Using &array with scanf

Discussion in 'C Programming' started by James Daughtry, Jan 5, 2006.

  1. char array[20];
    scanf("%19s", &array);

    I know this is wrong because it's a type mismatch, where scanf expects
    a pointer to char and gets a pointer to an array of 20 char. I know
    that question 6.12 of the C FAQ says that it's wrong for that very
    reason. What I don't know is where the standard tells me conclusively
    that it's wrong. What I also don't know is somewhere that this type
    mismatch will break in practice.

    A peer asked me recently why it was wrong when I told him that it was
    wrong, and I was very uncomfortable because I know it's wrong and I had
    no good answer when he asked me to prove it. So how do I prove
    something this to an exceptionally stubborn programmer who wants to
    have black and white proof as well as a real example that fails?

    Thanks!
    James Daughtry, Jan 5, 2006
    #1
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  2. James Daughtry said:

    > char array[20];
    > scanf("%19s", &array);
    >
    > I know this is wrong because it's a type mismatch, where scanf expects
    > a pointer to char and gets a pointer to an array of 20 char.


    Yup. It's also wrong because it doesn't check the return value of scanf.

    > I know that question 6.12 of the C FAQ says that it's wrong for that very
    > reason.


    Yup.

    > What I don't know is where the standard tells me conclusively that it's
    > wrong.


    The fscanf specification:

    s Matches a sequence of non-white-space characters. The corresponding
    argument shall be a pointer to the initial character of an array large
    enough to accept the sequence and a terminating null character, which
    will be added automatically.

    &array is not a pointer to the initial character of an array large enough to
    accept the sequence; it's a pointer to an entire array. Different type.
    That's a violation of a "shall" outside a constraint, so the behaviour is
    undefined.

    > What I also don't know is somewhere that this type
    > mismatch will break in practice.


    Irrelevant. A conforming implementation which breaks it could be released
    tomorrow.

    > A peer asked me recently why it was wrong when I told him that it was
    > wrong, and I was very uncomfortable because I know it's wrong and I had
    > no good answer when he asked me to prove it.


    Ask him to explain how he can possibly confuse a char (*)[20] and a char *,
    given that they are completely different types with completely different
    sizes.

    --
    Richard Heathfield
    "Usenet is a strange place" - dmr 29/7/1999
    http://www.cpax.org.uk
    email: rjh at above domain (but drop the www, obviously)
    Richard Heathfield, Jan 5, 2006
    #2
    1. Advertising

  3. Hi,

    At 5 Jan 2006 05:12:32 -0800,
    James Daughtry wrote:
    >
    > char array[20];
    > scanf("%19s", &array);
    >
    > A peer asked me recently why it was wrong when I told him that it was
    > wrong, and I was very uncomfortable because I know it's wrong and I had
    > no good answer when he asked me to prove it.


    &array is a pointer to the pointer to the first element of array, it is
    of type char**, not char*.
    You can use
    scanf ("%19s", array);
    or
    scanf ("%19s", &array[0]);


    Regards,
    Roland
    --
    Roland Csaszar ----------- \\\ /// -------------- +43 316 495 2129
    Software Development ------ \\\ /// ----------- http://www.knapp.com
    KNAPP Logistics Automation - \\V// - mailto:
    Roland Csaszar, Jan 5, 2006
    #3
  4. Richard Heathfield wisely wrote:
    > James Daughtry said:
    >
    > > char array[20];
    > > scanf("%19s", &array);
    > >
    > > I know this is wrong because it's a type mismatch, where scanf expects
    > > a pointer to char and gets a pointer to an array of 20 char.

    >
    > Yup. It's also wrong because it doesn't check the return value of scanf.
    >


    Yea, I didn't want to dilute the example with error checking.
    Fortunately, I have no intention of compiling that snippet, and I don't
    imagine it will do any harm when executed as a Usenet post. ;-)

    > > I know that question 6.12 of the C FAQ says that it's wrong for that very
    > > reason.

