typecasting?

This is a discussion on typecasting? within the C Programming forums, part of the General Programming Boards category; Sheesh I'm just full of questions. I got a question about typecasting in one particular instance. I know that normally ...

  1. #1
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    typecasting?

    Sheesh I'm just full of questions.

    I got a question about typecasting in one particular instance. I know that normally you'd typecast like this:

    (type)command

    Right? Well, they use a malloc command with a * next to the typecast, like this:

    struct list {
    int x;
    struct *next;
    };

    struct list *root;
    struct list *pointer;
    struct list *new;

    new = (list*) malloc (sizeof (struct list));

    Can anyone tell me what the * is next to the typecast for? Would the malloc command have worked even without a typecast?

  2. #2
    Registered User The Dog's Avatar
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    >> Can anyone tell me what the * is next to the typecast for?

    malloc() returns the address of the allocated memory (a pointer to the memory), so it is casted to the appropriate pointer type.

    >> Would the malloc command have worked even without a typecast?

    In C programming the malloc would've worked. In C you don't need to cast malloc().

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    If malloc only returns one type of data, can you think of any reason why the writer of that program would typecast malloc?

  4. #4
    Registered User The Dog's Avatar
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    It has to do with making your program type-safe. IOW, making sure that someting gets casted to the correct type and not relying on the compiler.

    >> can you think of any reason why the writer of that program would typecast malloc?

    It probably increases code readability. Other than that, it's just a style of programming IMO.

  5. #5
    Guest Sebastiani's Avatar
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    With a C++ compiler, due to stricter rules, malloc, and all other functions returning void* must explicitely cast the result. So the author might have been using such a compiler.
    Code:
    bool fun(bool value)
    {
        return std::pow(std::exp(1), std::complex<float>(0, 1) 
        * std::complex<float>(std::atan(1)*(1 << (value + 2))))
        .real() > 0;
    }

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    But malloc doesn't return void does it? I guess I can live with it just being the writer's own style, but I still don't understand what the * was doing next to the cast. Why not just go like this:

    (list)malloc(sizeof(struct list));

    ?

    I don't get what the dillyoh is with the *. If you're typecasting a pointer, wouldn't you put it like this:

    (*list)



    ???

  7. #7
    Registered User The Dog's Avatar
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    >> If you're typecasting a pointer, wouldn't you put it like this: (*list)

    No, whenever pointers are involved, the statement/expression/declaration/whatever is always read from right to left.

    Here are a few examples of how the code is read.
    Code:
    int *i_ptr;            //i_ptr is a pointer to an int
    float *f_ptr;         //f_ptr is a pointer to a float
    char **c_pptr;     //c_pptr is a pointer to a pointer to a char 
    void *function();  //function returns a pointer to void
    So, to cast malloc(), the indirection operator, *, is placed after the type. So in this case :
    (list*)malloc(sizeof(struct list));

    malloc() would be casted to a pointer to list.

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