Simple Issue with Arrays

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'}; 

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.

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

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.