i need help about Turbo C

Discussion in 'C Programming' started by pipito, Jul 17, 2006.

  1. pipito

    pipito Guest

    Hi...i am a beginner in C programming...
    my question is like this...
    im using the scanf in C to accept input data in output area,

    printf("name: ");scanf("%s",name);
    printf(Address: ");scanf("%s",address);
    printf("age: ");scanf("%d",age);

    when i run the program, i use a spacebar in the input data, the
    following string after i press the space bar doesnt read anymore or it
    skip and going to the next line

    name: pipito otso
    address: manila philippines
    age: 18

    i want to input like that one

    but it appear like this one

    name: pipito otso
    address:
    age: ------<<<<<the cursor was in this area...but i want to put data in
    address:

    what are the commands or the best way in order to make it

    pls help me experts....

    pipito....
    pipito, Jul 17, 2006
    #1
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  2. pipito

    Tom St Denis Guest

    pipito wrote:
    > Hi...i am a beginner in C programming...
    > my question is like this...
    > im using the scanf in C to accept input data in output area,
    >
    > printf("name: ");scanf("%s",name);


    You need to add a "\n" or the printf won't always show. The stream is
    only guaranteed to flush if you send a newline.

    Also your scanf doesn't eat the newline [that the user enters]. So
    when you get to the second scanf() you get just the newline.

    You need to flush the input stream [hint: don't use fflush on it].

    You may also want to try and use fgets() instead of scanf [hint:
    buffer overflows].

    And don't spell things with TXT'ing grammar. It's annoying and makes
    you look like a small retarded child who should be wearing a helmet.

    Tom
    Tom St Denis, Jul 17, 2006
    #2
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  3. pipito

    pemo Guest

    Tom St Denis wrote:
    > pipito wrote:
    >> Hi...i am a beginner in C programming...
    >> my question is like this...
    >> im using the scanf in C to accept input data in output area,
    >>
    >> printf("name: ");scanf("%s",name);

    >
    > You need to add a "\n" or the printf won't always show. The stream is
    > only guaranteed to flush if you send a newline.
    >
    > Also your scanf doesn't eat the newline [that the user enters]. So
    > when you get to the second scanf() you get just the newline.
    >
    > You need to flush the input stream [hint: don't use fflush on it].
    >
    > You may also want to try and use fgets() instead of scanf [hint:
    > buffer overflows].
    >
    > And don't spell things with TXT'ing grammar. It's annoying and makes
    > you look like a small retarded child who should be wearing a helmet.


    What, the OP used 'plse' - I think we ought to cut him some slack and save
    the 'panning' for those that *really* use /baby speak/.

    --
    ==============
    Not a pedant
    ==============
    pemo, Jul 17, 2006
    #3
  4. On 17 Jul 2006 06:12:43 -0700, "pipito" <> wrote:

    >Hi...i am a beginner in C programming...
    >my question is like this...
    >im using the scanf in C to accept input data in output area,
    >
    >printf("name: ");scanf("%s",name);
    >printf(Address: ");scanf("%s",address);
    >printf("age: ");scanf("%d",age);
    >
    >when i run the program, i use a spacebar in the input data, the
    >following string after i press the space bar doesnt read anymore or it
    >skip and going to the next line
    >
    >name: pipito otso


    The %s in your call to scanf stops at the first white space (in this
    case, the space between pipito and otso). However, the extra text
    (otso) remains in the buffer.

    Your next call to scanf will read the otso into address and stop at
    the '\n' which resulted from the enter key

    Your third call (the %d) will ignore the '\n', realize the buffer is
    empty, and wait for your input, which is now out of step with what you
    intended.

    >address: manila philippines
    >age: 18
    >
    >i want to input like that one
    >
    >but it appear like this one
    >
    >name: pipito otso
    >address:
    >age: ------<<<<<the cursor was in this area...but i want to put data in
    >address:
    >
    >what are the commands or the best way in order to make it
    >


    You should not use scanf with a %s in general (it can lead to buffer
    overflows) and you cannot use it if you want your input to contain
    spaces. I suggest you check the fgets function in your reference
    material. It solves both the overflow and space problems.


