pointer declarations and dereferencing 4

[ivanachukapawn2]

A tutorial says:

float* px;
defines px as a pointer to a float.

AND LATER

float *px;
defines px as a pointer to a float

I am a C newbie and find this very confusing. Can you confirm, are these 2 statements (defining px as a pointer) equivalent?

Is there one way of declaring a pointer predominately used and/or preferred?

Further on in the tutorial, I find this:

The content of the memory location referenced by a pointer is obtained using the ``*'' operator (this is called dereferencing the pointer). Thus, *px refers to the value of x.

This is confusing also. In a previous example, the tutorial states that *px declares px to be a pointer, but in this example *px is said to dereferenc the pointer. When I encounter *px in code how can I discern its meaning?

 

[ArkM]

The C declaration (informal) syntax:

type_name comma_separated_list_of_declarators;

Semicolon terminates declaration.
The simplest declaration:

type_name declarator;

Now we have in

float * px;

float - type name,
*px - declarator.

float, * and px are tokens (lexemes, i.e. lexical units of the C language). Remember: whitespaces are optional token separators. See, for example, six identical declarations:

/*1*/ float*px;
/*2*/ float * px;
/*3*/ float* px;
/*4*/ float *px;
/*5*/ float * px ;
/*6*/ float
*
px
;

There is one and only one name in the declarator (px in your snippet). Declaration meaning: if you see (below the declaration) declarator construct in a correct expression, it yields a value of the basic type. So after float*px this construct: *px must yield float value. In other words, px is a pointer to float because of unary * operator (dereferencing) has a pointer operand in the language (expression *name is correct if and only if name is a pointer).

No common preferred forms of declaration in this case. I like case #3, but be careful:

float* px, py;

declares px as a pointer to float but py is not a pointer variable! We have two declarators in this declaration: *px and py (not *py)...

That's funny the C language declaration syntax. By the way, see yet another equivalent of the px as a pointer to float declaration:

float(*px);

 

[Prattaratt]

And expanding on ArkM's answer, anytime you encounter *px outside of a variable declaration, it will be a pointer dereference.

 

[ArkM]

>...anytime you encounter *px outside of a variable declaration, it will be a pointer dereference.

Be careful, we can see *px in the snippet::

typedef struct { float* p; int junk; } S;
S px;
...
*px.p = 0.0; /* *px here... and px is not a pointer! */

Now px is NOT a pointer, it's a structure variable (bad name.
Let's set out parentheses (remember operator precedence):

*(px.p) = 0.0;

Now px.p must be a pointer, but px must be a structure variable name: operator . wants a structure as a left operand.

 

[tcltest]

float *px; // this only for the px to be floating pointer

float* x,y; // here both x and y become pointers.

SHRIKANT BADIGER
SJCE MYSORE

 

[ArkM]

float* x, y; /* px - pointer to float, y - float! */

(that's a common C trap for beginners.
There are two declarators: *x and y - same as two declarations:

float*x; /* pointer to float */
float y; /* float */

Compare with

typedef float* Pointer2float;
Pointer2float px, py; /* (That's OK;) */

Incidentally, avoid using float type in math calculations: it has insufficient precision (less than 7 decimal digits). Use double type in all cases (it's the other story where is float data natural habitat.