Add a 1 if user entered an “*”, or add a 0 if user entered a “ ” to an array in C

Question

I need to take the users input (asterisk or a space) and if the user entered an asterisk I would then insert a 1 into an array, if the user entered a space I would then insert a 0 into an array. Here is what I have so far:

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

void userInput(){
    printf("Enter Generation 0 (Length must be 10, * for 1 and " " for 0):\n");
    char c;                 /* int */
    int count;
    char arr[10];
    for(int i = 0; i < 10; i++){
        scanf("%c", &c);
        if(c == '*')
            arr[i] = '1';
        else
            arr[i] = '0';
    }
    for(int j = 0; j < 10; j++){
        printf("%c", arr[j]);
    }
}

int main(int argc, const char * argv[]) {
    userInput();
}

The problem is that the loop only allows the user to enter a character 6 times and then prints.

Output if I enter all *'s:

Enter Generation 0 (Length must be 10, * for 1 and  for 0):
*
*
*
*
*
1010101010

Output if I enter all spaces:

Enter Generation 0 (Length must be 80, * for 1 and  for 0):





0000000000

Answer 1

There are any number of ways to approach capturing the '*' and ' ' and translating them to 1 and 0 , respectively. You are essentially writing a little input handler that will read until your array is full, or the user terminates input with [ctrl + d] . You simply need to check to insure the array is not full and that you haven't reached EOF . Something simple will do:

#include <stdio.h>

#define MAXC 10

int main (void) {

    unsigned char arr[MAXC] = {0};
    int c = 0, i, idx = 0;

    while (idx < MAXC && (c = getchar()) != EOF) {
        if (c == '*') arr[idx++] = 1u;
        if (c == ' ') arr[idx++] = 0u;
    }

    for (i = 0; i < idx; i++)
        printf (" arr[%d] : %u\n", i, arr[i]);

    return 0;
}

It will take any form of input, act appropriately for '*' and ' ' , and ignore an additional characters until the array is full or EOF is generated:

Example Use/Output

$ ./bin/rdspcasterix
(**) (**)
  (**)
*
 arr[0] : 1
 arr[1] : 1
 arr[2] : 0
 arr[3] : 1
 arr[4] : 1
 arr[5] : 0
 arr[6] : 0
 arr[7] : 1
 arr[8] : 1
 arr[9] : 1

$ ./bin/rdspcasterix
The quick*brown fox*jumps over*a lazy*dog
**  **
 arr[0] : 0
 arr[1] : 1
 arr[2] : 0
 arr[3] : 1
 arr[4] : 0
 arr[5] : 1
 arr[6] : 0
 arr[7] : 1
 arr[8] : 1
 arr[9] : 1

There is no one-right way to do this, so take a look at all answers and choose the one that works best for your situation.

Answer 2

You could also use fgetc with STDIN as your file stream to read char by char

EDIT Added a solution with buffer validation

#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#define EXTRACHARS 2
#define SIZE 10

static void clear_buf(void)
{
  int ch;
  while(ch = getc(stdin), ch != '\n' && ch!=EOF);
  clearerr(stdin);
}
int main(void)
{
  int c = 0; 
  char seq[SIZE + EXTRACHARS];

  printf("Enter 10 * or ' ': ");  
  fgets(seq, SIZE+EXTRACHARS, stdin);
  if (seq[strlen(seq)- 1] != '\n')
    {
      printf("Buffer overflowed \n");
      clear_buf();
    }
  else
    {
      while(c < SIZE)
    {
      if(seq[c] == '*')
        {
          printf("1");
        }
      else if(seq[c] == ' ')
        {
          printf("0");
        }
      ++c;
    }
      printf("\n");
    }
  return EXIT_SUCCESS;
}

Answer 3

ok new code work real well for me tested on QT Ide Mingw Compiler

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

void userInput(){
    printf("Enter Generation 0 (Length must be 10, * for 1 and " " for 0):\n");
    char c;
    char temp;
    char arr[10];
    int i=0;
    do{
        printf("\nEnter number %d : ", i+1);
        c = getchar();
        while((temp = getchar()) != '\n' && c != EOF);    // this is alternative to fflush
        if(c == 42)     // 42 is ascii code to *
            arr[i] = '1';
        else
            arr[i] = '0';

        i++;
    }while((c!=EOF)&&(i<10));
    for(int j = 0; j < 10; j++){
        printf("%c\n", arr[j]);
    }
}

int main(int argc, const char * argv[]) {
    userInput();
}

this may not work on online compilers