    Remove del for email
    Barry Schwarz, Jul 17, 2006
    #4
  5. Tom St Denis wrote:
    > You need to add a "\n" or the printf won't always show. The stream is
    > only guaranteed to flush if you send a newline.
    >
    > Also your scanf doesn't eat the newline [that the user enters]. So
    > when you get to the second scanf() you get just the newline.


    At the second time, every input function will be fed with the newline
    left by the previous scanf.

    > You need to flush the input stream [hint: don't use fflush on it].
    >
    > You may also want to try and use fgets() instead of scanf [hint:
    > buffer overflows].


    What you said are really good points. Someone, e.g. Keith Thompson,
    recommended to use scanf for numbers. Do you agree?

    I write a trivial program, it maybe have many flaws. Could you comment
    on it..

    #include <stdio.h>
    #include <stdlib.h>
    #include <string.h>

    enum gender_e{
    male, /*0 for male*/
    female /*1 for female*/
    };

    struct student{
    char *name;
    enum gender_e gender;
    int age;
    char *address;
    };

    void disc(void);
    void delnl(char *s);

    #define STRLEN 200

    int main(void){
    struct student s1;
    int gender = 0;

    s1.name = malloc(STRLEN * sizeof *s1.name);
    s1.address = malloc(STRLEN * sizeof *s1.address);

    printf("Student Name: ");
    fgets(s1.name, STRLEN, stdin);
    delnl(s1.name);
    printf("Gender (0 for male; 1 for female): ");
    scanf("%d", &gender);
    s1.gender = (enum gender_e)gender;
    disc();
    printf("Age: ");
    scanf("%d", &s1.age);
    disc();
    printf("Address: ");
    fgets(s1.address, STRLEN, stdin);
    delnl(s1.address);
    printf("\n---\nstudent: %s, %s, %d, %s\n---\n",
    s1.name, s1.gender?"male":"female", s1.age, s1.address);
    free(s1.name);
    free(s1.address);

    return 0;
    }

    /*discard characters i.e. newline in stdin left by scanf*/
    void disc(void){
    int c;
    while ((c = getchar()) != '\n' && c != EOF)
    ;
    }

    /*delete the newline read into memory by fgets*/
    void delnl(char *s){
    char *p;
    if(p = strrchr(s, '\n'))
    *p = '\0';
    }

    $ cc -ansi -pedantic -Wall -W inputoutput.c
    inputoutput.c: In function `delnl':
    inputoutput.c:60: warning: suggest parentheses around assignment used
    as truth value
    $ ./a.out
    Student Name: dennis
    Gender (0 for male; 1 for female): 1
    Age: 30
    Address: bell

    ---
    student: dennis, male, 30, bell
    ---
    $

    lovecreatesbeauty
    lovecreatesbeauty, Jul 17, 2006
    #5
  6. lovecreatesbeauty wrote:
    > printf("\n---\nstudent: %s, %s, %d, %s\n---\n",
    > s1.name, s1.gender?"male":"female", s1.age, s1.address);


    sorry, this should be:
    s1.name, s1.gender?"female":"male", s1.age, s1.address);

    it might correct the errors in output.

    > $ ./a.out
    > Student Name: dennis
    > Gender (0 for male; 1 for female): 1


    Gender (0 for male; 1 for female): 0

    > Age: 30
    > Address: bell
    >
    > ---
    > student: dennis, male, 30, bell
    > ---
    lovecreatesbeauty, Jul 17, 2006
    #6
  7. pipito

    sarathy Guest

    pipito wrote:
    > Hi...i am a beginner in C programming...
    > my question is like this...
    > im using the scanf in C to accept input data in output area,
    >
    > printf("name: ");scanf("%s",name);
    > printf(Address: ");scanf("%s",address);
    > printf("age: ");scanf("%d",age);
    >
    > when i run the program, i use a spacebar in the input data, the
    > following string after i press the space bar doesnt read anymore or it
    > skip and going to the next line
    >
    > name: pipito otso
    > address: manila philippines
    > age: 18
    >


    use
    scanf ("%[^\n]",name);
    scanf ("%[^\n]",address);

    Now your name and address can contain any number of spaces..
    sarathy, Jul 17, 2006
    #7
  8. pipito

    Simon Biber Guest

    sarathy wrote:
    > use
    > scanf ("%[^\n]",name);
    > scanf ("%[^\n]",address);
    >
    > Now your name and address can contain any number of spaces..


