Question

Which part of this program is used for input one context free grammar? And if you want to give another different context free grammar, how to do that?

code:

#include<stdio.h>

#include<stdlib.h> #include<string.h>

#define MaxVtNum 20

#define MaxVnNum 20

#define MaxPLength 20

#define MaxSTLength 50

char stack[20]={'#','E'};

char input[MaxSTLength];

char termin[MaxVtNum]={'i','+','*','(',')','#'}; char non_termin[MaxVnNum]={'E','G','T','H','F'};

struct product{

char left;

char right[MaxPLength];

int length;

};

struct product E,T,G,G1,H,H1,F,F1;

struct product M[MaxVnNum][MaxVtNum];

int flag=1;

int top=1;

int l;

void print_stack(){

}

}

}

}

}

}

int i;

for(i=0;i<=top;i++)

printf("%c",stack[i]);

void print_input(){

int i;

for(i=0;i<flag-1;i++)

printf(" ");

for(i=flag-1;i<=l;i++)

printf("%c",input[i]);

int non_termin_row(char c){

int i;

for(i=0;i<(int)strlen(non_termin);i++)

if(c==non_termin[i])

return i;

printf("Error in non_termin_row()>%c\n",c);

exit(0);

int termin_col(char c){

int i;

for(i=0;i<(int)strlen(termin);i++)

if(c==termin[i])

return i;

printf("Error in termin_col()>%c\n",c);

exit(0);

bool isNT(char c){

int i;

for(i=0;i<(int)strlen(non_termin);i++)

if(c==non_termin[i])

return false;

return true;

bool isT(char c){

int i;

for(i=0;i<(int)strlen(termin)-1;i++)

if(c==termin[i])

return true;

return false;

void main(void){

int i,j=0,k=0;

int flag2=0;

char ch;

char X=' ';

struct product cha;

E.left='E';

strcpy(E.right,"TG");

E.length=2; T.left='T';

strcpy(T.right,"FH");

T.length=2;

G.left='G';

strcpy(G.right,"+TG");

G.length=3;

G1.left='G';

G1.right[0]='~';

G1.length=1;

H.left='H';

strcpy(H.right,"*FH");

H.length=3;

H1.left='H';

H1.right[0]='~';

H1.length=1;

F.left='F';

strcpy(F.right,"(E)");

F.length=3;

F1.left='F';

F1.right[0]='i';

F1.length=1;

M[0][0]=E;M[0][3]=E;

M[1][1]=G;M[1][4]=G1;M[1][5]=G1;

M[2][0]=T;M[2][3]=T;

M[3][1]=H1;M[3][2]=H;M[3][4]=M[3][5]=H1;

M[4][0]=F1;M[4][3]=F;

printf("this program do parsing just for strings consisting with

'i','+','*','(',')', and ended by'#',\n");

printf("please input the one string:");\

do{

scanf("%c",&ch);

if((ch!='i')&&(ch!='+')&&(ch!='*')&&(ch!='(')&&(ch!=')')&&(ch!='# ')){

printf("error\n");

exit(1);

}

input[k]=ch;

k++;

}while(ch!='#'); l=strlen(input);

ch=input[0];

printf("step\t stack\t input string\t rule \n");

while(1){

printf("\n(%d)\t",j++);

print_stack();

printf("\t");

print_input();

printf("\t"); if(flag2==1){

printf("%c->%s",cha.left,cha.right);

flag2=0;

}

printf("\t");

X=stack[top--];

if(X=='#'){

if(X==ch){

}

printf("\tAcc\n");

else

printf("\tERROR\n");

break;

}

else if(isT(X)){

ch=input[flag++];

}

else if(isNT(X)){

cha=M[non_termin_row(X)][termin_col(ch)];

flag2=1;

for(i=(cha.length-1);i>=0;i--)

stack[++top]=cha.right[i];

if(stack[top]=='~')

}

top--;

else{

printf("ERROR in main()>%c\n",X);

exit(0);

}

}

}

Gr. C:\windows\system32\cmd.exe 本程序只能对由'''''(','》构成的以“#”结束的字符串进行分析。 请输入要分析的字符串:#i+i 分析栈 输入串 所用规则 基E E->TG T->FH E->i H->FH F->i ii+i# #GT i*i+i# #GHF iji i# #Hi ii+i# #GH jifi# #GHF* i+i# #GHF i+i! #GHI i+i #SGH #G i# #GT #GT 主料 #GHP i# #GHi 1# #GH 1与 #G (16) | # 请按任意键继续... H- G->+TG T->FH F- 1 | H- G-> Acc 半:

Answer 1

do{

scanf("%c",&ch);

if((ch!='i')&&(ch!='+')&&(ch!='*')&&(ch!='(')&&(ch!=')')&&(ch!='# ')){

printf("error\n");

exit(1);

}

input[k]=ch;

k++;

}while(ch!='#'); l=strlen(input);

This part of the code is used to input context free grammer.

The same part can be changed to provide another input context free grammer.