variable not incrementing in c

fakie_flip · 09-10-2010, 08:10 AM

I am using gcc main.c -o main
It compiles fine, but gets stuck in the loop. z never increments. I was using a for loop originally but had the same problem so I decided to try a while loop.

Code:

/* Includes */
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

/* Constants */
#define MAX_DIGITS 6
#define MAX_INPUT 30

/* Function prototypes */
static void print_one_digit(int d);
static void print_error(char s[]);
static void remove_newline_if_present(char *str);

/* Main program */
int main() {

  printf("Enter numbers in their decimal form.\n");
  printf("To stop, enter a negative value.\n");

//  do {
    char str_num_arr[20]; 
    
    printf("Number: ");
    fgets(str_num_arr, sizeof str_num_arr, stdin);

    remove_newline_if_present(&str_num_arr);


    printf("%s\n", str_num_arr);

    int int_num_arr[20] = {};

    int z = 0;
    while(z<MAX_INPUT) {
      int_num_arr[z] = atoi(&str_num_arr[z]);
      char a = str_num_arr[z];
      printf("%d", int_num_arr[z]);
      z++;
    } 
    printf("does it ever get here?");


//@TODO Add breaks
//    do {

//    } while(1);

//  } while(1);
  return 0;
}

static void print_error(char s[]){ 
  printf("%s", s); abort(1);
}
static void print_one_digit(int d) {
  switch(d) {
    case 0: printf("zero"); break;
    case 1: printf("one"); break;
    case 2: printf("two"); break;
    case 3: printf("three"); break;
    case 4: printf("four"); break;
    case 5: printf("five"); break;
    case 6: printf("six"); break;
    case 7: printf("seven"); break;
    case 8: printf("eight"); break;
    case 9: printf("nine"); break;
    default: print_error("Illegal call to print_one_digit");
  }
}

static void remove_newline_if_present(char *str) {
  int b;
  b = strlen(str)-1;
  if(str[b] == '\n') 
    str[b] = '\0';
}

AlucardZero · 09-10-2010, 08:25 AM

Quote:

It compiles fine

You sure? What version GCC? I get:

Code:

a.c: In function ‘main’:
a.c:27: warning: passing argument 1 of ‘remove_newline_if_present’ from incompatible pointer type
a.c: In function ‘print_error’:
a.c:54: error: too many arguments to function ‘abort’

with gcc 4.3, 4.1, and 3.4.

graemef · 09-10-2010, 09:12 AM

MAX_INPUT is defined to be 30
Array sizes are 20

You possibly end up with one of the arrays overwriting the data in z which could result in the loop continuing for much longer than you were expecting.

imitheos · 09-11-2010, 12:25 PM

Quote:

Originally Posted by fakie_flip

I am using gcc main.c -o main
It compiles fine, but gets stuck in the loop. z never increments. I was using a for loop originally but had the same problem so I decided to try a while loop.

It is exactly as graemef said. You overwrite the value of z
because you write after the array.

Quote:

Originally Posted by fakie_flip

Code:

    int z = 0;
    while(z<MAX_INPUT) {

Even if you enlarge the array to 30 characters, this check is not correct.
You need to check the length of the string given (for example if i type
the number "50", you check 30 chars when you should check 2).

Quote:

Originally Posted by fakie_flip

Code:

      int_num_arr[z] = atoi(&str_num_arr[z]);

You want to get only one digit, right ? If yes, then this won't work.
You can get the digit easily with

Code:

      int_num_arr[z] = str_num_arr[z] - '0';