    %[^\n] is good advice, but you should also include a number to limit the
    number of characters it will attempt to read and put in the array.

    Otherwise if the user inputs a line that is too long, it will overflow
    the array and cause undefined behaviour - something bad might happen.

    For example:

    char name[128];
    scanf("%127[^\n]", name);

    Note how the number is one less than the size of the array, to allow for
    the null character to be stored.

    --
    Simon.
    Simon Biber, Jul 17, 2006
    #8
  9. sarathy wrote:

    > pipito wrote:
    > > Hi...i am a beginner in C programming...
    > > my question is like this...
    > > im using the scanf in C to accept input data in output area,
    > >
    > > printf("name: ");scanf("%s",name);
    > > printf(Address: ");scanf("%s",address);
    > > printf("age: ");scanf("%d",age);
    > >
    > > when i run the program, i use a spacebar in the input data, the
    > > following string after i press the space bar doesnt read anymore or it
    > > skip and going to the next line
    > >
    > > name: pipito otso
    > > address: manila philippines
    > > age: 18
    > >

    >
    >> Hi Pipito, This is manmeet Mittal, U can also write the code as follows :-
    >> printf("\n Name"); scanf("%s",name);
    >> printf("\n Address"); scanf("%s",address);
    >> printf("\n Age"); scanf("%d",&age);
    >> ' \n ' is an escape sequence which causes the output in new line. In input of age u can >> use ' & ' ampersand. U can also write code to input a string as follows :-
    >> puts("\n Name"); gets(name);


    >> I think it will serve your purpose. If u have any query then send ur queris.


    >> Name :-Kumar Gaurav
    >> address :-India
    >> age 19
    Manmeet Mittal, Jul 17, 2006
    #9
  10. pipito

    Simon Biber Guest

    Manmeet Mittal wrote:
    >>> Hi Pipito, This is manmeet Mittal, U can also write the code as follows :-
    > >> printf("\n Name"); scanf("%s",name);


    Bad idea. The %s has no limit to how many characters it will attempt to
    read and place in the name array. It might overflow the array. This kind
    of bug is the cause of most of the security issues in programs around
    the world. It's important to be vigilant and always consider buffer
    overflow, no matter how simple or trivial the program is.

    >>> printf("\n Address"); scanf("%s",address);
    >>> printf("\n Age"); scanf("%d",&age);
    >>> ' \n ' is an escape sequence which causes the output in new line.
    >>> In input of age u can use ' & ' ampersand.


    If age has type int, then you are right, you need the ampersand to give
    the address of age, so that scanf can store into your age variable.

    >>> U can also write code to input a string as follows :-
    >>> puts("\n Name"); gets(name);


    NO!!! NEVER USE gets!!!

    Didn't I just tell you above, how important it is to consider buffer
    overflows? The gets function is bad, evil and wrong. It is unsafe, and
    it cannot be made safe. Do you see why?

    One alternative is to use fgets:

    fgets(name, sizeof name, stdin);
    char *p = strchr(name, '\n');
    if(p != NULL) *p = 0;

    The two extra lines are to find and remove the newline character from
    the end of the string, since fgets doesn't do that for you.

    >>> I think it will serve your purpose. If u have any query then send ur queris.

    >
    >>> Name :-Kumar Gaurav
    >>> address :-India
    >>> age 19


    A man of your age should know better than to use juvenile abbreviations
    like ā€˜uā€™ when talking to adults.

    --
    Simon.

    (for fun, ...)
    Name :- Simon Biber
    Address :- Sydney Australia
    Age :- 23
    Simon Biber, Jul 17, 2006
    #10
  11. "lovecreatesbeauty" <> writes:
    > Tom St Denis wrote:
    >> You need to add a "\n" or the printf won't always show. The stream is
    >> only guaranteed to flush if you send a newline.
    >>
    >> Also your scanf doesn't eat the newline [that the user enters]. So
    >> when you get to the second scanf() you get just the newline.

