if string is a number (beginner)

Discussion in 'C Programming' started by fool, Dec 22, 2006.

  1. fool

    fool Guest

    Dear group,
    I want to check if the given string is a number. If any value is
    inputed other than a number then error. The following is not the correct
    solution. Can some one tell me any link for the above task? Pls just
    don't give the code. Thanks.

    #include<stdio.h>
    #include<stdlib.h>
    int main(void)
    {
    char *got;
    got = malloc(sizeof (char));
    if(got == NULL)
    {
    got = malloc(sizeof (char));
    if(got == NULL)
    exit(EXIT_FAILURE);
    }

    scanf("%s",got);
    while( (*got)++ != '\0')
    {
    if(*got != ('0' - '9'))
    {
    printf("error\n");
    break;
    }
    else
    printf("%s\n",got);
    }
    return 0;
    }
    fool, Dec 22, 2006
    #1
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  2. fool

    Default User Guest

    fool wrote:

    > Dear group,
    > I want to check if the given string is a number. If any value is
    > inputed other than a number then error. The following is not the
    > correct solution. Can some one tell me any link for the above task?
    > Pls just don't give the code. Thanks.
    >
    > #include<stdio.h>
    > #include<stdlib.h>
    > int main(void)
    > {
    > char *got;
    > got = malloc(sizeof (char));


    This allocates exactly one character's worth of space. That's pretty
    useless. Why are you even dynamically allocating memory?

    > if(got == NULL)
    > {
    > got = malloc(sizeof (char));


    If it failed before, try again? Why?

    > if(got == NULL)
    > exit(EXIT_FAILURE);
    > }
    >
    > scanf("%s",got);


    Your allocated memory does not enough room to store any string longer
    than the empty string. The scanf() function as used above has no way to
    limit the number of characters entered. If it's longer than the empty
    string, then Undefined Behavior results.

    > while( (*got)++ != '\0')


    You dereferenced the pointer, then incremented the result. I doubt
    that's what you wanted.

    > {
    > if(*got != ('0' - '9'))


    What do you think that minus sign does? Whatever, what it actually does
    is subract the numerical value of the two characters, which will result
    in -9.

    It doesn't matter, because if the loop executed at all (not the empty
    string) then you had UB before, and more now.

    > {
    > printf("error\n");
    > break;
    > }
    > else
    > printf("%s\n",got);
    > }
    > return 0;
    > }



    What book are you using?


    Here's a brief snippet (the comments are for you to think about):

    char buffer[80]; /* magic numbers are bad, how would you fix that? */
    int i;

    fgets(buffer, sizeof buffer, stdin);
    /* what if more than that is entered? */

    for(i = 0; buffer; i++) /* why this instead of the while? */
    {
    if (buffer < '0' || buffer > '9') /* can you decode this? */
    {
    printf("error\n");
    break;
    }
    }


    Brian
    Default User, Dec 22, 2006
    #2
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  3. fool wrote:
    > char *got;
    > got = malloc(sizeof (char));
    > if(got == NULL)
    > {
    > got = malloc(sizeof (char));
    > if(got == NULL)
    > exit(EXIT_FAILURE);
    > }


    This allocation looks strange. You retry without anything changing and
    expect that to then yield a result - it won't happen. If a malloc failure
    can't be recovered from, I'd suggest using xalloc. This is not a library
    function like malloc but a wrapper around it that tries to allocate the
    memory and, in case of failure, writes an error message to stderr and
    exit()s. Using this helps you remove those 6 lines of error-handling code
    that distracts from the real logic of the program.
    Also, you only allocate space for a single character and that size is 1, by
    definition (i.e. the standard says that sizeof (char)==1).

    > scanf("%s",got);


    Bad idea, you are reading an unbounded string into an array of size 1.
    Well, in fact this is a bad idea regardless of how long the array is, you
    should never use unbounded function parameters. You can tell scanf how
    long the target string is. Also, there is fgets() (be sure to read the
    warnings for gets(), which looks more convenient at first!) which only
    does input as a string.

    > while( (*got)++ != '\0')


    This is not what you want. Try this on a loop with a normal character
    literal and step through it with a debugger.

    > if(*got != ('0' - '9'))


    Sorry, but while this looks intuitive, C doesn't let you. The '0' and '9'
    are treated as integers, and the minus sign in between just yields the
    difference between them. What you need is a check that the character is at
    least zero and at most nine. There is also a library function isdigit(),
    but firstly that is difficult to use correctly and secondly it also
    includes the minus and plus signs, IIRC.

    Lastly, in the while() expression, it seems like you already moved to the
    next character. I suggest printing the current character as integer inside
    your loop. That way you will better understand what's going on.

