hi
is this correct
char teststr[100]="\0";
or is there another method to initialise;
thanks
lee
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG 10 43472
Yang Lee wrote: hi is this correct
char teststr[100]="\0";
or is there another method to initialise;
char teststr[100] = "" suffices, because the string literal already has a
NUL character at the end (in position 0, in this case).
Christian
Yang Lee wrote: hi is this correct
char teststr[100]="\0";
It works.
or is there another method to initialise;
The following both do the same thing, perhaps a touch more elegantly:
char teststr[100] = "";
char teststr[100] = {0};
infobahn wrote: Yang Lee wrote: hi is this correct
char teststr[100]="\0"; It works.
or is there another method to initialise;
The following both do the same thing, perhaps a touch more elegantly:
char teststr[100] = "";
This initializes the first element to a terminating null character and
does nothing to the rest of the array. char teststr[100] = {0};
This one does not do the same thing. This one initializes the entire
array to zeros.
gooch wrote: infobahn wrote: Yang Lee wrote: hi is this correct
char teststr[100]="\0";
It works.
or is there another method to initialise;
The following both do the same thing, perhaps a touch more elegantly:
char teststr[100] = ""; This initializes the first element to a terminating null character and does nothing to the rest of the array. char teststr[100] = {0}; This one does not do the same thing. This one initializes the entire array to zeros.
infobahn is right, gooch is wrong.
--
pete
pete wrote: gooch wrote:
infobahn wrote:
Yang Lee wrote:
hi is this correct
char teststr[100]="\0";
It works.
or is there another method to initialise;
The following both do the same thing, perhaps a touch more elegantly:
char teststr[100] = "";
This initializes the first element to a terminating null character and does nothing to the rest of the array.
char teststr[100] = {0};
This one does not do the same thing. This one initializes the entire array to zeros.
infobahn is right, gooch is wrong.
To clarify that: C99, 6.7.8#20 (Initializers)
"If there are fewer initializers in a brace-enclosed list than there are
elements or members of an aggregate, or fewer characters in a string
literal used to initialize an array of known size than there are
elements in the array, the remainder of the aggregate shall be
initialized implicitly the same as objects that have static storage
duration."
That is, the second behaviour gooch was talking about (rest initialised
to zero) is always what we can expect.
Cheers
Michael
--
E-Mail: Mine is a gmx dot de address. infobahn is right, gooch is wrong.
-- pete
I stand corrected. I seem to remember reading this in the standard at
one point but I went to my document after reading your reply and there
it is just as you say.
"Michael Mair" <Mi**********@i nvalid.invalid> wrote in message
news:36******** *****@individua l.net... char teststr[100] = {0};
This one does not do the same thing. This one initializes the entire array to zeros. infobahn is right, gooch is wrong.
To clarify that: C99, 6.7.8#20 (Initializers) "If there are fewer initializers in a brace-enclosed list than there
are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration."
That is, the second behaviour gooch was talking about (rest
initialised to zero) is always what we can expect.
What does this do:
char teststr[100] = {1,2,3};
Does teststr now have 1, 2, 3 then 97 zeros? Or 1, 2, and 98 threes? Or
what?
--
Mabden
In article <RZ************ ***@newssvr13.n ews.prodigy.com >,
Mabden <mabden@sbc_glo bal.net> wrote: "Michael Mair" <Mi**********@i nvalid.invalid> wrote in message news:36******** *****@individua l.net...
That is, the second behaviour gooch was talking about (rest initialised to zero) is always what we can expect.
What does this do:
char teststr[100] = {1,2,3};
Does teststr now have 1, 2, 3 then 97 zeros? Or 1, 2, and 98 threes? Or what?
Its 1, 2, 3 then 97 zeros. It uses the values provided and then fills the
rest of the array will zeros.
Kevin
Mabden wrote: "Michael Mair" <Mi**********@i nvalid.invalid> wrote in message news:36******** *****@individua l.net...
>char teststr[100] = {0};
This one does not do the same thing. This one initializes the entire array to zeros.
infobahn is right, gooch is wrong.
To clarify that: C99, 6.7.8#20 (Initializers) "If there are fewer initializers in a brace-enclosed list than there
are
elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration."
That is, the second behaviour gooch was talking about (rest
initialised
to zero) is always what we can expect.
What does this do:
char teststr[100] = {1,2,3};
Does teststr now have 1, 2, 3 then 97 zeros? Or 1, 2, and 98 threes? Or what?
Think hard, mock elsewhere.
Hint: Integer variables of static storage duration without an explicit
initialiser are initialised to 0, for arrays this applies elementwise.
-Michael
--
E-Mail: Mine is a gmx dot de address. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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