Problem of character array in C programming

[Amie]

# include <stdio.h>
void main ()
{
 char chr[50];
 int i,size;
 printf ("Enter the size of the array:\n");
 scanf ("%d",&size);
 printf ("Enter the characters of the array:\n");
 for (i=0; i<size; i++)
 scanf ("%c",&chr[i]); 
 printf ("The characters of the array are\n");
 for (i=0; i<size; i++)
 printf ("%c",chr[i]); 
 }

I am having the above code and I am really confused about this. In the program if I say size = 5, it should take 5 characters but it actually takes only 2 inputs. Can you help me what is the problem with my character array?

[absolute55]

I think this is because you have not entered any spaces or newline after each entered character and so C compiler and has taken 3 characters in first position and the next 2 characters in second position. This is the reason why you are having only 2 inputs. If you want to differentiate between entries then you have to separate the entries by either spaces or newline character (\n).

[void]

First of all you didn't defined main () as "int". The return type for main (0 should always be int and not void. A main () function always returns something and it is the basic requirement of most of the C compiler. Next after each scanf statement, there should be some delimiters that is needed to separate the inputs. You can use \t for tabbing, \n for new line or spaces. Other than that your code is perfectly OK.