arrays and strings misunderstanding...

Discussion in 'C Programming' started by Broeisi, Feb 16, 2006.

  1. Broeisi

    Broeisi Guest

    Hello,

    I wrote the tiny progam below just to understand arrays and strings
    better.

    I have 2 questions about arrays and strings in C.

    1. Why is it that when you want to assign a string to an character array
    that you must use the strcpy() function?

    Why doesn't a assignment like gender = "male" not work for a character
    array but for a pointer it works fine?

    2. I declared a character array for 10 characters including the NULL
    character, but as you see below I put more characters and it worked
    fine. Can somebody explain me that please....

    Thanks a lot in advance....

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

    int main(void)
    {
    char *name;
    char gender[10];

    name = "broeisi";
    strcpy(gender,"superb male");

    printf("%s is a %s.\n",name, gender);

    return 0;
    }

    This is the output of the program

    broeisi@kitana Misc $ gcc arraystring.c -o arraystring
    broeisi@kitana Misc $ ./arraystring
    broeisi is a superb male.

    I use gentoo linux with gcc version 3.4.4
    Broeisi, Feb 16, 2006
    #1
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  2. Broeisi wrote:
    > Hello,
    >
    > I wrote the tiny progam below just to understand arrays and strings
    > better.
    >
    > I have 2 questions about arrays and strings in C.
    >
    > 1. Why is it that when you want to assign a string to an character array
    > that you must use the strcpy() function?


    You cannot assign to a character array, but you can assign to the array
    elements. So if you want a char array to hold certain string, you need
    to copy the chars.

    > Why doesn't a assignment like gender = "male" not work for a character
    > array but for a pointer it works fine?


    An array is not a modifiable lvalue, so you cannot assign to it. A
    pointer is a modifiable lvalue, that's why in the case of a pointer the
    assignment works without problems. However if gender is a pointer the
    assignment does not magically copy the string "male" to some
    automagically allocated portion of memory. What it does make is storing
    the address of the unnamed string literal "male" in the gender
    variable.

    > 2. I declared a character array for 10 characters including the NULL
    > character, but as you see below I put more characters and it worked
    > fine. Can somebody explain me that please....


    You were unlucky. Stepping outside the boundaries of an array or
    outside a dinamically allocated block of memory invokes undefined
    behaviour, including seeming to work correctly. But the result could've
    been anything, from crushing your computer to sending you a thousand
    years into the past. (I'm tired of nasal demons... it's been a slow day
    at work.)

    HTH

    >
    > Thanks a lot in advance....
    >
    > #include <stdio.h>
    > #include <string.h>
    >
    > int main(void)
    > {
    > char *name;
    > char gender[10];
    >
    > name = "broeisi";
    > strcpy(gender,"superb male");
    >
    > printf("%s is a %s.\n",name, gender);
    >
    > return 0;
    > }
    >
    > This is the output of the program
    >
    > broeisi@kitana Misc $ gcc arraystring.c -o arraystring
    > broeisi@kitana Misc $ ./arraystring
    > broeisi is a superb male.
    >
    > I use gentoo linux with gcc version 3.4.4
    Antonio Contreras, Feb 16, 2006
    #2
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  3. Broeisi

    Broeisi Guest

    Antonio,

    Thank you very much for your explanation.
    It surely helped.

    Have a nice and thanks again for your help.

    Broeisi

    Antonio Contreras wrote:

    > Broeisi wrote:
    >> Hello,
    >>
    >> I wrote the tiny progam below just to understand arrays and strings
    >> better.
    >>
    >> I have 2 questions about arrays and strings in C.
    >>
    >> 1. Why is it that when you want to assign a string to an character array
    >> that you must use the strcpy() function?

    >
    > You cannot assign to a character array, but you can assign to the array
    > elements. So if you want a char array to hold certain string, you need
    > to copy the chars.
    >
    >> Why doesn't a assignment like gender = "male" not work for a character
    >> array but for a pointer it works fine?