    >
    > At the second time, every input function will be fed with the newline
    > left by the previous scanf.
    >
    >> You need to flush the input stream [hint: don't use fflush on it].
    >>
    >> You may also want to try and use fgets() instead of scanf [hint:
    >> buffer overflows].

    >
    > What you said are really good points. Someone, e.g. Keith Thompson,
    > recommended to use scanf for numbers. Do you agree?


    Perhaps you have me confused with someone else.

    Sure, you can use scanf for numbers. Doing so doesn't present the
    same overflow possibilities as an unbounded scanf("%s"). But I
    usually recommend using fgets() to read a line at a time, then
    sscanf() to parse the line once it's been read. (fgets() has its
    problems, but they're easier to work around than scanf()'s problems.)

    --
    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, Jul 17, 2006
    #11
  12. "Tom St Denis" <> writes:
    > pipito wrote:
    >> Hi...i am a beginner in C programming...
    >> my question is like this...
    >> im using the scanf in C to accept input data in output area,
    >>
    >> printf("name: ");scanf("%s",name);

    >
    > You need to add a "\n" or the printf won't always show. The stream is
    > only guaranteed to flush if you send a newline.

    [...]

    Another approach is to use fflush:

    printf("name: ");
    fflush(stdout);
    fgets(name, sizeof name, stdin);

    This isn't absolutely guaranteed to work, but it should work on any
    system where this kind of interactive prompting makes sense.

    --
    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, Jul 17, 2006
    #12
  13. In article <>, Keith Thompson <> writes:
    >
    > Sure, you can use scanf for numbers. Doing so doesn't present the
    > same overflow possibilities as an unbounded scanf("%s").


    But it does present the opportunity for UB (per C99 7.19.6.2 #10).

    And while this UB is in practice somewhat less frequently an
    exploitable defect than character buffer overflows, it often enough
    *is* exploitable. (Integer overflow exploits are well-understood
    and amply documented.)

    Unfortunately, unless an implementation makes stronger guarantees for
    the formatted-input functions than are required by the Standard,
    they're all unsafe. For nontrivial applications it's better to
    gather text with a safe function and parse it using strtol, etc.

    fscanf is hardly a high-performance function; it's a pity the
    committee didn't elect to err on the side of safety, and require at
    least unspecified but safe behavior for numeric conversions.

    --
    Michael Wojcik

    The penance was not building the field and bringing back Shoeless Joe
    Jackson, but rather tossing on the field with his father. -- Kevin Aug
    Michael Wojcik, Jul 18, 2006
    #13
  14. (Michael Wojcik) writes:
    > In article <>, Keith Thompson
    > <> writes:
    >>
    >> Sure, you can use scanf for numbers. Doing so doesn't present the
    >> same overflow possibilities as an unbounded scanf("%s").

    >
    > But it does present the opportunity for UB (per C99 7.19.6.2 #10).


    Oof, I hadn't realized that.

    For those without a copy of the standard handy:

    [...]
    If this object does not have an appropriate type, or if the result
    of the conversion cannot be represented in the object, the
    behavior is undefined.

    For example this:

    int i;
    sscanf("9999999999999999999999999999", "%d", &i);

    invokes undefined behavior (assuming 9999999999999999999999999999
    overflows type int).

    Ick.

    --
    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, Jul 18, 2006
    #14
  15. Keith Thompson wrote:
    > Perhaps you have me confused with someone else.
    >
    > Sure, you can use scanf for numbers. Doing so doesn't present the
    > same overflow possibilities as an unbounded scanf("%s"). But I
    > usually recommend using fgets() to read a line at a time, then
    > sscanf() to parse the line once it's been read. (fgets() has its
    > problems, but they're easier to work around than scanf()'s problems.)


    Oh, did I? Sorry if I did.

    Your next post shows that even sscanf is not safe. Does that `undefined
    behavior' (caused by *scanf) mean the control flow won't reach line 9
    in the following code?