    >
    > Yup.
    >
    > > What I don't know is where the standard tells me conclusively that it's
    > > wrong.

    >
    > The fscanf specification:
    >
    > s Matches a sequence of non-white-space characters. The corresponding
    > argument shall be a pointer to the initial character of an array large
    > enough to accept the sequence and a terminating null character, which
    > will be added automatically.
    >
    > &array is not a pointer to the initial character of an array large enough to
    > accept the sequence; it's a pointer to an entire array. Different type.
    > That's a violation of a "shall" outside a constraint, so the behaviour is
    > undefined.
    >


    Ah, now that's my problem. I made a beeline to that very paragraph to
    prove my point, and the result was a quickie program much like the
    following. He was trying to tell me through the output of the program
    that array and &array result in the same address, and the type doesn't
    matter because scanf will treat the same address like a pointer to
    char, and the type mismatch is irrelevant.

    #include <stdio.h>

    int main(void)
    {
    char array[20];

    printf("%p\n%p\n", (void*)&array, (void*)array);

    return 0;
    }

    Like I said, he's a stubborn little bugger.

    > > What I also don't know is somewhere that this type
    > > mismatch will break in practice.

    >
    > Irrelevant. A conforming implementation which breaks it could be released
    > tomorrow.
    >


    Unfortunately, he's the kind of person who uses the "it works for me"
    argument. I know he's wrong, you know he's wrong, but he refuses to
    admit that he's wrong until I can write a program that proves him
    wrong. :p

    > > A peer asked me recently why it was wrong when I told him that it was
    > > wrong, and I was very uncomfortable because I know it's wrong and I had
    > > no good answer when he asked me to prove it.

    >
    > Ask him to explain how he can possibly confuse a char (*)[20] and a char *,
    > given that they are completely different types with completely different
    > sizes.
    >


    I asking him almost the same question. I asked why it should work when
    pointers of different types aren't required to have the same
    representation even if they point to the same address, adding emphasis
    by pointing out the relevant parts of the standard. Holding his ground,
    he ran the program again and said that the addresses are the same, then
    ran an incorrect scanf example to prove that it worked the way he
    expected, and repeated that scanf will do an implicit conversion
    internally.

    > --
    > Richard Heathfield
    > "Usenet is a strange place" - dmr 29/7/1999
    > http://www.cpax.org.uk
    > email: rjh at above domain (but drop the www, obviously)
    James Daughtry, Jan 5, 2006
    #4
  5. Roland Csaszar wrote:
    > &array is a pointer to the pointer to the first element of array, it is
    > of type char**, not char*.


    Actually, it's of type char (*)[20]. The address-of operation is one of
    the three cases where an array name isn't converted to a pointer to the
    first element of the array.
    James Daughtry, Jan 5, 2006
    #5
  6. James Daughtry

    kaikai Guest

    "James Daughtry" <>
    ??????:...
    > char array[20];
    > scanf("%19s", &array);
    >
    > I know this is wrong because it's a type mismatch, where scanf expects
    > a pointer to char and gets a pointer to an array of 20 char. I know
    > that question 6.12 of the C FAQ says that it's wrong for that very
    > reason. What I don't know is where the standard tells me conclusively
    > that it's wrong. What I also don't know is somewhere that this type
    > mismatch will break in practice.
    >
    > A peer asked me recently why it was wrong when I told him that it was
    > wrong, and I was very uncomfortable because I know it's wrong and I had
    > no good answer when he asked me to prove it. So how do I prove
    > something this to an exceptionally stubborn programmer who wants to
    > have black and white proof as well as a real example that fails?
    >
    > Thanks!
    >


    scanf does not (also, could not) read arguments by their real type, that why
    it
    need a format string. All arguments will be reinterpreted from its value.
    Since
    array and &array do have the same value, the scanf function will produce
    same
    results. It is okay to write such code, but you'd better remember that it is
    a
    hack way.

    kaikai
    kaikai, Jan 5, 2006
    #6
  7. "James Daughtry" <> wrote in message
    news:...
    > Richard Heathfield wisely wrote:
    >> > A peer asked me recently why it was wrong when I told him that it was
    >> > wrong, and I was very uncomfortable because I know it's wrong and I had
    >> > no good answer when he asked me to prove it.

