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Differnce between C and Java returning an Object reference/pointer

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  1. #1
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    Question Differnce between C and Java returning an Object reference/pointer

    In C we know that the memory allocated locally inside a function will vanish when the function returns

    Code:
    char *getName()
    {
    char name[]="Sumit";
    return name
    }

    The caller of the function will get a pointer to the array.Although the pointer points a garbage unknown location since the array was local, it has now been destroyed.



    Now I have a doubt the same thing works in Java

    Code:
    public String getName()
    {
    String name="Sumit"; //name here is local as above;memory allocated for the object locally
    return name;
    }

    The caller of the method will correctly get the String returned.

    What is the difference in the mechanism of returning pointer in C and reference in Java?

    Hope I m made my point clear.
    Eagerly waiting for reply.

    Regards

  2. #2
    ATH0 quzah's Avatar
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    The difference here isn't in the way the pointer is returned really. The difference is what you are actually returning. Or rather, what type of object you are trying to return. In the first example, you are returning a local variable. If you were actually allocating some memory dynamically, copying your characters into that block of memory, and then returning that - then you would be like what your Java example does; assuming the Java example is actually correct.
    Code:
    char *getName( void )
    {
        char buf[ BUFSIZ ] = { "some name" };
        char *s = NULL;
        s = malloc( strlen( buf ) + 1 );
        strcpy( s, buf );
        return s;
    }
    Something like that. Of course you need to free that when you are done.


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  3. #3
    and the hat of wrongness Salem's Avatar
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    Do you understand why this C version works?
    Code:
    const char *getName()
    {
        const char *name="Sumit";
        return name;
    }
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  4. #4
    ATH0 quzah's Avatar
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    Quote Originally Posted by Salem View Post
    Do you understand why this C version works?
    Code:
    const char *getName()
    {
        const char *name="Sumit";
        return name;
    }
    I was over thinking it. I suppose we could even shorten that:
    Code:
    char *getName( void )
    {
        return "Sumit";
    }

    Quzah.
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    This C version does not work. It returns garbage (i.e the pointer points to garbage, not to memory conatainig the desired string)

    Quote Originally Posted by Salem View Post
    Do you understand why this C version works?
    Code:
    const char *getName()
    {
        const char *name="Sumit";
        return name;
    }

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    Thanks @quzah for your quick reply.

    You mean to say that the memory allocated this way is not dynamic
    Code:
    char name[]="Sumit";
    What we call this type of memory allocation.

    Moreover you mean to say that dynamic memory allocated does not have a restricted local (automatic scope)?
    Am i right in saying above statements.

    Regards



    Quote Originally Posted by quzah View Post
    The difference here isn't in the way the pointer is returned really. The difference is what you are actually returning. Or rather, what type of object you are trying to return. In the first example, you are returning a local variable. If you were actually allocating some memory dynamically, copying your characters into that block of memory, and then returning that - then you would be like what your Java example does; assuming the Java example is actually correct.
    Code:
    char *getName( void )
    {
        char buf[ BUFSIZ ] = { "some name" };
        char *s = NULL;
        s = malloc( strlen( buf ) + 1 );
        strcpy( s, buf );
        return s;
    }
    Something like that. Of course you need to free that when you are done.


    Quzah.

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    Registered User manasij7479's Avatar
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    Moreover you mean to say that dynamic memory allocated does not have a restricted local
    In general, when you allocate memory, it is done on the heap.
    But variable declarations get memory on the stack, making them local in scope.
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    1.None of the other participants are fast and steady.
    2.The fast and unsteady suddenly falls asleep while running !



  8. #8
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by sumit180288
    Now I have a doubt the same thing works in Java
    (...)
    The caller of the method will correctly get the String returned.
    You could say that the local String reference variable is destroyed, but the String object itself is not necessarily destroyed when the method returns. It may be destroyed at some point when it is clear that there are no references that refer to the String object. In general, object lifetime in Java is not deterministically associated with the scope of a variable.