    #include <stdio.h>

    int main(void){
    int age = 0, c;
    while (age < 1 || age > 200){
    printf("Your age (1 ~ 200): ");
    fflush(stdout);
    scanf("%d", &age);
    while ((c = getchar()) != '\n' && c != EOF) ; /*line 9*/
    }
    printf("Sdudent: dennis, age: %d, address: bell\n", age);
    }

    $ cc -std=c99 -Wall -W ioexercise.c
    $ ./a.out
    Your age (1 ~ 200): HOWOLDAREYOU?
    Your age (1 ~ 200): 999999999999999999999999999999999999999999999999
    Your age (1 ~ 200): 29
    Sdudent: dennis, age: 29, address: bell
    $
    lovecreatesbeauty, Jul 19, 2006
    #15
  16. "lovecreatesbeauty" <> writes:
    > Keith Thompson wrote:
    >> Perhaps you have me confused with someone else.
    >>
    >> Sure, you can use scanf for numbers. Doing so doesn't present the
    >> same overflow possibilities as an unbounded scanf("%s"). But I
    >> usually recommend using fgets() to read a line at a time, then
    >> sscanf() to parse the line once it's been read. (fgets() has its
    >> problems, but they're easier to work around than scanf()'s problems.)

    >
    > Oh, did I? Sorry if I did.


    S'awright.

    > Your next post shows that even sscanf is not safe. Does that `undefined
    > behavior' (caused by *scanf) mean the control flow won't reach line 9
    > in the following code?

    [snip]

    Um, undefined behavior means the behavior is undefined.

    --
    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, Jul 19, 2006
    #16
  17. pipito

    av Guest

    On 18 Jul 2006 21:20:56 -0700, "lovecreatesbeauty"
    <> wrote:
    >Keith Thompson wrote:
    >> Perhaps you have me confused with someone else.
    >>
    >> Sure, you can use scanf for numbers. Doing so doesn't present the
    >> same overflow possibilities as an unbounded scanf("%s"). But I
    >> usually recommend using fgets() to read a line at a time, then
    >> sscanf() to parse the line once it's been read. (fgets() has its
    >> problems, but they're easier to work around than scanf()'s problems.)

    >
    >Oh, did I? Sorry if I did.
    >
    >Your next post shows that even sscanf is not safe. Does that `undefined
    >behavior' (caused by *scanf) mean the control flow won't reach line 9
    >in the following code?
    >
    >#include <stdio.h>
    >
    >int main(void){
    > int age = 0, c;
    > while (age < 1 || age > 200){
    > printf("Your age (1 ~ 200): ");
    > fflush(stdout);
    > scanf("%d", &age);
    > while ((c = getchar()) != '\n' && c != EOF) ; /*line 9*/
    > }
    > printf("Sdudent: dennis, age: %d, address: bell\n", age);
    >}
    >
    >$ cc -std=c99 -Wall -W ioexercise.c
    >$ ./a.out
    >Your age (1 ~ 200): HOWOLDAREYOU?
    >Your age (1 ~ 200): 999999999999999999999999999999999999999999999999


    my little function for get integers unsigned like above return UNS_MAX
    set a flag for overflow and read the number until the last "9"
    save the char after 9 ungetc it

    >Your age (1 ~ 200): 29
    >Sdudent: dennis, age: 29, address: bell
    >$
    av, Jul 19, 2006
    #17
  18. In article <>, "lovecreatesbeauty" <> writes:
    >
    > Your next post shows that even sscanf is not safe. Does that `undefined
    > behavior' (caused by *scanf) mean the control flow won't reach line 9
    > in the following code?


    It's *undefined*, which means it says nothing about the behavior after
    that point.

    It could cause Scott Nudds to come flying out a demon's nose and
    reformat the hard drive of your AS/400.[1]

    scanf could refuse to convert the input data at that point in the
    stream and return a value indicating it had only performed the
    specified conversions up to that point in the format string. (Note
    that scanf returns the number of conversions it successfully
    performed.)

    Or it could do anything else. Since the behavior is undefined, it's
    purely a matter of Quality of Implementation (QoI).

    IMO, the highest-quality result would be for scanf to refuse to
    convert and to return the proper return code at that point.

    However, a naive implementation could perform, say, %d conversions by
    reading characters into a buffer until it reached a character that
    wasn't part of a valid sequence for a %d conversion; and it could use
    a fixed-sized buffer for that purpose; and it could overflow that
    buffer if too many digits were supplied. That would be lousy QoI,
    but it wouldn't violate the standard.