    >>
    >> Ask him to explain how he can possibly confuse a char (*)[20] and a char
    >> *,
    >> given that they are completely different types with completely different
    >> sizes.
    >>

    >
    > I asking him almost the same question. I asked why it should work when
    > pointers of different types aren't required to have the same
    > representation even if they point to the same address, adding emphasis
    > by pointing out the relevant parts of the standard. Holding his ground,
    > he ran the program again and said that the addresses are the same, then
    > ran an incorrect scanf example to prove that it worked the way he
    > expected, and repeated that scanf will do an implicit conversion
    > internally.
    >


    You can't prove something beyond proving it :) The simple fact that the
    standard renders the behavior undefined is certainly proof enough that it's
    plain wrong. As far as the C language is concerned, there is no "work" and
    "not work", or "correct" and "incorrect" once you enter the domain of UB.
    Anything at all can happen, including something that someone might think is
    "correct" (but, as I said before, there is no "correct", so he's wrong again
    :)

    There really is no point in trying to go beyond a simple quote to the
    relevant paragraphs of the standard.

    --
    Ivan Budiselic
    ICQ# 104044323
    IRC: buda @ #gamer.hr@quakenet
    remove 'remove' for reply
    Ivan Budiselic, Jan 5, 2006
    #7
  8. James Daughtry said:

    > Richard Heathfield wisely wrote:
    >>
    >> &array is not a pointer to the initial character of an array large enough
    >> to accept the sequence; it's a pointer to an entire array. Different
    >> type. That's a violation of a "shall" outside a constraint, so the
    >> behaviour is undefined.
    >>

    >
    > Ah, now that's my problem. I made a beeline to that very paragraph to
    > prove my point, and the result was a quickie program much like the
    > following. He was trying to tell me through the output of the program
    > that array and &array result in the same address,


    But they don't. What they result in is [SFX - takes huge breath] a sequence
    of printable characters which, were it read back into scanf using a %p
    format specifier, would allow the retrieval of a pointer value which would
    refer to the same object as originally pointed to by the pointer value
    passed to printf.

    > and the type doesn't
    > matter because scanf will treat the same address like a pointer to
    > char,


    The Standard does not say this will happen, so what makes your friend so
    sure?

    > and the type mismatch is irrelevant.


    On the contrary, the type mismatch means the behaviour is undefined.

    > #include <stdio.h>
    >
    > int main(void)
    > {
    > char array[20];
    >
    > printf("%p\n%p\n", (void*)&array, (void*)array);


    This program doesn't actually demonstrate anything useful.

    > Unfortunately, he's the kind of person who uses the "it works for me"
    > argument.


    Well, there's a certain amount to be said for such an argument! But I fail
    to see what it gains him. He acknowledges that scanf requires a char *, and
    he knows he's passing a char (*)[20] instead. He knows that omitting the &
    is a simple enough operation which will make the code squeaky-clean, and
    which is quicker to type than the wrong version. So he must have some very
    powerful motivation for typing that &. Perhaps you would do better to ask
    him what the & wins that pays for the type-wrongness of the code.

    Please note that the "it means I don't have to worry about whether to put an
    & on the front" is not a good-enough reason, because there are plenty of
    cases where it matters a lot whether char * or char (*)[] is supplied (not
    least when we start messing about with multi-dimensional arrays in argument
    expressions), so he can't just think "always use a &"; as a paradigm it
    doesn't work.

    So - what does being wrong /buy/ him? Let him answer that.

    > he ran the program again and said that the addresses are the same, then
    > ran an incorrect scanf example to prove that it worked the way he
    > expected, and repeated that scanf will do an implicit conversion
    > internally.