    Quote Originally Posted by sumit180288
    This C version does not work. It returns garbage (i.e the pointer points to garbage, not to memory conatainig the desired string)
    Look more carefully at the type of name and what name refers to before jumping to that conclusion.
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    Registered User gardhr's Avatar
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    Quote Originally Posted by sumit180288 View Post
    This C version does not work. It returns garbage (i.e the pointer points to garbage, not to memory conatainig the desired string)
    It does work, and here's why: the string literal "Sumit" is effectively a global/static variable, so returning it's address (stored in the pointer 'name', that is) it perfectly safe. Now had you declared it:

    Code:
    char name[ ] = "Sumit";
    Here, the string literal is copied into a local array of bytes and then a pointer to the latter returned to the caller, which is of course unsafe.

    Dynamic allocation (such as using malloc) is similar to the first case, except now we are using a library routine that serves up pieces of a global/static chunk of data which can be safely returned to the caller.
    Last edited by gardhr; 09-07-2011 at 12:14 AM. Reason: clarification

  10. #10
    and the hat of wrongness Salem's Avatar
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    > This C version does not work. It returns garbage (i.e the pointer points to garbage, not to memory conatainig the desired string)
    What crap are you smoking?
    Code:
    $ cat baz.c
    #include <stdio.h>
    const char *getName()
    {
        const char *name="Sumit";
        return name;
    }
    int main ( ) {
      printf("%s\n",getName());
      return 0;
    }
    $ gcc -W -Wall -ansi -pedantic baz.c
    $ ./a.out 
    Sumit
    $
    This is NOT the same as your array code in your first post.

    If you're still under the illusion that arrays and pointers are the same thing in C, then read this.
    Arrays and Pointers
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
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    Code:
    char *f()
            {
                    char buf[10];
                    /* ... */
                    return buf;
            }
    According to a link http:///www.lysator.liu.se/c/c-faq/c-3.html, the above set of statements will result in garbage set of values returned.

    Quote Originally Posted by Salem View Post
    > This C version does not work. It returns garbage (i.e the pointer points to garbage, not to memory conatainig the desired string)
    What crap are you smoking?
    Code:
    $ cat baz.c
    #include <stdio.h>
    const char *getName()
    {
        const char *name="Sumit";
        return name;
    }
    int main ( ) {
      printf("%s\n",getName());
      return 0;
    }
    $ gcc -W -Wall -ansi -pedantic baz.c
    $ ./a.out 
    Sumit
    $
    This is NOT the same as your array code in your first post.

    If you're still under the illusion that arrays and pointers are the same thing in C, then read this.
    Arrays and Pointers

  12. #12
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by sumit180288
    According to a link http:///www.lysator.liu.se/c/c-faq/c-3.html, the above set of statements will result in garbage set of values returned.
    Effectively yes, though strictly speaking there is undefined behaviour if the return value is used. It is good that you recognise this, but you should also recognise that your claim that "this C version does not work" is not backed up by this example, because the essential part of the code is different.
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    OK i got ur point
    Can you tell me what is the difference between these two codes:
    Code:
    char *f()
    {
    char *name="Sumit";
    return name;
    }
    and
    Code:
    char *f()
    {
    char name[]="Sumit";
    return name;
    }
    WHY the FIRST VERSION WORKS???


    Quote Originally Posted by Salem View Post
    > This C version does not work. It returns garbage (i.e the pointer points to garbage, not to memory conatainig the desired string)
    What crap are you smoking?
    Code:
    $ cat baz.c
    #include <stdio.h>
    const char *getName()
    {
        const char *name="Sumit";
        return name;
    }
    int main ( ) {
      printf("%s\n",getName());
      return 0;
    }
    $ gcc -W -Wall -ansi -pedantic baz.c
    $ ./a.out 
    Sumit
    $
    This is NOT the same as your array code in your first post.

    If you're still under the illusion that arrays and pointers are the same thing in C, then read this.
    Arrays and Pointers

  14. #14
    and the hat of wrongness Salem's Avatar
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    > According to a link http:///www.lysator.liu.se/c/c-faq/c-3.html, the above set of statements will result in garbage set of values returned.
    Like I said, until you recognise that char* and char[] are different things, you're just going to be confused.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

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    @Salem. Sorry to say i didn't see the pointer version. I thought u are asking about the array version.
    I am also not knowing why this verison works?
    Anyone plz clear this doubt.

    Regards

    Quote Originally Posted by Salem View Post
    Do you understand why this C version works?
    Code:
    const char *getName()
    {
        const char *name="Sumit";
        return name;
    }

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