question about nul termination

for a char array define as below,

char buff[6] = "hello";

when cout << buff; the result is:
hello

right?

when we debug that , we found that:
buff[0] = h
buff[1] = e
buff[2] = l
buff[3] = l
buff[4] = o
buff[5] = 0 (nul terminater)
buff[6] = ? (unknown character)

i would like to ask:
1.) we can see "hello" after running the program but not "hello0?", is it because when the nul terminater reached, the progaram will stop printing the character?

2.) for readfile example like below:

char buff2[256]
readfile ( handle, buff2, 256, &size,NULL)

i.e. i read 256 characters to the array buff2 from the com port, will the program add the nul terminater to the char array??

char buff[6] = "hello";

That statement makes an array with 6 characters. buff[6] is beyond the size of the array.

Why'd you start a new thread for this?

when a \0 null terminating character is reached cout stops printing the string and in your case the last element of the array of chars is ignored. You are correct.

The length of the string is dictated by the amount of elements in the array before the \0 not the size of the array.

The number of elements in the array -1 simply dictates the maximum length of string that may be stored.

As for the second question, look at the docs for the function.

1) the null terminator has a value of 0 and is unprintable.Well you'll never see 1. '0' has an value of 40ish and is obviously different. In your example you do realise that buf[6] is not in the bounds of your array dont you.
2) no readfile does not null terminate.plenty of info on api funcs at msdn

one thing that you should remember is that even though you have declared your string as buff[6] the string "hello\0" requires only 6 (0-5) elements so the array will be initialised as buff[5]

daniel