Question

I am having a conceptual doubt.

char ch[20]="some string";

I wanted to know how ch is being stored i.e. whether 20 bytes are allocated or just the length of the string assigned to it? Can we access something like ch[18] here?

Answer 1

I wanted to know how ch is being stored i.e. whether 20 bytes are allocated or just the length of the string assigned to it?

It creates an array by name ch which is 20 characters long and initalizes it with “some string”.
Yes, the array is 20 bytes in size.

Can we access something like ch[18] here?

Yes We can. And even modify the contents.

Good Read:

What is the difference between char a[] = “string”; and char *p = “string”;

---

To answer Q in comment:

     +---+---+---+---+---+---+----+----+----+----+----+----+----+      +----+----+
  ch:| s | o | m | e |   | s |  t |  r |  i |  n |  g | \0 |    |      | b  | \0 |
     +---+---+---+---+---+---+----+----+----+----+----+----+----+      +----+----+
       0   1   2   3   4   5   6    7    8     9   10   11   12  ......  18   19

When you do,

 ch[18]='b';   

The modification has happend only that you cannot see it.
printf determines the end of string by detecting \0. The \0 was placed at the end of the string when you initialized it.The rule in C/C++ while declaring an array whenever initializers are given, any uninitialized elements are automatically set to 0. As you see in above the diagram depiction the modified character is placed after what printf thinks is end of string and hence you cannot see it in output of printf.

If you output the string each character by a for-loop by iterating over it, You can see your modification.