& *

[Lanarcam]

Hi,

I couldn't find the answer in the archives to this question :

Does the C language define this instruction?

int j;
int * pi = &j;

int k = & * pi;

 

[Lanarcam]

Le 02/04/2012 20:38, Lanarcam a écrit :

Hi,

I couldn't find the answer in the archives to this question :

Does the C language define this instruction?

int j;
int * pi = &j;

int k = & * pi;

Sorry, one must read:

int * pk = & * pi;

 

[James Kuyper]

Le 02/04/2012 20:38, Lanarcam a écrit :

Sorry, one must read:

int * pk = & * pi;

The C standard defines the behavior of that construct: "The unary &
operator yields the address of its operand. ... If the operand is the
result of a unary * operator, neither that operator nor the & operator
is evaluated and the result is as if both were omitted, except that the
constraints on the operators still apply and the result is not an
lvalue." (6.5.2p3).

In other words, so long as pi is a pointer variable, &*pi is equivalent
to pi itself, except that &*pi can't be used as an lvalue. In this
context, that means that &*pi == &j is true, but that &*pi = &j would be
a constraint violation.

 

[Ben Bacarisse]

Le 02/04/2012 20:38, Lanarcam a écrit :

Sorry, one must read:

int * pk = & * pi;

Yes, though it's not an instruction in a very real sense. The language
definition says that when the operand of & is a * expression (I'm
abbreviating the wording) both operators are ignored. Your final
declaration is that same as

int *pk = pi;

 

[Lanarcam]

Le 02/04/2012 21:18, James Kuyper a écrit :

The C standard defines the behavior of that construct: "The unary&
operator yields the address of its operand. ... If the operand is the
result of a unary * operator, neither that operator nor the& operator
is evaluated and the result is as if both were omitted, except that the
constraints on the operators still apply and the result is not an
lvalue." (6.5.2p3).

In other words, so long as pi is a pointer variable,&*pi is equivalent
to pi itself, except that&*pi can't be used as an lvalue. In this
context, that means that&*pi ==&j is true, but that&*pi =&j would be
a constraint violation.

Thanks a lot. That question comes from a tool provider which generates
C code, we have a nasty bug and thought that could be the reason, but
from your answer it appears that we must look elsewhere.

(unless the compiler dosn't respect the C standard of course)

 

[Lanarcam]

Le 02/04/2012 21:21, Ben Bacarisse a écrit :

Yes, though it's not an instruction in a very real sense. The language
definition says that when the operand of& is a * expression (I'm
abbreviating the wording) both operators are ignored. Your final
declaration is that same as

int *pk = pi;

Thanks

 

[jacob navia]

Le 02/04/12 21:18, James Kuyper a écrit :

The C standard defines the behavior of that construct: "The unary&
operator yields the address of its operand. ... If the operand is the
result of a unary * operator, neither that operator nor the& operator
is evaluated and the result is as if both were omitted, except that the
constraints on the operators still apply and the result is not an
lvalue." (6.5.2p3).

Make that 6.5.3.2, or maybe you are quoting from an older standard?