12-16 实验四 递归下降语法分析程序设计

对于给定的文法G[E],消除左递归后的文法是：

E→TE'

E'→+TE'|-TE'|ε

T→FT'

T'→*FT'|/FT'|ε

F→(E)|i

 源程序：

#include<stdio.h>
char curr;
char LL1[1000];
int N=-1;
 
void Top();
void Top1();
void Top2();
void top();
void top1();
void error();
void scaner();
 
void main()
{    char ch;
    int i=0;
    printf("请输入需要分析的curr语法：(以#结束)
");
    do{
        scanf("%c",&ch);
        LL1[i] = ch;
        i++;
    }while(ch != '#');
    scaner();
    Top1();
    if(curr == '#')
        printf("成功！
");
    else{
        printf("文法有误！！
");
        error();
    }
 
}
 
void scaner(){   
 
    N++;
    if(LL1[N] == ' '){
        N++;
    }else{
        curr = LL1[N];
    }
}
 
void Top(){
    Top2();
    top();
}
 
void Top1(){
    Top();
    top1();
}
 
void Top2(){
    if(curr == '('){
        scaner();
        Top1();
        if(curr == ')'){
            scaner();
        }
        else
        {
            error();
        }
        else
        {
        
        while(1){
            if(curr >= 'a' && curr<='z' || curr >= 'A' && curr <= 'Z' || curr >= '0' && curr <= '9'){
                scaner();
            }
            else
            {
                break;
            }
        }
    }
}
 
void top(){
    if(curr == '*'){
        scaner();
        Top2();
        Top();
    }else if(curr == '/'){
        scaner();
        Top2();
        Top();
    }else if(LL1[N+1] == '#' || LL1[N+1] == '+' || LL1[N+1] == '/' || LL1[N+1] == '*' || LL1[N+1] == '-'){
        if(LL1[N] != '(' && LL1[N] != ')')
        error();
    }
}
 
void top1(){
    if(curr == '+'){
        scaner();
        Top();
        top1();
    }
    else if(curr == '-'){
        scaner();
        Top();
        top1();
    }else if(LL1[N+1] == '#' || LL1[N+1] == '+' || LL1[N+1] == '/' || LL1[N+1] == '*' || LL1[N+1] == '-'){
        if(LL1[N] != '(' && LL1[N] != ')')
        error();
    }
}
 
void error(){
    printf("%d %c之后的文法有错！
",N,curr);
}
　　

运行结果：