Thread: Function Pointers - What does this block do?

Hey. I've been researching into operating system programming.
I am trying to figure out what exactly what each line of code does.

For example, the function below copies chunks of memory from a source to a destination. In order for me to understand how the code works, It is necessary to decipher each line individually.

Code:

unsigned char *memcpy(unsigned char *dest, const unsigned char *src, int count)
{
    const char *sp = (const char *)src;
    char *dp = (char *)dest;
    for(; count != 0; count--) *dp++ = *sp++;
    return dest;
}

From what I know already, the following code takes the source's addresses in memory, and maps it onto the destination memory address, thereby copying it.

Code:

 for(; count != 0; count--) *dp++ = *sp++;
 return dest;

However I do not understand some things.

Firstly,

Code:

 char *dp = (char *)dest;

I've never seen anything like the above code. What I do not understand is what the parenthesis does. Common sense would dictate that this is simply assigning the destination to a pointer, however as I do not understand what the purpose of the parentheses is, I cannot fully understand the purpose of this code. Could someone kindly explain this to me?

The other thing I do not understand is at the declaration of the function, the name of the function has a * before it. I do not understand what this does. Does it return the pointer of the return value instead of the actual return value?

Thanks in Advance,


~ShadowzReloaded

Originally Posted by shadowzreloaded

I've never seen anything like the above code. What I do not understand is what the parenthesis does. Common sense would dictate that this is simply assigning the destination to a pointer, however as I do not understand what the purpose of the parentheses is, I cannot fully understand the purpose of this code. Could someone kindly explain this to me?

The other thing I do not understand is at the declaration of the function, the name of the function has a * before it. I do not understand what this does. Does it return the pointer of the return value instead of the actual return value?

Thanks in Advance,


~ShadowzReloaded

No offense, but rather than have us teach you the basics of C here, you should probably be studying that sort of thing on your own. I'd recommend reading at least two books on the subject before you start asking questions, actually.

Originally Posted by shadowzreloaded

I've never seen anything like the above code. What I do not understand is what the parenthesis does. Common sense would dictate that this is simply assigning the destination to a pointer, however as I do not understand what the purpose of the parentheses is, I cannot fully understand the purpose of this code. Could someone kindly explain this to me?

Search the Web for "C typecast".

Originally Posted by shadowzreloaded

The other thing I do not understand is at the declaration of the function, the name of the function has a * before it. I do not understand what this does. Does it return the pointer of the return value instead of the actual return value?

The return type of the function is a pointer type.

Originally Posted by shadowzreloaded

What I really meant was a pointer to a function, not a function pointer.

The terms are synonymous.

Originally Posted by shadowzreloaded

Please post something constructive. I never usually post questions. But this is an exception, I was unable to find anything on this subject, also, since when were function pointers basic C?

Nothing you posted had anything to do with function pointers, actually.
So, you see, when you don't even know the basics of the language, it just makes it difficult for all parties involved. Do your part, for everyone's sake. Pretty please.

Originally Posted by shadowzreloaded

My function name is a pointer, so it is a variation of a function pointer.

It is true that you do not need to use the address-of operator in order to obtain the address of a function, i.e., a function pointer. Basically, a function is convertible to a function pointer. But a function is not a function pointer.

Originally Posted by shadowzreloaded

A function pointer itself however would have parenthesis around the *name of the function?

No, but you are probably talking about the name of the function pointer, not the name of the function.

Originally Posted by shadowzreloaded

I see. Thank you for clarifying that for me.

Go to the library and get a basic "Learn C in 7 days" book, or do some basic C tutorials on line. You are totally wasting your time trying to decypher things this way when you don't understand even very basic syntax such as this (that a function returning a pointer means there is a pointer to that function).

I am not saying that to be mean, I am saying it because it is true. If you are trying to learn C, doing it by "researching into operating system programming" is just plain stupid (sorry). If you do not understand C, which you do not, then "researching into operating system programming" is not going to go very well for you.