这是一个能识别加法和乘法的语法分析器，例如在提示输入后输入一个算式：3+5*2*（2+1）这类的，那么就会给出匹配过程，并判断是不是成功，请赏析！由于构造了first集，follow集，select集之类，所以比较占篇幅，惭愧！ #include<string.h> int isterminal(char a){if((a=='i')||(a=='+')||(a=='*')||(a=='(')||(a==')')||(a=='$')) return(1); else return(0);} int notin(char a,char *b){char *c; for(c=b;(*c)!='\0';c++) {if((*c)==a) break; } if((*c)=='\0') return(1); else return(0);} void add(char *g,char *q,char *a){char *tem; tem=a; for(;(*tem)!='\0';tem++) {if(notin((*tem),g))  {(*q)=(*tem);   q++;   (*q)='\0';  } }} int maybenull(char E){char product[8][5]={"SBA","A+BA","A","BDC","C*DC","C","D(S)","Di"}; int i; for(i=0;i<8;i++) {if((product[i][0]==E)&&(product[i][1]=='\0'))  break; } if(i<8) return(1); else if(i>=8) return(0);} char *first(char E){char product[8][5]={"SBA","A+BA","A","BDC","C*DC","C","D(S)","Di"}; char group[200],*q; int i,j; q=group; if(isterminal(E)) {(*q)=E;  q++;  (*q)='\0'; } else {for(i=0;i<8;i++)  {if(product[i][0]==E)   {if(isterminal(product[i][1])&&notin(product[i][1],group))    {(*q)=product[i][1];     q++;     (*q)='\0';    }    else if((!isterminal(product[i][1]))&&((product[i][1])!='\0'))    {add(group,q,first(product[i][1]));     for(j=1;j<(strlen(product[i]));j++)     {if(maybenull(product[i][j])&&(product[i][j+1]!='\0'))      add(group,q,first(product[i][j+1]));     }    }   }  } } return(group);} char *follow(char E){char product[8][5]={"SBA","A+BA","A","BDC","C*DC","C","D(S)","Di"}; char group[200],*q,*tem; int i,j,k; q=group; if(E=='S') {(*q)='$';  q++;  (*q)='\0'; } for(i=0;i<8;i++) {for(j=1;j<strlen(product[i]);j++)  /*startfor1*/  {if(product[i][j]==E)  /*startif*/   {if(isterminal(product[i][j+1]))  /*startif1*/    {(*q)=product[i][j+1];     q++;     (*q)='\0';    }  /*endif1*/    else if(!isterminal(product[i][j+1]))  /*startif2*/    {if(product[i][j+1]=='\0')  /*startif3*/     {if(product[i][0]!=E)      add(group,q,follow(product[i][0]));     }  /*startif3*/     else if(product[i][j+1]!='\0')  /*startif4*/     {tem=first(product[i][j+1]);      for(;(*tem)!='\0';tem++)      {if(notin((*tem),group))       {(*q)=(*tem);        q++;        (*q)='\0';       }      }      for(k=j+1;k<(strlen(product[i])-1);k++)  /*startfor2*/      {if(maybenull(product[i][k]))       {add(group,q,first(product[i][k+1]));       }       else break;      }  /*endfor2*/      if(k==(strlen(product[i])-1))add(group,q,follow(product[i][k]));     }  /*endif4*/    }  /*endif2*/   }  /*endif*/  }  /*endfor1*/ } return(group);} char *select(int n){char product[8][5]={"SBA","A+BA","A","BDC","C*DC","C","D(S)","Di"}; char group[200],*q; int i=1,j=0; q=group; if(product[n][i]!='\0') {add(group,q,first(product[n][1]));  for(j=i;j<(strlen(product[n]));j++)  {if(maybenull(product[n][j]))   {if(product[n][j+1]!='\0')    add(group,q,first(product[n][j+1]));    else if(product[n][j+1]=='\0') add(group,q,follow(product[n][0]));   }   else break;  } } else if(product[n][1]=='\0') add(group,q,follow(product[n][0])); return(group);} void print(char *a){char *b=a; if(strlen(a)==1)printf("%c->null",*b); else {printf("%c->",*b);  for(b++;(*b)!='\0';b++)printf("%c",*b); }} main(){char product[8][5]={"SBA","A+BA","A","BDC","C*DC","C","D(S)","Di"}; char s[200],*p,sentence[40][7],type[40],stack[15]="$S"; int count=0,i,j,k=0; printf("Please write in the string you want to analysis:\n"); gets(s); p=s; while((*p)!='\0')  /*while1*/ {if(((*p)=='+')||((*p)=='*')||((*p)=='(')||((*p)==')'))  {sentence[count][0]=(*p);   type[count]=(*p);   type[count+1]='$';   type[count+2]='\0';   sentence[count][1]='\0';   p++;   count++;  }  else if((*p)>='0'&&(*p)<='9')  {for(i=0;(*p)>='0'&&(*p)<='9';i++)   {sentence[count][i]=(*p);    p++;   }   sentence[count][i]='\0';   type[count]='i';   type[count+1]='$';   type[count+2]='\0';   count++;  }  else  {printf("Fail to analyse!");   break;  } }  /*endwhile1*/ sentence[count][0]='$'; sentence[count][1]='\0'; count++; printf("The process of analysis is below:\n\n"); printf("Stack\t\t\tInput String\t\tOutput\n\n"); printf("%s\t\t\t%s$\n",stack,s); p=type; while((*p)!='\0') {if(stack[strlen(stack)-1]=='$')break;  if(!isterminal(stack[strlen(stack)-1]))  {for(i=0;i<8;i++)   {if((product[i][0]==stack[strlen(stack)-1])&&(!(notin((*p),select(i)))))    {stack[strlen(stack)-1]='\0';     if(strlen(product[i])>1)     {for(j=strlen(product[i])-1;j>=1;j--)      stack[strlen(stack)]=product[i][j];     }     printf("%s\t\t\t",stack);     for(j=k;j<count;j++)printf("%s",sentence[j]);     printf("\t\t\t");     print(product[i]);     printf("\n");     break;    }    else continue;   }   if(i>=8)   {printf("\nError analysis!");    break;   }  }  else if(isterminal(stack[strlen(stack)-1]))  {if(stack[strlen(stack)-1]==(*p))   {stack[strlen(stack)-1]='\0';    p++;    printf("%s\t\t\t",stack);    for(j=k;j<count;j++)printf("%s",sentence[j]);    printf("\n");    k++;   }   else printf("Error analysis!");  } } if(((*p)=='$')&&stack[strlen(stack)-1]=='$') printf("Analyse Successfully!"); getch();}

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