cprog integer calculator

[aismm]

Basically the user enters one of the following + - * / % | < > c q
followed by one or more spaces, followed by an integer, and then the newline key to get the answer and the current vaule is set to 0 then it keeps on adding up until the user enter c for clear or q for quit.
however, when I run the program, it goes straight to error message instead of compiling, what am I doing wrong? Thanks.

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

int main ()
{
	int b; //result
	int i; 
	int n;
	int a; //number to be calculated
	char f; //operator
	char s[33]; // number follows c
	char *c;

	printf("ECE15 Lab2 \n");
	printf("Name: Sherry Li\n");
	printf("You may enter one of the allowed operations, + - * / % & | < > c q\n");

	printf("Press c for clear.\n");
	printf("Press q for quit. \n\n");

	while (gets(s)!=0)
{
		//strncpy(s,f,1);
		c = strpbrk(s, "+-*/%&|<>cq");
		if (c != 0)
		*c = ' ' ; //zero out the operation 
		a = atoi(s);
          	switch (a) 
	{
		case '+': 
			b=b+a;
       	       break;           
   	 	case '-':
			b=b-a;
			break;
  	  	case '*':        
			b=b*a;
			break;             
  	  	case '/':
			b=b/a;
			break;
		case '%':
			b=b%a;
			break;
		case '&':
			b=b&a;
			break;
		case '|':
			b=b|a;
			break;
		case '<':
			b=b<a;
			break;
		case '>':
			b=b>a;
			break;
		case 'c':
			a = 0;
			break;
		case 'q':
			return 0;
			break;
   		 case '\n':                
      printf("The dec answer= %d\n", b);
      printf("The hex answer= %x\n", b);
      printf("The oct answer= %o \n\n", b);
      break;
    default:
      printf("error: unknown command %s\n", s);
	}
}
}

[John A]

Don't use gets , because it never checks the buffer size and is consider to be unsafe -- fgets is what you're supposed to use.

[WaltP]

Output values that are key to the operation of the code (like a, b, and c) at important decision points to see what values they contain.

[Nick Evan]

it goes straight to error message instead of compiling

what compiler are you using, because on VS2005 pro, your program compiles fine..

[aismm]

ok so I changed gets to fgets, however, it's still not working?? What I mean is when I try to enter calculation such as + 3 it goes straight to default in switch function -- "error: unknown command", did I do something wrong there?

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


int main ()
{
	int b; //result
	int d; //binary to integer 
	int i; //integer to be converted
	int n;
	int a; //number to be calculated
	//char f; 
	char s[34]; //entered value
	char *c;
	char *p;


	printf("ECE15 Lab2 \n");
	printf("Name: Sherry Li\n");
	printf("You may enter one of the allowed operations, + - * / % & | < > c q\n");

	printf("Press c for clear.\n");
	printf("Press q for quit. \n\n"); 

	printf("Decimal \t Octal \t Hex\t Binary\t Op Integer: ");     // while(gets(s)!=0)
	fgets(s, sizeof(s), stdin);
    	// check for and remove trailing \n
    	if ((p = strchr(s,'\n')) != NULL)
   	 {
    	  *p = '\0';
   	 }
		c = strpbrk(s, "+-*/%&|<>cq"); //strncpy(s,f,1);
		if (c != 0)
		*c = ' ' ; //zero out the operation 
		a = atoi(s);
		//d = b2i(c); //binary to integer 
          	switch (a)
	{
		case '+': 
			b=b+a;
       	       break;           
   	 	case '-':
			b=b-a;
			break;
  	  	case '*':    
			b=b*a;
			break;             
  	  	case '/':
			b=b/a;
			break;
		case '%':
			b=b%a;
			break;
		case '&':
			b=b&a;

			break;
		case '|':
			b=b|a;
			break;
		case '<':
			b=b<a;
			break;
		case '>':
			b=b>a;
			break;
	
   		 case '\n':
			printf("The dec answer= %d \t The hex answer= %x\t The oct answer= %o \n\n", b, b, b);                
			break;
		case 'c':
			b = 0;
			break;
		case 'q':
			break;
    default:
      printf("error: unknown command %s\n", s);
			break;
	}
}

[WaltP]

I repeat:

Output values that are key to the operation of the code (like a, b, and c) at important decision points to see what values they contain.