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ethdr = (ETHERNET_FRAME*)&tmp[0];
I thought pointer (whether declearing or referencing it), * should be on left side of the word.
One of the code I was studying, I saw code like above where * shows up on right side of struct.
Is this not a pointer ? what is this?
Can someone explain please?
It's a "cast". It means the variable is to be converted to this type -- in this case, an ETHERNET_FRAME pointer. This makes sense, since &tmp[0] would be the address of (&) some variable.
And as we know, pointers store the address of some variable of a particular type. Although void pointers can be used to refer to ANY type.
Hi, thanks for replying on this issue.
I guess I have a problem understanding the syntax of it.
Wouldn't Ethernet_Frame pointer be --> *Ethernet_Frame ?
but instead, I have Ethernet_Frame* ........ ??
Well it shouldn't really matter where it goes, it's correct either way. HOWEVER, logically, IMO it should go with the type rather than the variable because the symbol in that case is identifying type.
Not in the slightest. That would be multiplying by Ethernet_Frame (if Ethernet_Frame was a variable and not a type), or dereferencing Ethernet_Frame (ditto). Even when you declare the variable, that's what you use:Quote:
Originally Posted by convenientstore
Code:Ethernet_Frame* this_is_a_pointer_variable;
When you declare a pointer you doQuote:
Originally Posted by convenientstore
So when you want to declare that a variable is a pointer you put the type first and the symbol * after that.Code:type * name; //int * myInt
ETHERNET_FRAME is a type, not a variable...
EDIT: Oops too slow
Nevermind, you gave a very clear description that I'm sure hammers the point home.Quote:
Originally Posted by C_ntua
Talking of hammers, think of it this way ;)
RoundHole = (hammer*)SquarePeg;