2D char array (Storing text with the spacebar in an one char array element?)

Question

Acc. to this post, the most used method to store text in a 2D array is by using %s approach. But, this approach has a downfall, i.e. whenever you press spacebar, the text which is typed after goes into the second element of array. for e.g. you typed

Input:-

1st element of char array = Hi everyone

Output:-

1st element of char array = Hi

2nd element of char array = everyone

Expected output:-

1st element of char array = Hi everyone

So, i want to ask why the below written approach cannot be used to enter text into a 2D array?

#include <stdio.h>
int main()
{
     char ch[20];
     printf("Enter name:");  
     scanf("%19[^\n]", ch);
     printf("Your name is: %s", ch);
     return 0;
}

If the above approach can be used, then please let me know how?

Please do not introduce pointer concepts/code in answer to this post. This is a question to understand why the above written approach fails.

Consider this as the code which fails:-

#include <stdio.h>

int main()
{
    char name[4][20];
    int i;
    printf("Enter names:-\n");
    for(i=0; i<4; i++)
    {
            printf("Enter name %d: ", i);
            scanf("%19[^\n]", name[i]);
            printf("\n");   
    }
    for(i=0; i<4; i++)
    {
        printf("Entered name %d: %s", i, name[i]);
        printf("\n");
    }
    return 0;
}

The above program compiles without any error or warning, but fails during runtime.

Answer 1

The problem in the example you provided is that the newline is left in the buffer. You can discard this leading whitespace by changing this:

scanf("%19[^\n]", &name[i]);

To this

scanf(" %19[^\n]", &name[i]);

Note the space before %19. With this, your program prints:

Enter names:-

Enter name 0: foo

Enter name 1: bar

Enter name 2: baz

Enter name 3: qux

Entered name 0: foo

Entered name 1: bar

Entered name 2: baz

Entered name 3: qux

This is because the %[^\n] specifier tells it to take everything but the newline. So the newline will be left in the buffer, and when scanf is called again, it is the first thing in the buffer, so those additional calls can't take any input.

That leading space in the scanf fixes the problem, because it tells scanf to discard any trailing whitespace, which includes that newline left in the buffer.

This is what a reference says about it:

Whitespace character: the function will read and ignore any whitespace characters encountered before the next non-whitespace character (whitespace characters include spaces, newline and tab characters -- see isspace). A single whitespace in the format string validates any quantity of whitespace characters extracted from the stream (including none).

Answer 2

Figured out another approach to get it done.

#include <stdio.h>

int main()
{
    char name[4][20];
    int i;
    printf("Enter names:-\n");
    for(i=0; i<4; i++)
    {
            printf("Enter name %d: ", i);
            scanf("%19[^\n]", name[i]);     
            fflush(stdin);
            printf("\n");   
    }
    for(i=0; i<4; i++)
    {
        printf("Entered name %d: %s", i, name[i]);
        printf("\n");
    }
    return 0;
}