    Uli
    Ulrich Eckhardt, Dec 22, 2006
    #3
  4. fool

    Malcolm Guest

    "fool" <> wrote in message
    news:...
    > Dear group,
    > I want to check if the given string is a number. If any value is
    > inputed other than a number then error. The following is not the correct
    > solution. Can some one tell me any link for the above task? Pls just
    > don't give the code. Thanks.
    >
    > #include<stdio.h>
    > #include<stdlib.h>
    > int main(void)
    > {
    > char *got;
    > got = malloc(sizeof (char));
    > if(got == NULL)
    > {
    > got = malloc(sizeof (char));
    > if(got == NULL)
    > exit(EXIT_FAILURE);
    > }
    >
    > scanf("%s",got);
    > while( (*got)++ != '\0')
    > {
    > if(*got != ('0' - '9'))
    > {
    > printf("error\n");
    > break;
    > }
    > else
    > printf("%s\n",got);
    > }
    > return 0;
    > }


    strtol() for integers, or strtod, for floating-point numbers, are your
    friend.
    Unfortunately for newbies they take a pointer to a pointer to indicate the
    end of the digits that can be interpreted as numbers.
    However the idea is not too complicated. if the end pointer points to 0, the
    end of string character, you have a valid number. If it is a non-digit, you
    might want to throw the number out. If the end pointer is at the beginning
    of the string, there were no digits at the beginning of the string so you
    must throw the input out.
    Watch for whitespace, by the way.
    --
    www.personal.leeds.ac.uk/~bgy1mm
    freeware games to download.
    Malcolm, Dec 22, 2006
    #4
  5. fool

    CBFalconer Guest

    fool wrote:
    >
    > I want to check if the given string is a number. If any value is
    > inputed other than a number then error. The following is not the
    > correct solution. Can some one tell me any link for the above
    > task? Pls just don't give the code. Thanks.


    You don't need any intermediate string. Just getc, together with
    routines to skip blanks and the tests available in ctype.h. If you
    want a demonstration see txtinput.c in txtio.zip, routine readxwd,
    available at:

    <http://cbfalconer.home.att.net/download/>

    Some things in that zip are undergoing revision, but that routine
    is solid. The changes have to do with signed input and arranging
    for readxwd to follow normal unsigned behaviour for overflows. At
    exit readxwd returns the termination char, so you can easily check
    that is an allowable one. In particular if it is a '.' or a 'e'
    (or 'E') you may have found a real value (double or float).

    --
    Chuck F (cbfalconer at maineline dot net)
    Available for consulting/temporary embedded and systems.
    <http://cbfalconer.home.att.net>
    CBFalconer, Dec 22, 2006
    #5
  6. fool

    Scorpio Guest

    fool wrote:
    > Dear group,
    > I want to check if the given string is a number. If any value is
    > inputed other than a number then error. The following is not the correct
    > solution. Can some one tell me any link for the above task?


    Why not just use the function atoi from<stdlib.h> ?
    Scorpio, Dec 22, 2006
    #6
  7. fool

    jaysome Guest

    On 22 Dec 2006 00:58:53 -0800, "Scorpio" <> wrote:

    >
    >fool wrote:
    >> Dear group,
    >> I want to check if the given string is a number. If any value is
    >> inputed other than a number then error. The following is not the correct
    >> solution. Can some one tell me any link for the above task?

    >
    >Why not just use the function atoi from<stdlib.h> ?


    Because with atoi() there is not a guaranteed way to distiguish
    between an integer value of 0 and an invalid integer value--both may
    return 0.

    #include<stdio.h>
    #include<stdlib.h>
    int main(void)
    {
    printf("atoi(\"0\") == %d\n", atoi("0"));
    printf("atoi(\"A\") == %d\n", atoi("A"));
    return 0;
    }

    Others have properly suggested the use of the strto* functions.

    Happy Holidays
    --
    jay
    jaysome, Dec 22, 2006
    #7
  8. fool

    Richard Bos Guest

    "Scorpio" <> wrote:

    > fool wrote:
    > > I want to check if the given string is a number. If any value is
    > > inputed other than a number then error. The following is not the correct
    > > solution. Can some one tell me any link for the above task?

    >
    > Why not just use the function atoi from<stdlib.h> ?


    Because strtol() is better - it doesn't have undefined behaviour on
    overflow.

    Richard
    Richard Bos, Dec 22, 2006
    #8
  9. fool

    CBFalconer Guest

    Richard Bos wrote:
    > "Scorpio" <> wrote:
    >> fool wrote:
    >>
    >>> I want to check if the given string is a number. If any value is
    >>> inputed other than a number then error. The following is not the
    >>> correct solution. Can some one tell me any link for the above task?

    >>
    >> Why not just use the function atoi from<stdlib.h> ?

    >
    > Because strtol() is better - it doesn't have undefined behaviour on
    > overflow.


