having problems

[happy]

i still have problem

i tried to solve the first function which checks if there is an article or not in the sentence as follow

 #include<stdio.h>
#include<conio.h>
#include<ctype.h>
#include<string.h>

typedef char* token_t ;
typedef token_t  token;

int get_next_token();




int main()
{int y;
 y=get_next_token();
 if(y==0)
 printf("No Article \n");
 else
 printf("GOOD PHRASE ");
 return(0);
 }
/*******************************/


int get_next_token()
{
 token string[30];
 token str1[10];
 int i=0,x;
 printf("Enter the sentence\n");
 scanf("%s",string);

 while(isspace((string[i])) )
 { str1[i]=string[i] ;
   i++;
 }


 x=article(str1);
 if(x==0)
 	return(0);

 else
	return(0);
}


/* the following function returns 1 if there is article otherwise it returns 0*/
int article (token str)
{
if((strcmp(str,"The")==0) ||(strcmp( str, "A")==0)||
   (strcmp(str,"the")==0)||(strcmp(str,"a")==0))
   return(1);
   else
   return(0);
}

but the compiler gave me errors such as could not convert
' char * ' to int

[The Junglist]

I don't know what it is you want to do, so I only got rid of the warnings/errors.

#include<stdio.h>
#include<conio.h>
#include<ctype.h>
#include<string.h>

typedef char* token_t ;
typedef token_t token;

int get_next_token();
int article (token str);

int count=0;

int main()
{
	int y;
	y = get_next_token();
	
	if(y==0)
		printf("No Article \n");
	else
		printf("GOOD PHRASE %d",count);
	
	return(0);
}
/*******************************/


int get_next_token()
{
	token string[30];
	token str1[10];
	int i=0,x;
	printf("Enter the sentence\n");
	scanf("%s",string);

	while( isspace((int)string[i]) )
	{
		str1[i] = string[i] ;
		i++;
	}

	x = article( (char*) str1 );
	
	if(x==0)
		return(0);
	else
		return(0);
}


/* the following function returns 1 if there is article otherwise it returns 0*/
int article (token str)
{
	if( (strcmp(str,"The")==0) || (strcmp( str, "A") == 0) || (strcmp(str,"the")==0) || (strcmp(str,"a") ==0))
		return(1);
	else
		return(0);
}

[Monster]

What are you trying to do here?

if(x==0)
   return(0);
else
   return(0);

Just return x

Don't name your variable string (it's a reserved word in C++).

[happy]

I asked to write a program works as parser This program implements an interpreter in the C language,which will read a sentence at atime from the user and print :Good:<n> words when the input noun phrase is grammatically correct,and <n> is the total number of words in the phrase.If the input is grammatically incorrect,it will print :Bad phrase to the user.

The sentence is correct(according to this program) if it contains
1.article(a,the)
2.adjective {hungry,black,white , thirsty}
3.noun {cat,dog}

the sentence may contain more than one objective or no objective at all and the program must contain the foolowing function getnexttoken(); noun_phrase();
article(); noun();
example:


input output
The cat Good<2>words
The black dog. Good<3>words
The white hungry cat. Good<4>words
a hungry Bad phrase

some members suggested to use small steps and solve the easiest problem an so on .

[happy]

sorry. i did mistake
the second return must be return(1)