    What a trusting soul he is. Does he really think undefined behaviour will
    manifest its nastiness at the most convenient moment for /him/? Hah! It
    lurks. It waits in the background. It bides its time. And, when your boss
    is proudly looking on as you demo the code to your best customer...

    --
    Richard Heathfield
    "Usenet is a strange place" - dmr 29/7/1999
    http://www.cpax.org.uk
    email: rjh at above domain (but drop the www, obviously)
    Richard Heathfield, Jan 5, 2006
    #8
  9. kaikai said:

    > Since
    > array and &array do have the same value,


    But they don't. The values have different types, so how can they be the
    same?

    Are $3.14 and 3.14kg the same? Of course not.
    Are 3.14% and 3.14km the same? Of course not.

    The type /matters/.

    > It is okay to write such code,


    No, it isn't.

    --
    Richard Heathfield
    "Usenet is a strange place" - dmr 29/7/1999
    http://www.cpax.org.uk
    email: rjh at above domain (but drop the www, obviously)
    Richard Heathfield, Jan 5, 2006
    #9
  10. Richard Heathfield wrote:
    > James Daughtry said:
    >
    > > Richard Heathfield wisely wrote:
    > >>
    > >> &array is not a pointer to the initial character of an array large enough
    > >> to accept the sequence; it's a pointer to an entire array. Different
    > >> type. That's a violation of a "shall" outside a constraint, so the
    > >> behaviour is undefined.
    > >>

    > >
    > > Ah, now that's my problem. I made a beeline to that very paragraph to
    > > prove my point, and the result was a quickie program much like the
    > > following. He was trying to tell me through the output of the program
    > > that array and &array result in the same address,

    >
    > But they don't. What they result in is [SFX - takes huge breath] a sequence
    > of printable characters which, were it read back into scanf using a %p
    > format specifier, would allow the retrieval of a pointer value which would
    > refer to the same object as originally pointed to by the pointer value
    > passed to printf.
    >


    :-D

    > > and the type doesn't
    > > matter because scanf will treat the same address like a pointer to
    > > char,

    >
    > The Standard does not say this will happen, so what makes your friend so
    > sure?
    >


    Good question.

    > > and the type mismatch is irrelevant.

    >
    > On the contrary, the type mismatch means the behaviour is undefined.
    >


    To some people, "undefined" is just a word, and not even a scary word.
    I'm sure that one day he'll get burned by UB and start to understand.
    But until then, I have to work with the guy and suffer his "it works
    for me" attitude.

    > > #include <stdio.h>
    > >
    > > int main(void)
    > > {
    > > char array[20];
    > >
    > > printf("%p\n%p\n", (void*)&array, (void*)array);

    >
    > This program doesn't actually demonstrate anything useful.
    >


    That's what I thought. Sure, when he ran it the same two values were
    printed, but it in no way proves that a type mismatch doesn't cause
    problems.

    > > Unfortunately, he's the kind of person who uses the "it works for me"
    > > argument.

    >
    > Well, there's a certain amount to be said for such an argument! But I fail
    > to see what it gains him. He acknowledges that scanf requires a char *, and
    > he knows he's passing a char (*)[20] instead. He knows that omitting the &
    > is a simple enough operation which will make the code squeaky-clean, and
    > which is quicker to type than the wrong version. So he must have some very
    > powerful motivation for typing that &. Perhaps you would do better to ask
    > him what the & wins that pays for the type-wrongness of the code.
    >
    > Please note that the "it means I don't have to worry about whether to put an
    > & on the front" is not a good-enough reason, because there are plenty of
    > cases where it matters a lot whether char * or char (*)[] is supplied (not
    > least when we start messing about with multi-dimensional arrays in argument
    > expressions), so he can't just think "always use a &"; as a paradigm it
    > doesn't work.
    >
    > So - what does being wrong /buy/ him? Let him answer that.
    >


    I'll ask, that's a very good question.

