fgets() doesn't recognize all Inputs. Any ideas what im doing wrong?

Question

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

int main()
{
    int size = 0;
    do
    {
        printf("Put in size of the list (Range[1;2^16-1]): ");
        scanf("%d", &size);//input
        if (size <= 0)
            printf("Put in a correct length!\n");
    } while (size <= 0);
    char temp[size+1];
    while (fgets(temp, size, stdin))//break with ^Z or ^D
    {
    }
    for (int i = 0; i < size; i++)//output
    {
        printf("%c", temp[i]);
        printf("%d", i);
    }
    return 0;
}

Output:

Put in size of the list (Range[1;2^16-1]): 5
pizza
^Z(my powershell input)
a012z34

I tried to use the fgets() function but it doesnt recognize all chars, only the last two chars. So I did some research but I couldn't find anything.

Answer 1

You have a few issues:

1. You do not consume the trailing \n from your scanf call. That leads to your first fgets reading only "\n" . This is clearly not what you would want but as you run fgets in a loop it does not cause problems.

2. You provide a size to fgets that does not match your buffer and cannot hold your input. Your buffer is 6 bytes but you only provide 5 to fgets . 5 Bytes are not enough to hold "pizza" as there is no room for \0 . It also does not hold the \n for that line. This means you will first get "pizz" and immediately afterwards you will get "a\n\0" . The remaining 3 bytes from the buffer will be unchanged ans still hold "z\0" from previous call.

3. After you end with ^D you print the content. But for values \n , \0 etc. you will not see something that you typed. That is why you don't see anything between your numbers.

You could make all input visible by printing like this: printf("%d: '%c' (%d)\n", i, temp[i], temp[i]);

Solution: If you want to get input with up to 5 characters, you need a buffer of 7 bytes and you must provide that size to fgets .