Output in C not as expected on winows in regards to characters

Question

I have written a program in C and it is not outputting how I expected. Im used to linux and so I am confused why for this basic program the output is not as I expected. I am using c std 99. I am also relativley new to C so if it is a issue in my code also please let me know

for input: A B C D E F G H I J K L M N O

I get output: -A-- --B-- --C-- --D-- --E-- --F-- --G-- --H-- --I-- --J-- --K-- --L-- --M-- --N-- --O-- -

expected output -A--B--C--D--E--F--G--H--I--J--K--L--M--N--O_

int main() {
  char *arr = safeMalloc(sizeof(char) * 32);
  int sizeArr = 32;
  int cou = 0;
  while (scanf("%c", &arr[cou]) != EOF || cou == 10) {
    cou++;
    if(cou == sizeArr-2) {
      arr = realloc(arr, 2*sizeArr);
      sizeArr*2;
    }
  }
  for (int i = 0; i < cou; ++i) {
    printf("-%c-", arr[i]);
  }
  printf("\n");
}

`scanf("%c", &arr[cou])` -> `scanf(" %c", &arr[cou])` You need a space before the `%c` to skip leading whitespace. https://godbolt.org/z/Mss9P4864 — Retired Ninja, Oct 24 '22 at 16:24
Each of those spaces is a character that `%c` will read unless suppressed. — Weather Vane, Oct 24 '22 at 16:25
`sizeArr*2;` is essentially a no op. You meant `sizeArr *= 2;` — Avi Berger, Oct 24 '22 at 16:26
Hi all thanks for the help. That answers my question. the cou 10 was just so that it accepted 10 letters. — MaxVK, Oct 24 '22 at 16:30

score 1 · Answer 1 · answered Oct 24 '22 at 16:23

1

The program is considering the spaces as input characters. Changing the for loop to this should fix it:

for (int i = 0; i < cou; ++i) {
    if(arr[i] == ' ') continue;
    printf("-%c-", arr[i]);
}

answered Oct 24 '22 at 16:23

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Output in C not as expected on winows in regards to characters

1 Answers1