    > > he ran the program again and said that the addresses are the same, then
    > > ran an incorrect scanf example to prove that it worked the way he
    > > expected, and repeated that scanf will do an implicit conversion
    > > internally.

    >
    > What a trusting soul he is. Does he really think undefined behaviour will
    > manifest its nastiness at the most convenient moment for /him/? Hah! It
    > lurks. It waits in the background. It bides its time. And, when your boss
    > is proudly looking on as you demo the code to your best customer...
    >


    Indeed. :) Thanks for your comments, they were very helpful.
    James Daughtry, Jan 5, 2006
    #10
  11. James Daughtry

    Mike Wahler Guest

    "James Daughtry" <> wrote in message
    news:...
    > Richard Heathfield wisely wrote:
    >> James Daughtry said:
    >>
    >> > char array[20];
    >> > scanf("%19s", &array);
    >> >
    >> > I know this is wrong because it's a type mismatch, where scanf expects
    >> > a pointer to char and gets a pointer to an array of 20 char.

    >>
    >> Yup. It's also wrong because it doesn't check the return value of scanf.
    >>

    >
    > Yea, I didn't want to dilute the example with error checking.
    > Fortunately, I have no intention of compiling that snippet, and I don't
    > imagine it will do any harm when executed as a Usenet post. ;-)
    >
    >> > I know that question 6.12 of the C FAQ says that it's wrong for that
    >> > very
    >> > reason.

    >>
    >> Yup.
    >>
    >> > What I don't know is where the standard tells me conclusively that it's
    >> > wrong.

    >>
    >> The fscanf specification:
    >>
    >> s Matches a sequence of non-white-space characters. The corresponding
    >> argument shall be a pointer to the initial character of an array large
    >> enough to accept the sequence and a terminating null character, which
    >> will be added automatically.
    >>
    >> &array is not a pointer to the initial character of an array large enough
    >> to
    >> accept the sequence; it's a pointer to an entire array. Different type.
    >> That's a violation of a "shall" outside a constraint, so the behaviour is
    >> undefined.
    >>

    >
    > Ah, now that's my problem. I made a beeline to that very paragraph to
    > prove my point, and the result was a quickie program much like the
    > following. He was trying to tell me through the output of the program


    Someone who tries to prove program correctness (as defined by
    the language standard) by using the behavior of a particular
    implementation as evidence, is imo not really worthy of the
    title 'programmer'.

    > that array and &array result in the same address,


    Yes, the same 'value', but certainly not the same type.
    Also note that while the 'value' might be the same,
    the representation is not required to be the same.

    >and the type doesn't
    > matter



    According to the standard, it *does* matter, very much.

    > because scanf will treat the same address like a pointer to
    > char,


    Nowhere does the language standard make this assertion.

    > and the type mismatch is irrelevant.


    It's *very* relevant, because it results in undefined behavior.

    >
    > #include <stdio.h>
    >
    > int main(void)
    > {
    > char array[20];
    >
    > printf("%p\n%p\n", (void*)&array, (void*)array);


    This is valid because the types of the arguments being passed
    have been converted to types which *are* valid for specifier
    '%p'

    >
    > return 0;
    > }
    >
    > Like I said, he's a stubborn little bugger.


    Taking your description as valid, he seems to be to be
    both ignorant and arrogant. Very dangerous combination.
    I sincerely hope he's not involved in creating critical
    software (e.g. that controlling hospital equipment, etc.).

    >
    >> > What I also don't know is somewhere that this type
    >> > mismatch will break in practice.

    >>
    >> Irrelevant. A conforming implementation which breaks it could be released
    >> tomorrow.
    >>

    >
    > Unfortunately, he's the kind of person who uses the "it works for me"
    > argument.


    And I predict he'll be very surprised when suddenly it
    ceases to work, and he has no idea why.

    >I know he's wrong, you know he's wrong, but he refuses to
    > admit that he's wrong until I can write a program that proves him
    > wrong. :p


    That's a losing battle. No program can prove him right or wrong,
    since it's not implementations that define the language, but the
    ISO standard document (ISO 9899).

