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Top 100 Interview Questions & Answers in a convenient and easy to read book!

“I bought this guide a few days ago to prepare for my interview with Oracle. Many of the questions they asked me were from this guide. I found this book absolutely great!”

– Ravi, California

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  1. Submitted By: Khisar — October 6, 2006
    -16 votes
      + -

    void *malloc(size_t number_of_bytes);

  2. Submitted By: octav_timofte — October 6, 2006
    +29 votes
      + -

    I think the answer is simply
    void* p;

    malloc is just the library function called to allocated some memory and of course a void pointer will be returned , but it is the declaration of a void pointer.

  3. Submitted By: Veerbhan — November 10, 2006
    +2 votes
      + -

    Ofcourse declaring a void pointer means void* var;
    we will use malloc to assign memory to var before using it ie before assigning it a value of any type.

  4. Submitted By: Jim Stevens — November 14, 2006
    -2 votes
      + -

    void near * n_ptr;

  5. Submitted By: raushan — December 14, 2006
    +1 votes
      + -

    Better is to use void *p.
    Even you can associate the * or & to the type instead of the variable like void* p; But this may lead to a confusion when you go for this kind of declaration void* p, q.
    By looking at this you may think q is also a void pointer but it is not so it becomes a ordinary variable.

  6. Submitted By: siva — May 24, 2007
    -2 votes
      + -

    To raushan:

    we can declare a pointer of type void,but we cant declare a variable of type void bcoz datatype void doesnt exist.

  7. Submitted By: Talmeez ul Hassan — September 17, 2008
    -1 votes
      + -

    Void pointers
    At first glance, a void pointer seems to be of limited, if any, use. However, when combined with the ability to cast such a pointer to another type, they turn out to be quite useful and flexible.

    Consider the example of the previous section, where we constructed a function pointer to a function of type void and argument int. Such a function pointer in this form could not be used for a void function with a different type of argument (for example, float). This can be done, however, through the use of void pointers, as the following example illustrates.
    #include
    void use_int(void *);
    void use_float(void *);
    void greeting(void (*)(void *), void *);

    int main(void) {
    char ans;
    int i_age = 22;
    float f_age = 22.0;
    void *p;
    printf(”Use int (i) or float (f)? “);
    scanf(”%c”, &ans);
    if (ans == ‘i’) {
    p = &i_age;
    greeting(use_int, p);
    }
    else {
    p = &f_age;
    greeting(use_float, p);
    }
    return 0;
    }
    void greeting(void (*fp)(void *), void *q) {
    fp(q);
    }
    void use_int(void *r) {
    int a;
    a = * (int *) r;
    printf(”As an integer, you are %d years old.\n”, a);
    }
    void use_float(void *s) {
    float *b;
    b = (float *) s;
    printf(”As a float, you are %f years old.\n”, *b);
    }
    Although this requires us to cast the void pointer into the appropriate type in the relevant subroutine (use_int or use_float), the flexibility here appears in the greeting routine, which can now handle in principle a function with any type of argument.

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