Guennadi Liakhovetski wrote:
On Thu, 2 Aug 2007, Andi Kleen wrote:
Guennadi Liakhovetski <[EMAIL PROTECTED]> writes:
char c[4] = "0123";
and - a wonder - no warning.
It's required by the C standard.
6.7.8.14 of C99:
``
An array of character type may be initialized by a character string literal,
optionally
enclosed in braces. Successive characters of the character string literal
(including the
terminating null character if there is room or if the array is of unknown size)
initialize the
elements of the array.
''
Note the "if there is room".
I believe the rationale is that it still allows to conveniently initialize
non zero terminated strings.
Right, I accept that it will compile, but I don't understand why "01234"
produces a warning and "0123" doesn't? Don't think C99 says anything about
Because 5 characters will not fit in a 4 character array, even without
the null terminator.
that. And, AFAIU, using structs with fixed-size char array we more or less
rely on the compiler warning us if anyone initializes it with too long a
string.
Also interesting, that with
char c[4] = "012345";
the compiler warns, but actually allocates a 6-byte long array...
Thanks
Guennadi
--
Robert Hancock Saskatoon, SK, Canada
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