    >
    >> > A peer asked me recently why it was wrong when I told him that it was
    >> > wrong, and I was very uncomfortable because I know it's wrong and I had
    >> > no good answer when he asked me to prove it.

    >>
    >> Ask him to explain how he can possibly confuse a char (*)[20] and a char
    >> *,
    >> given that they are completely different types with completely different
    >> sizes.
    >>

    >
    > I asking him almost the same question. I asked why it should work when
    > pointers of different types aren't required to have the same
    > representation even if they point to the same address, adding emphasis
    > by pointing out the relevant parts of the standard. Holding his ground,
    > he ran the program again and said that the addresses are the same,



    He can run it until doomsday, yet that will still prove nothing,
    except perhaps a few things about a particular implementation.

    > then
    > ran an incorrect scanf example to prove that it worked the way he
    > expected,


    That only proves that a particular implementation worked the
    way he expected. It proves nothing at all about the correctness
    of the code.


    > and repeated that scanf will do an implicit conversion
    > internally.


    Ask him to cite from where a guarantee of such a conversion
    comes. Certainly not from ISO 9899.

    -Mike
    Mike Wahler, Jan 5, 2006
    #11
  12. James Daughtry

    Mike Wahler Guest

    "Roland Csaszar" <> wrote in message
    news:87lkxukdrr.wl%...
    > Hi,
    >
    > At 5 Jan 2006 05:12:32 -0800,
    > James Daughtry wrote:
    >>
    >> char array[20];
    >> scanf("%19s", &array);
    >>
    >> A peer asked me recently why it was wrong when I told him that it was
    >> wrong, and I was very uncomfortable because I know it's wrong and I had
    >> no good answer when he asked me to prove it.

    >
    > &array is a pointer to the pointer


    It is not. It is simply a pointer.

    > to the first element of array,


    No. It's a pointer to the entire array.

    > it is
    > of type char**, not char*.


    No. It's type is char(*)[[20] (pointer to array of 20 char).

    > You can use
    > scanf ("%19s", array);
    > or
    > scanf ("%19s", &array[0]);


    Correct, since both expressions 'array' and '&array[0]'
    yield the same type object.

    -Mike
    Mike Wahler, Jan 5, 2006
    #12
  13. "James Daughtry" <> writes:
    > Richard Heathfield wisely wrote:

    [snip]
    >> Ask him to explain how he can possibly confuse a char (*)[20] and a char *,
    >> given that they are completely different types with completely different
    >> sizes.

    >
    > I asking him almost the same question. I asked why it should work when
    > pointers of different types aren't required to have the same
    > representation even if they point to the same address, adding emphasis
    > by pointing out the relevant parts of the standard. Holding his ground,
    > he ran the program again and said that the addresses are the same, then
    > ran an incorrect scanf example to prove that it worked the way he
    > expected, and repeated that scanf will do an implicit conversion
    > internally.


    scanf doesn't do a conversion, implicit or otherwise. It takes
    whatever bits you give it and *assumes* that they constitute a valid
    representation of an object of the correct type. There are
    open-source/freeware implementations of scanf; ask him to show you the
    code that does this "implicit conversion".

    On a machine where hardware addresses point to words, it's plausible
    that a byte pointer (char*) will have a different representation than
    a word pointer (such as int*); a word pointer could be a machine
    address, while a byte pointer could be a machine address plus an
    offset specifying one byte within the word. (The C compilers on Cray
    vector machines almost do this, but they put the offset into the
    otherwise unused high-order bits of the word pointer.)

    He ran the program again *on the same platform* and claimed that he
    had proven that it works correctly. Pick a platform that he doesn't
    have access to and ask him to "prove" that his code works correctly on
    that platform.

