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    SubjectRe: gcc fixed size char array initialization bug - known?
    On Thu, 2 Aug 2007, Robert Hancock wrote:

    > Because 5 characters will not fit in a 4 character array, even without the
    > null terminator.

    On Fri, 3 Aug 2007, Stefan Richter wrote:

    > How should gcc know whether you actually wanted that char foo[len] to
    > contain a \0 as last element?

    Robert, Stefan, I am sorry, I think, you are VERY wrong here. There is no
    "even" and no guessing. The "string" DOES include a terminating '\0'. It
    is EQUIVALENT to {'s', 't', 'r', 'i', 'n', 'g', '\0'}. And it contains
    SEVEN characters. Please, re-read your K&R. Specifically, the Section
    "Initialization" in the "Function and Program Structure" chapter (section
    4.9 in my copy), the paragraph about initialization with a string, which I
    quoted in an earlier email.

    And, Stefan, there is a perfect way to specify a "0123" without the '\0' -
    {'0', '1', '2', '3'}.

    Guennadi Liakhovetski
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      Last update: 2007-08-03 00:39    [W:0.019 / U:34.284 seconds]
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