Why parenthesis around function parameter?

Question

I'm studying section 2.1 from Computer Systems A Programmer's Perspective and I understand what the code is doing but I don't understand why there are parenthesis around the formal parameter "byte_pointer" followed by &x?

typedef unsigned char *byte_pointer;

void show_bytes(byte_pointer start, int len) {
    int i;
    for (i = 0; i < len; i++)
        printf(" %.2x", start[i]);
    printf("\n");
}

void show_float(float x) {
    show_bytes((byte_pointer) &x, sizeof(float));
}

void show_int(int x) {
    show_bytes((byte_pointer) &x, sizeof(int));
}

void show_pointer(void *x) {
    show_bytes((byte_pointer) &x, sizeof(void *));
}

`byte_pointer` is not a formal parameter (at call point you only have effective parameters). 1) `byte_pointer` is a type, 2) expression like `(type)value` is a cast, a conversion if you prefer. Thus address of `float x` is converted to address of *byte*. — Jean-Baptiste Yunès, Jan 16 '18 at 08:30
And if you don't have a grasp on C concepts yet, I think you need a slightly more [basic book](https://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list). — StoryTeller - Unslander Monica, Jan 16 '18 at 08:30
@StoryTeller C++ has its own casts and learning C is not by any means a prerequisite. — Aluan Haddad, Jan 16 '18 at 08:31
@AluanHaddad: I'd still advise it though. Learn the low-level stuff via C. — Bathsheba, Jan 16 '18 at 08:31
@AluanHaddad - I don't see any C++ in this post. And I agree with Christian's removal of the C++ tag. This is C, nothing wrong with it. And it's best to learn it without conflating it with C++. — StoryTeller - Unslander Monica, Jan 16 '18 at 08:33
Well, after wasting years in university in applied C++ courses taught by professors who never moved beyond C, I disagree as it often leads to stagnation and misconception. — Aluan Haddad, Jan 16 '18 at 08:33
@AluanHaddad - I don't get what you are disagreeing with, exactly. That one should learn C if they intend to work in C? — StoryTeller - Unslander Monica, Jan 16 '18 at 08:35
I'm working my way through a community college CS sequence and we never took any C classes, only C++. I will try to learn C on my own, but now this class followed my C++ sequence. So if I understand correctly is (byte_pointer) casting the formal parameter to 1 byte? Sorry if I'm asking noob questions, but I've been searching endlessly on google. — David, Jan 16 '18 at 08:36
@StoryTeller I apologize for being vague. I disagree with the sentiment that you should learn C first, or the low level parts in C, if you are studying C++ — Aluan Haddad, Jan 16 '18 at 08:37
@AluanHaddad - I disagree with that too (heck, my profile even contains a link to [this fantastic lecture](https://www.youtube.com/watch?v=YnWhqhNdYyk)). I never suggested it. — StoryTeller - Unslander Monica, Jan 16 '18 at 08:38
@StoryTeller Ah, OK. It was my misunderstanding then. I've seen that lecture and it is indeed fantastic! Thanks for linking it. — Aluan Haddad, Jan 16 '18 at 08:40
A book that tells you to hide pointers behind a typedef is a very bad one. It is cold outside, light a fire and put that book to some good use! — Lundin, Jan 16 '18 at 10:52

user2736738 · Answer 1 · 2018-01-16T08:49:31.347

5

You are typecasting a float pointer to an unsigned char*. And then you will access each of the float variable's bytes. As pointer arithmetic is dictated by the type of thing it points to - the increment or dcrement on it will be directbed by sizeof(unsigned char) or 1 byte. You will access the float byte by byte. By casting we are basically saying that you need to consider this as unsigned char* even if it the float* that we are passing. There is a change in type information - and we are making the compiler aware that we are aware of it by providing the casting.

edited Jan 16 '18 at 08:49

answered Jan 16 '18 at 08:30

user2736738

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So if I understand correctly is (byte_pointer) casting the formal parameter to 1 byte? – David Jan 16 '18 at 08:40
@David.: casting to `unsigned char*` and as `sizeof(unsigned char)` is `1` yes now it accesses byte by byte – user2736738 Jan 16 '18 at 08:41
I'm using VS 2015 and I removed the type casting and only included &x, and it still printed each byte. Did the compiler type cast the parameter? – David Jan 16 '18 at 08:46
1

No, the cast does not cast anything to 1 byte. It does not cast something to `char`. It casts a pointer to `float` into a **pointer to `char`**. While `char` is 1 byte, pointer to `char` is not 1 byte. – Gerhardh Jan 16 '18 at 08:53

Why parenthesis around function parameter?

1 Answers1