    Show him the following program:

    #include <stdio.h>
    int main(void)
    {
    int before[10];
    int array[10];
    int after[10];
    array[-1] = array[10] = 42;
    printf("array[-1] = %d\n", array[-1]);
    printf("array[10] = %d\n", array[10]);
    return 0;
    }

    You and I know that it invokes undefined behavior. On most systems,
    I'd be surprised if the manifestation of that undefined behavior were
    something other than producing the following output:

    array[-1] = 42
    array[10] = 42

    Compile and run the program several times, showing him the consistent
    output each time. Ask him if this "proves" that accessing elements
    outside an array is valid.

    --
    Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
    San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
    We must do something. This is something. Therefore, we must do this.
    Keith Thompson, Jan 5, 2006
    #13
  14. Roland Csaszar <> writes:
    > At 5 Jan 2006 05:12:32 -0800,
    > James Daughtry wrote:
    >>
    >> char array[20];
    >> scanf("%19s", &array);
    >>
    >> A peer asked me recently why it was wrong when I told him that it was
    >> wrong, and I was very uncomfortable because I know it's wrong and I had
    >> no good answer when he asked me to prove it.

    >
    > &array is a pointer to the pointer to the first element of array, it is
    > of type char**, not char*.


    Nope. Time to re-read section 6 of the comp.lang.c FAQ.

    --
    Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
    San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
    We must do something. This is something. Therefore, we must do this.
    Keith Thompson, Jan 5, 2006
    #14
  15. Keith Thompson <> writes:
    [snip]
    > He ran the program again *on the same platform* and claimed that he
    > had proven that it works correctly. Pick a platform that he doesn't
    > have access to and ask him to "prove" that his code works correctly on
    > that platform.


    I suggest the IBM AS/400, especially if he's not familiar with it.
    I'm not very familiar with it myself, but it seems to be a canonical
    example of a system with a conforming C implementation that breaks a
    lot of assumptions if you don't carefully stick to what the standard
    actually guarantees; it's second only to the DS9K.

    --
    Keith Thompson (The_Other_Keith) <http://www.ghoti.net/~kst>
    San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
    We must do something. This is something. Therefore, we must do this.
    Keith Thompson, Jan 5, 2006
    #15
  16. James Daughtry

    Chris Torek Guest

    In article <>
    James Daughtry <> wrote:
    >Ah, now that's my problem. I made a beeline to that very paragraph to
    >prove my point, and the result was a quickie program much like the
    >following. He was trying to tell me through the output of the program
    >that array and &array result in the same address, and the type doesn't
    >matter because scanf will treat the same address like a pointer to
    >char, and the type mismatch is irrelevant.


    The problem with this argument lies in the phrase "the type mismatch
    is irrelevant". That may be the case on *his* machine, but it is
    definitely not the case on *every* machine.

    There are machines (Burroughs A-series) in which integers are stored
    in the same format as floating-point. (The compiler just makes sure
    that the fractional part is always zero.) On such a machine, if you
    do this:

    double x;

    ret = scanf("%d", (int *)&x);
    if (ret == 1)
    printf("I got %f\n", x);

    the code actually works! But it fails on most machines, because the
    type is *not* irrelevant on most machines.

    For a machine on which the type matters, find a 1960s era PR1ME, or
    perhaps even a 1980s Data General MV/10000 (aka Eclipse). (I am not
    sure whether the type "pointer to array N of char" on the Eclipse
    uses a word pointer, but if it does, the &array-with-scanf call will
    in fact fail.)

    >Unfortunately, he's the kind of person who uses the "it works for me"
    >argument. I know he's wrong, you know he's wrong, but he refuses to
    >admit that he's wrong until I can write a program that proves him
    >wrong. :p


    If he is familiar with basketball, try the argument Steve Summit once
    repeated here (I am not sure where he got it):

    Someone told me that, in basketball, you have to bounce the
    ball off the floor; you can't hold the ball and run around the
    court. But I tried it and it works just fine. Obviously he
    does not understand basketball!