    Not quite. strtoul won't detect the out-of-range input of "-1",
    for example. It makes the mistake of incorporating the - into the
    number parsing. I want to know if the user has tried to stuff
    something outside the principal range into the variable.

    --
    Chuck F (cbfalconer at maineline dot net)
    Available for consulting/temporary embedded and systems.
    <http://cbfalconer.home.att.net>
    CBFalconer, Dec 22, 2006
    #9
  10. fool

    Chris Torek Guest

    >Richard Bos wrote:
    >>... strtol() is better [than atoi()] - it doesn't have undefined
    >>behaviour on overflow.


    In article <>
    CBFalconer <> wrote:
    >Not quite. strtoul won't detect the out-of-range input of "-1",
    >for example.


    This is not "out of range", by definition. By strtoul()'s definition,
    admittedly. :) If this disagrees with your own definition, it is
    easy enough to prohibit a minus sign.

    If you want to allow '-':

    char buf[N];
    char *ep;
    int base;
    unsigned long result;
    ...
    errno = 0;
    result = strtoul(buf, &ep, base);
    if (result == ULONG_MAX && errno == ERANGE)
    overflow();
    else if (*ep != NULL)
    there_was_trailing_stuff();
    else
    all_is_well();

    To prohibit the '-', augment the strtoul() call with, e.g.:

    if (buf[strspn(buf, " \t\n\r\b\v")] == '-')
    there_is_a_leading_minus();
    else if ((result = strtoul(buf, &ep, base)) == ULONG_MAX && errno == ERANGE)
    ... as before ...

    or you can check for a '-' character after calling strtoul():

    char *minusp;
    ...
    result = strtoul(buf, &ep, base);
    minusp = strchr(buf, '-');
    if (minusp != NULL && (ep == NULL || minusp < ep))
    there_was_a_leading_minus();
    else if (result == ULONG_MAX && errno == ERANGE)
    ... as before ...

    Of course, strtoul() could have been defined to prohibit signs
    (plus and/or minus), and callers could explicitly allow them; or
    it could even have taken flags indicating how to treat signs; but
    this is the situation we have now, so code like the above will
    handle it.
    --
    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, Dec 22, 2006
    #10
  11. fool

    Ben Pfaff Guest

    Chris Torek <> writes:

    > CBFalconer <> wrote:
    >>Not quite. strtoul won't detect the out-of-range input of "-1",
    >>for example.

    >
    > This is not "out of range", by definition. By strtoul()'s definition,
    > admittedly. :) If this disagrees with your own definition, it is
    > easy enough to prohibit a minus sign.


    I suspect it's more difficult outside the C locale.
    --
    "Your correction is 100% correct and 0% helpful. Well done!"
    --Richard Heathfield
    Ben Pfaff, Dec 22, 2006
    #11
  12. fool

    CBFalconer Guest

    Chris Torek wrote:
    > CBFalconer <> wrote:
    >> Richard Bos wrote:
    >>
    >>> ... strtol() is better [than atoi()] - it doesn't have undefined
    >>> behaviour on overflow.

    >>
    >> Not quite. strtoul won't detect the out-of-range input of "-1",
    >> for example.

    >
    > This is not "out of range", by definition. By strtoul()'s
    > definition, admittedly. :) If this disagrees with your own
    > definition, it is easy enough to prohibit a minus sign.
    >

    .... snip ...
    >
    > Of course, strtoul() could have been defined to prohibit signs
    > (plus and/or minus), and callers could explicitly allow them; or
    > it could even have taken flags indicating how to treat signs; but
    > this is the situation we have now, so code like the above will
    > handle it.


    I have written code to do just that, and input from streams (no
    buffer needed). It will handle any line that a stream can supply.
    I expect to modify it to use unsigned long, as that will be more
    flexible eventually and the overhead is negligible. A wrapper
    handles signed ints. skipwhite, unlike ignoreblanks, pushes back
    the exit char. Thus ignoreblanks is static and local to this
    module. The count of leading zeroes is only limited by the ability
    of the stream itself. The following is an extract without
    #includes.

    /* -------------------------------------------------------------
    * Skip to non-blank on f, and return that char. or EOF The next
    * char that getc(f) will return is unknown. Local use only.
    */
    static int ignoreblks(FILE *f)
    {
    int ch;

    do {
    ch = getc(f);
    } while ((' ' == ch) || ('\t' == ch));
    /* while (isblank(ch)); */ /* for C99 */
    return ch;
    } /* ignoreblks */

    /*--------------------------------------------------------------
    * Skip all blanks on f. At completion getc(f) will return
    * a non-blank character, which may be \n or EOF
    *
    * Skipblks returns the char that getc will next return, or EOF.
    */
    int skipblks(FILE *f)
    {
    return ungetc(ignoreblks(f), f);
    } /* skipblks */

