我在c中访问错误的（越界）char数组索引

Question

Suppose we have a character array in c like,

char a[20];

Can we access index 20 to put terminating NULL like this.

a[20]='\0';

But in my algo. ("which is integer to char array converter") this is necessary to put terminating NULL if last index is,smaller then the size of char array for example,

If size of my int is 4 ("1421") then i have to put '1' at index 0, '4' at index 1, '2' at index 2 and '1' at index 3.

And finally terminating NULL at index 4

index=4;
a[index]='\0';

Another way to fix the same code,

if(index<20)  (Will increase one condition)
    a[index]='\0';

But i just wanna know is it possible......to put terminating NULL at index 20.

OK I GOT THIS EVERYONE THANK-U VERY MUCH FOR YOUR HELP.

Answer 1

An array char a[20] has space for 20 characters, at indexes 0 through 19. Writing to a[20] is writing outside the array and will have unpredictable consequences. You are limited to 20 characters including any terminating NULL character. If you need space for 20 characters plus a terminating NULL, you need to declare your array as char a[21]; . Also, declaring a[20] does not put a NULL anywhere.

Answer 2

char a[20] : 20 means you can have elements from 0-19 . and it's assumed that you will use only upto 19.if you fill upto 20 then you will get an error.

change it to :

a[20+1] => a[21] now you can use 20 to put \\0 . a[20]='\\0'

Code for @AnkeshKushwah

int main()
{
char arr[4]="hell";  
printf("%c",arr[4]); // here you will see garabage. 
printf("%c",arr[5]); // here is the terminating character. after 4.
char arr[]="hell";   
printf("%c",arr[4]); // here you will see terminating character.Because hell will take 
                        0-3 and 4 contains \0 
getch();
}

Answer 3

Consider initializing your char array with all zeroes: char a[20] = { 0 };

That way, no matter how many chars you write (up to 19), you're always null terminated.