CODEWITHSUNDEEP
c++pointersstruct
Roronoa Zoro
Roronoa Zoro

Reputation: 1013

Is there a difference between these two stuct-related declarations?

I have the following structure

typedef struct _LSHFunctionT 
{
    double *a;
    double b;
} LSHFunctionT, *PLSHFunctionT;

My question is; is there a difference between these two declarations

PLSHFunctionT myPointer1;

and

LSHFunctionT *myPointer2;

and if not, then why do people explicitly use two of them (LSHFunctionT and *PLSHFunctionT). Why not just use LSHFunctionT.

Does it go the same way for the following two declarations

PLSHFunctionT *myPointer3;

and

LSHFunctionT **myPointer3;

Upvotes: 3

Views: 122

Answers (7)

perreal
perreal

Reputation: 97948

The difference is in emphasis. Generally not explicitly writing the * may indicate that the PLSHFunctionT is designed to be used as a handle (without knowing/accessing structure elements ). If * is explicitly written, as in LSHFunctionT *myPointer, it might indicate an array or a structure that is to be used to access the values.

Upvotes: 1

LihO
LihO

Reputation: 42083

typedef struct _LSHFunctionT 
{
    double *a;
    double b;
} LSHFunctionT, *PLSHFunctionT;

Yes, PLSHFunctionT x; is equal to LSHFunctionT* x;

And yes, PLSHFunctionT* x; is equal to LSHFunctionT** x;

The purpose of typedef is to assign new names to existing types. You can define typedef int lol; and declare variable lol i;, but compiler will consider it int anyway.

You should also check these questions:
When should I use typedef in C++?
Why should structure names have a typedef?

Hope this helps.

Upvotes: 2

Mark B
Mark B

Reputation: 96241

There's no difference between the two on the surface. However, with the pointer typedef there's no way to declare it as pointer-to-const, just const-pointer-to-non-const.

For example you can say

const LSHFunctionT* const_ptr; but const PLSHFunctionT const_ptr2; makes the pointer const, NOT the pointee.

Finally note that in C++ the whole thing is of questionable legality because names starting with _<capital> are reserved for the implementation, and that typedefs are almost never used in such a way.

Upvotes: 1

Krzysztof Kozielczyk
Krzysztof Kozielczyk

Reputation: 5937

Those declarations are the same. The pointer-typedef approach improves code readability in some cases. One could argue that this:

PLSHFunctionT calls[];

is easier to read than this:

LSHFunctionT *calls[];

Upvotes: 0

Sarfaraz Nawaz
Sarfaraz Nawaz

Reputation: 361402

Yes. They are exactly same, even in the second case.

I personally prefer using * explicitly if I want to declare a pointer. It makes the code readable, in most cases. Usage of typedef of pointer-type reduces readability usually, though sometimes it may increases readability especially when you work with, say, Windows API.

Upvotes: 0

Mark Ransom
Mark Ransom

Reputation: 308158

Yes, they are identical.

One good reason to define a pointer type is for complicated expressions. If for example you have a function that takes a reference to a pointer, which do you find easier to understand?

void foo(PLSHFunctionT & ref);

void foo(LSHFunctionT * (&ref));

I'm not even sure I got the syntax correct for the second one!

Upvotes: 2

mikithskegg
mikithskegg

Reputation: 816

I seem there is no difference. There are different programming styles.

Upvotes: 0

Related Questions