    /*--------------------------------------------------------------
    * Skip all whitespace on f, including \n, \f, \v, \r. At
    * completion getc(f) will return a non-blank character, which
    * may be EOF
    *
    * Skipwhite returns the char that getc will next return, or EOF.
    */
    int skipwhite(FILE *f)
    {
    int ch;

    do {
    ch = getc(f);
    } while (isspace(ch));
    return ungetc(ch, f);
    } /* skipwhite */

    /*--------------------------------------------------------------
    * Read an unsigned value. Signal error for overflow or no
    * valid number found. Returns 1 for error, 0 for noerror, EOF
    * for EOF encountered before parsing a value.
    *
    * Skip all leading blanks on f. At completion getc(f) will
    * return the character terminating the number, which may be \n
    * or EOF among others. Barring EOF it will NOT be a digit. The
    * combination of error, 0 result, and the next getc returning
    * \n indicates that no numerical value was found on the line.
    *
    * If the user wants to skip all leading white space including
    * \n, \f, \v, \r, he should first call "skipwhite(f);"
    *
    * Peculiarity: This specifically forbids a leading '+' or '-'.
    */
    int readxwd(unsigned int *wd, FILE *f)
    {
    unsigned int value, digit;
    int status;
    int ch;

    #define UWARNLVL (UINT_MAX / 10U)
    #define UWARNDIG (UINT_MAX - UWARNLVL * 10U)

    value = 0; /* default */
    status = 1; /* default error */

    ch = ignoreblks(f);

    if (EOF == ch) status = EOF;
    else if (isdigit(ch)) status = 0; /* digit, no error */

    while (isdigit(ch)) {
    digit = ch - '0';
    if ((value > UWARNLVL) ||
    ((UWARNLVL == value) && (digit > UWARNDIG))) {
    status = 1; /* overflow */
    value -= UWARNLVL;
    }
    value = 10 * value + digit;
    ch = getc(f);
    } /* while (ch is a digit) */

    *wd = value;
    ungetc(ch, f);
    return status;
    } /* readxwd */

    Not too complicated. No use of errno. Straightforward for
    interactive use. Follows the conventions for wrapping of unsigned
    values.

    --
    Chuck F (cbfalconer at maineline dot net)
    Available for consulting/temporary embedded and systems.
    <http://cbfalconer.home.att.net>
    CBFalconer, Dec 23, 2006
    #12
  13. fool

    Richard Bos Guest

    Ben Pfaff <> wrote:

    > Chris Torek <> writes:
    >
    > > CBFalconer <> wrote:
    > >>Not quite. strtoul won't detect the out-of-range input of "-1",
    > >>for example.

    > >
    > > This is not "out of range", by definition. By strtoul()'s definition,
    > > admittedly. :) If this disagrees with your own definition, it is
    > > easy enough to prohibit a minus sign.

    >
    > I suspect it's more difficult outside the C locale.


    I suspect not, because Chris overlooked the simple, obvious answer.

    Call strtol() first. If the answer is negative, the buffer has a minus
    sign or a locale-specific equivalent in it. This is safe even if the
    answer is positive and too large for a signed long while fitting in an
    unsigned one, because of the way in which strtol() must handle overflow
    - i.e., it's not UB and, significantly, it will not wrap around to
    negative.
    Only if strtol()'s answer was positive or zero, call strtoul() to get
    your real answer.

    Richard
    Richard Bos, Dec 27, 2006
    #13
  14. On Fri, 22 Dec 2006 07:56:18 +0100, Ulrich Eckhardt
    <> wrote:
    <snip other good points>
    > > if(*got != ('0' - '9'))

    >
    > Sorry, but while this looks intuitive, C doesn't let you. The '0' and '9'
    > are treated as integers, and the minus sign in between just yields the
    > difference between them. What you need is a check that the character is at
    > least zero and at most nine. There is also a library function isdigit(),
    > but firstly that is difficult to use correctly and secondly it also


    I don't think it's difficult to use, although people's MMV. The only
    tricky bit about the ctype.h is* and to* functions is that you must
    give them a value in the range of unsigned char, or EOF (necessarily
    negative and usually -1). If you are getting thevalue from a plain
    char, including the common case of a string which is an array of plain
    char (with a terminating null), on a system where plain char is (or
    really is like) signed, that value might be negative and must be dealt
    with, most simply by casting to u-char. In some cases, like this one,
    a negative plain (or signed) char value can just be rejected, because
    digits are in the basic charset and must be positive in plain char.

    > includes the minus and plus signs, IIRC.
    >

    That definitely not. ato* and strto* and also *scanf %d and i and even
    u x o accept a leading sign, but isdigit and isxdigit do not.

    - David.Thompson1 at worldnet.att.net
    Dave Thompson, Jan 3, 2007
    #14
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