Subexpression types

[Kirk Zurell]

I'm working through an (older) copy of the C standard to learn if there
is any minute difference between

char * ptr;
/* ... */
*ptr++ = 0x23;

and

char * ptr;
/* ... */
*(ptr)++ = 0x23;

owing to promotion or something obscure I don't know about.

Here's what I've concluded:

An identifier describing an object is an lvalue (6.5.1 2).

A parenthesized expression is an lvalue. Its type and value are
identical to the unparenthesized expression (6.5.1 5).

The operand of the * operator shall have pointer type (6.5.3.2 2) and is
an lvalue (6.5.3.2 4).

So, in *(ptr)++ it's required that the (ptr) expression have pointer
type. Based on this, I think *ptr++ and *(ptr)++ are identical.

Anyone spot me on this?

Thanks
Kirk

 

[Stefan Ram]

A parenthesized expression is an lvalue.

Then, the expression »(4)« would be an lvalue?

I think *ptr++ and *(ptr)++ are identical.

This depends on the notion of »identity« used, »identical« is
a vague term without further specification. Obvioulsy, those
two expression differ in their length and structural depth:
So a compiler with a certain implementation limit regarding
expression complexity might accept one but reject the other in
some context.

 

[Ben Bacarisse]

I'm working through an (older) copy of the C standard to learn if there
is any minute difference between

char * ptr;
/* ... */
*ptr++ = 0x23;

and

char * ptr;
/* ... */
*(ptr)++ = 0x23;

owing to promotion or something obscure I don't know about.

Here's what I've concluded:

An identifier describing an object is an lvalue (6.5.1 2).

A parenthesized expression is an lvalue.

In this case, yes, but only because ptr is an lvalue.

Its type and value are
identical to the unparenthesized expression (6.5.1 5).

The operand of the * operator shall have pointer type (6.5.3.2 2) and is
an lvalue (6.5.3.2 4).

The operand of * is not (ptr) -- it is (ptr)++ -- but that does not make
much difference in this case.

So, in *(ptr)++ it's required that the (ptr) expression have pointer
type. Based on this, I think *ptr++ and *(ptr)++ are identical.

I think you only need to know that the parentheses have no effect on
the value, type or "kind" of the expression. By "kind" I mean
whether the expression is an lvalue, a function designator or a void
expression.

 

[Kirk Zurell]

Then, the expression »(4)« would be an lvalue?

Oh bother.

[A] compiler with a certain implementation limit regarding
expression complexity might accept one but reject the other in
some context.

That's a good point. Thanks.

Kirk.

 

[Kirk Zurell]

In this case, yes, but only because ptr is an lvalue.

Double bother. Thanks.

I think you only need to know that the parentheses have no effect on
the value, type or "kind" of the expression. By "kind" I mean
whether the expression is an lvalue, a function designator or a void
expression.

That's what I was hoping to confirm.
Thanks
Kirk.