    >
    > An array is not a modifiable lvalue, so you cannot assign to it. A
    > pointer is a modifiable lvalue, that's why in the case of a pointer the
    > assignment works without problems. However if gender is a pointer the
    > assignment does not magically copy the string "male" to some
    > automagically allocated portion of memory. What it does make is storing
    > the address of the unnamed string literal "male" in the gender
    > variable.
    >
    >> 2. I declared a character array for 10 characters including the NULL
    >> character, but as you see below I put more characters and it worked
    >> fine. Can somebody explain me that please....

    >
    > You were unlucky. Stepping outside the boundaries of an array or
    > outside a dinamically allocated block of memory invokes undefined
    > behaviour, including seeming to work correctly. But the result could've
    > been anything, from crushing your computer to sending you a thousand
    > years into the past. (I'm tired of nasal demons... it's been a slow day
    > at work.)
    >
    > HTH
    >
    >>
    >> Thanks a lot in advance....
    >>
    >> #include <stdio.h>
    >> #include <string.h>
    >>
    >> int main(void)
    >> {
    >> char *name;
    >> char gender[10];
    >>
    >> name = "broeisi";
    >> strcpy(gender,"superb male");
    >>
    >> printf("%s is a %s.\n",name, gender);
    >>
    >> return 0;
    >> }
    >>
    >> This is the output of the program
    >>
    >> broeisi@kitana Misc $ gcc arraystring.c -o arraystring
    >> broeisi@kitana Misc $ ./arraystring
    >> broeisi is a superb male.
    >>
    >> I use gentoo linux with gcc version 3.4.4
    Broeisi, Feb 16, 2006
    #3
  4. Broeisi

    Guest

    Broeisi wrote:
    > I wrote the tiny progam below just to understand arrays and strings
    > better.
    >
    > I have 2 questions about arrays and strings in C.
    >
    > 1. Why is it that when you want to assign a string to an character array
    > that you must use the strcpy() function?
    >
    > Why doesn't a assignment like gender = "male" not work for a character
    > array but for a pointer it works fine?


    Because the original language designers simply didn't want to put that
    in there. This stems from the fact that "strings" are not first-class
    values in the C language. Characters are primitives, but strings are
    not. Strings are basically "simulated" in the C language (and not very
    well at that.)

    The only sort of "primitive" support that the C language has for
    strings is that inline strings (sequences of C source parsable chars
    delimited by double quote characters) are interpreted in the form of
    this simulation with its own implicit storage.

    > 2. I declared a character array for 10 characters including the NULL
    > character, but as you see below I put more characters and it worked
    > fine. Can somebody explain me that please....
    >
    > Thanks a lot in advance....
    >
    > #include <stdio.h>
    > #include <string.h>
    >
    > int main(void)
    > {
    > char *name;
    > char gender[10];
    >
    > name = "broeisi";


    This is not an array assignment, but a pointer assignment. The key
    difference is that the storage for the string is declared by the value,
    not the variable. So you cannot, for example, modify the value
    "broeisi" by following this with a like like:

    name[0] = 'd'; /* Ill defined, and will crash in some
    environments. */

    This is because the storage used comes from the value which is
    considered unmutable (essentially, its implicitely const).

    > strcpy (gender, "superb male");


    Here there are two strings. gender starts out uninitialized but this
    is overwritten by the contents of the string "superb male". You have
    an additional problem here because gender has only storage for 10
    characters, but "superb male" is actually the following twelve
    characters: { 's', 'u', 'p', 'e', 'r', 'b', ' ', 'm', 'a', 'l', 'e',
    '\0' }; This is a buffer overflow, which you can learn all about here:

    http://en.wikipedia.org/wiki/Buffer_Overflow

    So long as the C language is just "simulating" strings rather than
    providing a real primitive for them, there is an oppotunity for anyone
    to write their own "simulation" for strings in the C language. "The
    Better String Library" is one such example, and you can find the URL
    for it at the bottom of this post. Using this library things are a
    little more clear:

    #include <stdio.h>
    #include "bstrlib.h"

    int main () {
    struct tagbstring name = bsStatic ("broeisi");
    bstring gender = blk2bstr (bsStaticBlkParms ("superb male"));

    printf ("%s claims to be a %s.\n", name.data, bdatae (gender,
    "??"));
    bdestroy (gender);
    return 0;
    }

    For a problem this small, the code isn't really shorter or simpler,
    however the storage issue is somewhat clearer. Inline strings are
    wrapped in bsStatic* macros which makes their storage clear. Also
    issues like "buffer overflows" essentially disappear, since dynamic
    storage container size issues are dealt with automatically in the
    Better String Library (notice there is no value 10 anywhere in this
    sample code), which is closer to what you find in other modern
    languages.

    --
    Paul Hsieh
    http://www.pobox.com/~qed/
    http://bstring.sf.net/
    , Feb 16, 2006
    #4
  5. Broeisi

    Pedro Graca Guest

    Broeisi wrote:
    > I wrote the tiny progam below just to understand arrays and strings
    > better.
    >
    > 2. I declared a character array for 10 characters including the NULL
    > character, but as you see below I put more characters and it worked
    > fine. Can somebody explain me that please....



    > #include <stdio.h>
    > #include <string.h>
    >
    > int main(void)
    > {
    > char *name;
    > char gender[10];


    Reserve 10 bytes of memory for gender,
    so the memory goes something like this

    ...........##########..........
    aaaaaa<-GENDER->bbbbbb

    Where "aaaaaa" and "bbbbbb" is memory reserved for something else or not
    reserved at all (reserved for future use).

    >
    > name = "broeisi";
    > strcpy(gender,"superb male");


    Now the memory goes something like

    ...........superb male0........
    aaaaaa<-GENDER->bbbbbb

    So you've used memory space that doesn't belong to gender.
    You might have overwritten something that was already there or that
    final 'e0' may be overwritten by something else in your program removing
    the terminating 0 from the string gender.

    --
    If you're posting through Google read <http://cfaj.freeshell.org/google>
    Pedro Graca, Feb 16, 2006
    #5
  6. On Thu, 16 Feb 2006 11:37:07 -0600, in comp.lang.c , Broeisi
    <> wrote:

    >1. Why is it that when you want to assign a string to an character array
    >that you must use the strcpy() function?


    Because an array is an array, not a pointer. This is almost certainly
    a FAQ by the way.

    >Why doesn't a assignment like gender = "male" not work for a character
    >array but for a pointer it works fine?


    It depends what you mean by "work". In both cases, if you want to copy
    the data, you must copy it, not assign a pointer to point to it.

    >2. I declared a character array for 10 characters including the NULL
    >character, but as you see below I put more characters and it worked
    >fine. Can somebody explain me that please....


    You were unlucky. If your compiler and OS had been more robust, they
    would have stopped your programme executing before it wrotte over
    memory that did not belong to it.


    By teh way, that was three questions, not two... :)
    Mark McIntyre
    --
    "Debugging is twice as hard as writing the code in the first place.
    Therefore, if you write the code as cleverly as possible, you are,
    by definition, not smart enough to debug it."
    --Brian Kernighan

    ----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
    http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
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    Mark McIntyre, Feb 16, 2006
    #6
  7. Broeisi wrote:

    <snip Broeisi question and my answer>

    > Antonio,
    >
    > Thank you very much for your explanation.
    > It surely helped.
    >
    > Have a nice and thanks again for your help.
    >
    > Broeisi


    You're wellcome. Glad I could help.

    BTW, try to avoid top-posting. Other groups may have a more relaxed
    attitude concerning netiquette, but c.l.c. has literally hundreds of
    posts a day and top-posts makes reading them harder.
    Antonio Contreras, Feb 16, 2006
    #7
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