可以将字符串视为FILE*的函数(例如，参见Posix标准
That won't help you directly because it is based on fscanf, which means that it works on an input stream. Most C libraries contain functions which let you treat a character string as a FILE* (see, for example, the Posix standard fmemopen). Failing that, you'd have to replace the getc and scanf calls with string based alternatives, which means you will need to keep track of a buffer and input pointer somewhere. strtoul (or strtod) will prove useful because the second argument helps you keep track of how much of the string was used by the number.
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