    A good implementor wouldn't commit such an obvious error, but there
    are implementors who aren't all that good, and even the best
    implementors make mistakes once in a while.[2]

    Now, for practical C programming, you have to assume a certain QoI
    for the implementation(s) you will use; and most C programmers will
    also limit themselves to implementations that they know will adhere
    to certain behaviors not required by the standard. So it is often
    acceptable to say "this C program requires that, in addition to the
    requirements of the standard, the implementation behave reasonably
    when performing conversions of numeric data using the scanf family".

    Generally speaking, vendors who care about conformance to the
    standard also care about QoI; or to put it another way, if a vendor
    refuses to produce a quality implementation, I'd be suspicious of
    their conformance to the standard anyway.

    Just be aware that if you disregard the possibility of adverse
    behavior from scanf's numeric conversions, you are imposing
    additional restrictions on the implementations you use. Many of
    us do that every day, as professional C programmers, because
    perfect vigilance is too expensive; but it is a trade-off.


    [1] I think that covers all the standard c.l.c metaphors.

    [2] I'll bet that even Chris Torek and P. J. Plauger have made at
    least one mistake each.

    --
    Michael Wojcik

    Thanatos, thanatos! The labourer, dropping his lever,
    Hides a black letter close to his heart and goes,
    Thanatos, thanatos, home for the day and for ever. -- George Barker
    Michael Wojcik, Jul 20, 2006
    #18
  19. On 19 Jul 2006 23:22:54 GMT, (Michael Wojcik)
    wrote:

    >
    > In article <>, "lovecreatesbeauty" <> writes:
    > >
    > > Your next post shows that even sscanf is not safe. Does that `undefined
    > > behavior' (caused by *scanf) mean the control flow won't reach line 9
    > > in the following code?

    >
    > It's *undefined*, which means it says nothing about the behavior after
    > that point.

    <snip>
    > However, a naive implementation could perform, say, %d conversions by
    > reading characters into a buffer until it reached a character that
    > wasn't part of a valid sequence for a %d conversion; and it could use
    > a fixed-sized buffer for that purpose; and it could overflow that
    > buffer if too many digits were supplied. That would be lousy QoI,
    > but it wouldn't violate the standard.
    >

    It would if it failed for e.g. -0000000000000000000000000000012345 .
    The UB is allowed only for (converted) value out of range.

    You _could_ correctly parse (or buffer?) whitespace, sign, and leading
    zeros and then inadequately buffer other digits -- but IMNVHO that
    goes beyond naive into just silly. Or perverse, of course.

    > A good implementor wouldn't commit such an obvious error, but there
    > are implementors who aren't all that good, and even the best
    > implementors make mistakes once in a while.[2]
    >
    > Now, for practical C programming, you have to assume a certain QoI
    > for the implementation(s) you will use; <snip>
    > Generally speaking, vendors who care about conformance to the
    > standard also care about QoI; or to put it another way, if a vendor
    > refuses to produce a quality implementation, I'd be suspicious of
    > their conformance to the standard anyway.
    >

    And even if it somehow does conform I wouldn't want to use it.

    > Just be aware that if you disregard the possibility of adverse
    > behavior from scanf's numeric conversions, you are imposing
    > additional restrictions on the implementations you use. Many of
    > us do that every day, as professional C programmers, because
    > perfect vigilance is too expensive; but it is a trade-off.
    >

    Realistically, you can easily write conversion code that traps or
    returns garbage for overflow on some machines. I can, just barely,
    imagine a very limited machine, about the 8080 or 6502 or PDP-8 level,
    where multiplication must be done 'by hand' and the (most) available
    primitive for multiplication (by 10?) cleverly(?) handles both word
    and more-than-word (for some choice of word) in such a way that
    unexpected and unallowed-for overflow clobbers something vital.

    > [2] I'll bet that even Chris Torek and P. J. Plauger have made at
    > least one mistake each.


    What, you don't think they'd be willing to share? <G>

    - David.Thompson1 at worldnet.att.net
    Dave Thompson, Jul 31, 2006
    #19
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