C: scanf to array

Question

I\'m absolutely new to C, and right now I am trying master the basics and have a problem reading data from scanf straight into an array.

Right now the code looks lik

Answer 1

int main()
{
  int array[11];
  printf("Write down your ID number!\n");
  for(int i=0;i<id_length;i++)
  scanf("%d", &array[i]);
  if (array[0]==1)
  {
    printf("\nThis person is a male.");
  }
  else if (array[0]==2)
  {
    printf("\nThis person is a female.");
  }
  return 0;
}

Answer 2

The %d conversion specifier will only convert one decimal integer. It doesn't know that you're passing an array, it can't modify its behavior based on that. The conversion specifier specifies the conversion.

There is no specifier for arrays, you have to do it explicitly. Here's an example with four conversions:

if(scanf("%d %d %d %d", &array[0], &array[1], &array[2], &array[3]) == 4)
  printf("got four numbers\n");

Note that this requires whitespace between the input numbers.

If the id is a single 11-digit number, it's best to treat as a string:

char id[12];

if(scanf("%11s", id) == 1)
{
  /* inspect the *character* in id[0], compare with '1' or '2' for instance. */
}

Answer 3

Use

scanf("%d", &array[0]);

and use == for comparision instead of =

Answer 4

if (array[0]=1) should be if (array[0]==1).

The same with else if (array[0]=2).

Note that the expression of the assignment returns the assigned value, in this case if (array[0]=1) will be always true, that's why the code below the if-statement will be always executed if you don't change the = to ==.

= is the assignment operator, you want to compare, not to assign. So you need ==.

Another thing, if you want only one integer, why are you using array? You might want also to scanf("%d", &array[0]);