> If i give the size of the array as 15, like "unsigned char n[15] =
> {'a', 'b', 'c','d'};" , then it is appending '\0'.
> But if the size of the array is not given "unsigned char n[] ", then
> it is not appending '\0'.
>
> Does that mean, that if the size of the array is specified, it appends
> '\0' and if it is not specified then it does not append '\0'?
> Can you/anyone clarify this point?
I does more than just append a '\0'. It first zeroes the entire array,
then stores your characters there, one at a time.
Here is the (64-bit) assembly language for the beginning of your main
function. I have annotated it to show what's happening to the array.
main:
.LFB3:
pushq %rbp # save caller's base pointer
.LCFI2:
movq %rsp, %rbp # establish our base pointer
.LCFI3:
subq $48, %rsp # get memory for local variables
.LCFI4:
movq %fs:40, %rax # these three instructions are used to
movq %rax, -8(%rbp) # check for stack boundary violation
xorl %eax, %eax
# The array is in the stack frame, starting -32 from the base pointer
movq $0, -32(%rbp) # zero first 8 bytes of array
movl $0, -24(%rbp) # zero next 4 bytes of array
movw $0, -20(%rbp) # zero next 2 bytes of array
movb $0, -18(%rbp) # zero next byte of array
# Now all 15 bytes of the array have been zeroed.
movb $97, -32(%rbp) # n[0] = 'a';
movb $98, -31(%rbp) # n[1] = 'b';
movb $99, -30(%rbp) # n[2] = 'c';
movb $100, -29(%rbp) # n[3] = 'd';
leaq -32(%rbp), %rdi # load address of array
call slen
movl %eax, -36(%rbp) # t = slen(n);
leaq -32(%rbp), %rdi # load address of array
call slen
movl %eax, g(%rip) # g = slen(n);
movl g(%rip), %edx # load g
movl -36(%rbp), %esi # load t
movl $.LC0, %edi # address of "\n t = %d, g = %d\n"
movl $0, %eax # no SSE arguments
call printf
movl $0, %eax # return 0;
movq -8(%rbp), %rdx # these three instructions
xorq %fs:40, %rdx # check for stack boundary
je .L8
call __stack_chk_fail # violation
.L8:
leave # undo stack set up
ret # return to caller
Bob
