Recursive descent analysis

For a given grammar g[e]:

E→e+t| E-t| T
T→t*f| T/f| F
f→ (E) |i

The grammar to eliminate left recursion is:
E→te '

E ' →+te ' |-te ' |∑

T→ft '

T ' →*ft ' |/ft ' |∑

f→ (E) |i

is ll (1) grammar?

Select (E→te ') =first (TE ') ={(, i}
Select (E ' →+te ') =first (+te ') ={+}
Select (E ' →-te ') =first (-te ') ={-}
Select (E ' →∑) =follow (e ') ={), #}
Select (t→ft ') =first (FT ') ={(, i}
Select (T ' →*ft ') =first (*ft ') ={*}
Select (T ' →/ft ') =first (/ft ') ={/}
Select (t ' →∑) =follow (t ') ={+,-,), #)
Select (f→ (e)) =first ((e)) ={(}
Select (f→i) =first (i) ={i}

From the above analysis, learned that this grammar satisfies ll (1) grammar

The code is as follows

1#include <stdio.h>2#include <string>3 Charstr[Ten];//record the string to parse4 intx=0;//Record first character5  6 voidE (); 7 voidX (); 8 voidT (); 9 voidY ();Ten voidF (); One   A intMain () - { -     intLen; theprintf"Please enter an arithmetic expression:"); -scanf"%s", str); -len=strlen (str); -str[len]='#'; +str[len+1]=' /'; - E (); +printf"\ nthe correct!\n"); Astrcpy (str,""); atx=0; -     return 0; - } -   - voidE () - { in T (); - X (); to } +   - voidX () the { *     if(str[x]=='+'|| str[x]=='-') $     {Panax NotoginsengX + +; - T (); the X (); +     }  A } the   + voidT () - { $ F (); $ Y (); - } -   the voidY () - {Wuyi     if(str[x]=='*'|| str[x]=='/') the     { -X + +; Wu F (); - Y (); About     } $ } -   - voidF () - { A     if(str[x]>='a'&&str[x]<='Z') +     { theX + +; -     } $     Else if(str[x]>=0&&str[x]<=9) the     { theX + +; the     } the     Else if(str[x]=='(') -     {      inX + +; the E (); the         if(str[x]==')') About         { theX + +;  the         } the         Else +         { -printf"\ n Error!\n"); theExit0);Bayi         } the     }  the     Else -     { -printf"\ n Error r!\n");  theExit0); the  } the}

Recursive descent analysis