    Or, for Boston drivers, remember the rule about driving the wrong
    way on a one-way street ("it's OK as long as you're in reverse!").
    --
    In-Real-Life: Chris Torek, Wind River Systems
    Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
    email: forget about it http://web.torek.net/torek/index.html
    Reading email is like searching for food in the garbage, thanks to spammers.
    Chris Torek, Jan 5, 2006
    #16
  17. James Daughtry

    eerok Guest

    Richard Heathfield wrote:
    > kaikai said:


    >> Since
    >> array and &array do have the same value,


    > But they don't. The values have different types, so how can they be the
    > same?
    >
    > Are $3.14 and 3.14kg the same? Of course not.
    > Are 3.14% and 3.14km the same? Of course not.
    >
    > The type /matters/.


    >> It is okay to write such code,


    > No, it isn't.


    Just to illustrate that it's *not* okay to get sloppy with
    types, here's an example that I discovered the other day.


    #include <stdio.h>

    char *t1 = "This can produce a segfault";
    char t2[] = "Best to do it this way ....";

    int
    main (void)
    {
    int ia = 3, ib = 2;

    /* no apparent problem (on my compiler) when I
    * index into either string */
    printf ("\nt1: '%s', 4th char: '%c'", t1, t1[ia]);
    printf ("\nt2: '%s', 4th char: '%c'\n", t2, t2[ia]);

    /* this is of course fine since we've declared
    * the string is an array of char */
    t2[ib] = t2[ia];
    printf ("\nNo segfault here.\n");

    /* this will produce a segfault with my compiler */
    t1[ib] = t1[ia];
    printf ("\nMaybe you'll get this far, but I don't.\n");

    return 0;
    }


    One might think that "a string is a string is a string" --
    that a pointer to a string is the same as the address of the
    first char of an array of chars. It usually seems to work
    this way, but the example above shows this isn't true.

    --
    "The secret of being boring is to say everything." - Voltaire
    eerok, Jan 5, 2006
    #17
  18. On 5 Jan 2006 05:40:08 -0800, in comp.lang.c , "James Daughtry"
    <> wrote:

    >Unfortunately, he's the kind of person who uses the "it works for me"
    >argument.


    There's little point debating with such people. Give them the
    information, explain exactly once why its incorrect to do it
    otherwise, and walk away. If still they clean their gun while loaded,
    eventually Darwin will come to your rescue.
    Mark McIntyre
    --

    ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
    http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
    ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
    Mark McIntyre, Jan 5, 2006
    #18
  19. On 5 Jan 2006 09:36:23 -0800, in comp.lang.c , "James Daughtry"
    <> wrote:

    >I'm sure that one day he'll get burned by UB and start to understand.
    >But until then, I have to work with the guy and suffer his "it works
    >for me" attitude.


    You could complain to your manager that his code is badly written and
    potentially dangerous, pointing to the sections of the Standard which
    are relevant. Even complete idiot managers tend to rely on
    documentation and will accept that over "it works for me".....

    Mark McIntyre
    --

    ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
    http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
    ----= East and West-Coast Server Farms - Total Privacy via Encryption =----
    Mark McIntyre, Jan 5, 2006
    #19
  20. eerok said:

    > Just to illustrate that it's *not* okay to get sloppy with
    > types, here's an example that I discovered the other day.


    The example you give doesn't fail because of a type issue, but because you
    attempt to write to storage that you're not supposed to write to. Had t1
    been pointing at writeable storage, you'd have had no segfault. For
    example, if you'd done this:

    > #include <stdio.h>
    >
    > char *t1 = "This can produce a segfault";
    > char t2[] = "Best to do it this way ....";


    char s[] = "This won't produce a segfault";
    t1 = s;

    you'd have had no problem, even though the type of t1 hasn't changed.



    --
    Richard Heathfield
    "Usenet is a strange place" - dmr 29/7/1999
    http://www.cpax.org.uk
    email: rjh at above domain (but drop the www, obviously)
    Richard Heathfield, Jan 5, 2006
    #20
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