Syntax analyzer design and implementation

1. The purpose of the experiment

Write a recursive descent analysis program that implements the sequence of words provided by the lexical analyzer

grammar checking and structural analysis.

2. Experimental content

Use C  language to compile recursive descent analysis program, and carry out grammatical analysis of simple language.

2.1 The grammar of the simple language to be analyzed

Expressed in extended BNF  as follows:

⑴<program>::=begin<statement string>end

⑵<statement string>::=<statement>{;<statement>}

⑶ <statement>::=<assignment statement>

⑷<assignment statement>::=ID:=<expression>

⑸<expression>::=<item>{+<item> | -<item>}

⑹<item>::=<factor>{*<factor> | /<factor>

⑺<factor>::=ID|NUM|(<expression>)

2.2 Description of experimental requirements

Enter the word string, ending with "#", if it is a grammatically correct sentence, output the success message,

print "success," otherwise output "error."

E.g:

Enter begin a:=9; x:=2*3; b:=a+x end #

output success

Enter x:=a+b*c end #

output error

#include<stdio.h>
#include<string.h>
int id(char a[],int i)
{
	if(a[i]<='z'&&a[i]>='a')
	{
		for(i++;(a[i]<='z'&&a[i]>='a')||(a[i]>='0'&&a[i]<='9');i++);
		if(a[i]==':'&&a[i+1]=='=')
		  return i+2;
		else
		  return -1;
	}
	else
	  return -1;
}
int yinzi(char e[])
{
	int i=0;
	if(e[i]>='a'&&e[i]<='z')
	{
	  	for(i++;e[i]<='z'&&e[i]>='a'||e[i]>='0'&&e[i]<='9';i++);
	  	if(i!=strlen(e))
	  	  return -1;
	  	else
	  	  return 1;
	}  
	else if(e[i]>='0'&&e[i]<='9')
	{
		for(i++;e[i]<='9'&&e[i]>='0';i++);
	  	if(i!=strlen(e))
	  	  return -1;
	  	else
	  	  return 1;
	}
	else
	  return -1;
}
int biaodashi(char d[],int i);
int kuohao(char d[],int i)
{
	int kz=1,ky=0,j=0;
	char e [200] = {};
	for(i+=1;d[i]!='\0'&&kz>ky;i++)
	{
	    if(d[i]=='(')
	    {
			kz++;
			e[j++]=d[i];
		}
		else if(d[i]==')')
		{
			ky++;
			if(kz>ky)
			e[j++]=d[i];
		}
		else
		    e[j++]=d[i];  
	}
	if(kz!=ky)
	  return -1;
	else
	{
		j=0;
		if(biaodashi(e,j)>0)
		  return i;
		else
		  return -1;
	}
}
int jiequ (char d [], int i)
{
	int j=0;
	char e [200] = {};
	while(!(d[i]=='+'||d[i]=='-'||d[i]=='*')&&i!=strlen(d))
		e[j++]=d[i++];
	int k = yinzi (e);
	if(k<0)
	  return -1;
	else
	  return i;
}
int biaodashi(char d[],int i)
{
	int k=0;
	if(d[i]=='(')
	  k=kuohao(d,i);
	else
	  k=jiequ(d,i);
	if(k<0)
	  return -1;
	else if(k==strlen(d))
	  return 1;
	else
	  i=k;
	if(d[i]=='+'||d[i]=='-'||d[i]=='*'||d[i]=='/')
	{
		i++;
	    if(biaodashi(d,i)<0)
	      return -1;
	    else
	      return 1;
	}
	else if(d[i]==strlen(d))
	  return 1;
	else
	  return -1;
}
intmain()
{
	char a[200]={},b[10]={},c[10]={};
	scanf("%s ",b);
	gets(a);
	int n=0,i=0,j=0,k=0;
	n=strlen(a);
	for(i=n-6;i<=n-1;i++)
	   c[j++]=a[i];
	if((strcmp("begin",b)!=0)||(strcmp(" end #",c)!=0))
		printf("error\n");
	else	
	{
		for(i=0;i<=n-7;)
	    {
		 k=id(a,i);
		 if(k<0)
		   break;	
		 else
		 {
			i=k;
			k=0;
		 }
		 char d[200]={};
		 j=0;
		 while(a[i]!=';'&&i<=n-7)
		 	  d[j++]=a[i++];
		 i+=2;
		 k=biaodashi(d,0);
		 if(k<0)
		   break;
		 else
		   k=0;
	    }
	    if(k<0)
	      printf("error\n");
	    else
	      printf("success\n");
	}
	return 0;
}