C-strings are null-terminated which means that in a char-array, the char at index strlen()
is a byte with all bits set to 0. I've seen code where, instead of '\0'
, the integer 0
is used. But since sizeof(int) > sizeof(char)
, this might actually write beyond the allocated space for the array - am I wrong? Or does the compiler implicitely cast 开发者_开发技巧a an int
to char
in such a situation?
When storing the value 0
in an array of char
s, then the int will be implicitly converted to char.
Actually, the type of '\0'
is int
(contrary to what one may expect), so somechararray[x]=0
and somechararray[x]='\0'
are semantically equivalent.
Personally, I prefer to go with '\0'
, to make it more clear that I'm dealing with a string.
Yes you are wrong - the zero value will be truncated. After all, if you write:
char c = 0;
you would not expect the compiler to write beyond the bounds of the variable c, I hope. If you write something like this:
char c = 12345;
then the compiler should warn you of the truncation. GCC produces:
c.c:2: warning: overflow in implicit constant conversion
but still nothing will be written beyond the bounds of the variable.
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