Simple Issue with Arrays - c

Question

I was just wondering why the output for this code I have below is abcdef def instead of abc def.
main()
{
char array1[3]="abc";
array1[3]='\0';
char array2[3]="def";
array2[3]='\0';
printf("%s %s", array1, array2);
}

Answer 1

char array1[3]="abc";
array1[3]='\0';
char array2[3]="def";
array2[3]='\0';
array1[3]='\0'; and array2[3]='\0'; statements access arrays out of bounds and invoke undefined behavior. The last element of an array of 3 elements is array1[2] not array[3].
To fix your program, declare your arrays as:
char array1[]="abc";
char array2[]="def";
And don't manually add the null terminator as it will be already included in the declaration above.
EDIT:
some other answers incorrectly assume that
char array1[3]="abc";
would write a trailing null character outside the array. Actually no trailing null character is written in this initialization. The declaration is equivalent to:
char array1[3]= {'a', 'b', 'c'};

Answer 2

When you do:
char array1[3]="abc";
Then array1[3] is out of bounds.
You should do:
char array1[4]="abc"; //Remember the '\0'
^^^
Note that the '\0' will be added in this way.
Also please note that when you have array of size N, then the indexes are from 0 to N - 1.

Answer 3

In your code
main()
{
char array1[3]="abc"; //Undefined ,when you access this using printf() with %s
array1[3]='\0'; //here you are storing value which is out of bound in nature
char array2[3]="def"; //same as above
array2[3]='\0'; //same as above
printf("%s %s", array1, array2);
}
memory:
---------------
| a | b | c |\0|
----------------
The last \0 what you are putting is more than the space you have allocated.
solution:
main()
{
char array1[4]="abc";
//array1[3]='\0'; //no need
char array2[4]="def";
// array2[3]='\0'; //no need
printf("%s %s", array1, array2);
}
It will give desired output

Answer 4

This is such a bad code! What is happening is the following:
Compiler allocated space for 2x3 chars.
main()
{
char array1[3]="abc"; //Will write "abc" in array 1 AND \0 in array2[0]
array1[3]='\0'; //Out of array!!! (Re)writing \0 in array2[0]
char array2[3]="def"; //Will write "def" in array2 AND \0 after array2 space, POTTENCIALLY corrupting code!
array2[3]='\0'; //\0 (re)wrote after array2, POTTENCIALLY corrupting code!
What you have in memory is 7 bytes (just 6 were allocated): "abcdef\0".
Pointes still ok: array1 points "a" and array2 points "d".
In C standard libraries, strings sizes are determined by ending zero terminator. So when you printf array1, it will read "abcdef" until \0.