2-D character array - c

Question

#include<stdio.h>
void main()
{
char a[10][5] = {"hi", "hello", "fellow"};
printf("%s",a[0]);
}
Why this code printing only hi
#include<stdio.h>
void main()
{
char a[10][5] = {"hi", "hello", "fellow"};
printf("%s",a[1]);
}
While this code is printing "hellofellow"

Answer 1

Why this code printing only hi
You've told printf to print the string stored at a[0], and that string happens to be "hi".
While this code is printing "hellofellow"
This one is by coincidence, in fact your code ought to be rejected by the compiler due to a constraint violation:
No initializer shall attempt to provide a value for an object not contained within the entity being initialized.
The string "fellow", specifically the 'w' at the end of it does not fit within the char[5] being initialised, and this violates the C standard. Perhaps also by coincidence, your compiler provides an extension (making it technically a non-C compiler), and so you don't see the error messages that I do:
prog.c:3:6: error: return type of 'main' is not 'int' [-Werror=main]
void main()
^
prog.c: In function 'main':
prog.c:5:37: error: initializer-string for array of chars is too long [-Werror]
char a[10][5] = {"hi", "hello", "fellow"};
^
Note that the second error message is complaining about "fellow", but not "hello". Your "hello" initialisation is valid by exception:
An array of character type may be initialized by a character string literal or UTF-8 string literal, optionally enclosed in braces. Successive bytes of the string literal (including the terminating null character if there is room or if the array is of unknown size) initialize the elements of the array.
The emphasis is mine. What the emphasised section states is that if there isn't enough room for a terminal '\0' character, that won't be used in the initialisation.

Answer 2

Your code:
char a[10][5] = {"hi", "hello", "fellow"};
Allocates 10 char [5]
"hello" takes up 5 so there is no room for the terminating \0, so it runs into "fellow"
If you try it, a [3] should be "w" because "fellow" is too big and the "w" runs over from a[2] to a[3]
Aside from being undefined behavior, it is confusing what you were trying to do

Answer 3

It will give undefines behaviour as string are null-terminated.
And element hello has length of 5.
Declare your array as a[10][7] then you will get intended output.
See here -https://ideone.com/c2zUs0
Why this code printing only hi
Because a[0][2] is null indicating termination thus giving you hi.

Answer 4

This is undefined behavior due to insufficient space to store \0 character.
Please note that the memory allocated is 5bytes per string in your array of strings. Thus, for the a[1] there is not sufficient memory to store the \0 character as all five bytes are assigned with "hello".
Thus, the subsequent memory is read until the \0 character is found.
Thus, you can change the line:
char a[10][5] = {"hi", "hello", "fellow"};
to
char a[][7] = {"hi", "hello", "fellow"};
Why this code printing only hi
This is because the \0 character is already encountered at a[0][2] and thus the reading of the characters is stopped.

Answer 5

What Your Code Does:
Look at the following statement:
char a[10][5] = {"hi", "hello", "fellow"};
It allocates 10 rows. 5 characters are allocated for each index of a.
What is the Problem:
Strings are Null Terminated there is always a null-terminator needed to be stored except for the given characters, so basically the used size of array is numOfCharacters+1, the extra one byte is for the null terminator. When you are initializing the array with exactly size number of characters, the null terminator is skipped. Normally the character array value is printed until the first \0(null terminator) is not found. Please also have a look at this.
The Solution:
No need to worry about this problem, all you need to do is just to set the size equal to the numOfCharactersInString + 1. You can use the following statement:
char a[10][7] = {"hi", "hello", "fellow"};
Since the largest string is "fellow" which contains 6 characters, you need to set the size 6 + 1 that is why the statement should use char a[10][7] instead of char a[10][5]
Hope it helps.

Answer 6

When you declare a 2-D character array as
char a[10][5] = {"hi", "hello", "fellow"};
char a[10][5] reserves memory to store 10 strings each of length 5 which means 4 characters + 1 '\0' character. A point to note is that the array elements are stored in contiguous memory locations.
a[0] points to the first string, a[1] to the second and so on.
Also when you initialize an array partially the other uninitialized elements become 0 instead of being garbage values.
Now in your case,after initialization if you try to visualize the array it would be something like
hi\0\0\0hellofello\0\0...
Now the command
printf("%s",a[0]);
prints characters starting from 'h' of "hi" and stops printing when a '\0' is encountered so "hi" is printed.
Now for the second case,
printf("%s",a[1]);
characters are printed starting from the 'h' of "hello" till a '\0' is encountered.Now the '\0' character is encountered only after printing "hellofello" and hence the output.