Just value copied?

[Garma]

In this code,

int main(){
int a = 47;
int b;
int* p;

p = &b;
*p = a;
return 0;
}

Is there any problem of the code in C?
In statement of "*p = a;", is just value of 'a' copied to the memory address
of 'p' or 'p' re-pointed to address of 'a'?
Thanks for your comments!

[Alex Monjushko]

Garma <Ga*****@nosysserv.com> wrote:

In this code, int main(){
int a = 47;
int b;
int* p; p = &b;
*p = a;
return 0;
} Is there any problem of the code in C?
No problem.
In statement of "*p = a;", is just value of 'a' copied to the memory address
of 'p' or 'p' re-pointed to address of 'a'? p = &b;
This assigns the address of 'b' to 'p'.
*p = a;

This deferences 'p' and assigns the value of 'a' to whatever
'p' points to, namely 'b'. Hence, the value of 'b' changes but
'p' itself remains unchanged since it still contains the address
of 'b'.

--
Alex Monjushko (mo*******@hotmail.com)

[Horst Kraemer]

On Fri, 16 Jan 2004 00:03:16 GMT, "Garma" <Ga*****@nosysserv.com>
wrote:

In this code,

int main(){
int a = 47;
int b;
int* p;

p = &b;
*p = a;
return 0;
}

Is there any problem of the code in C?
No.
In statement of "*p = a;", is just value of 'a' copied to the memory address
of 'p'
it isn't stored to the memory address *of* 'p' but to the memory
address *contained* in p, i.e. the value of 'a' is stored in fact to
'b'.
or 'p' re-pointed to address of 'a'?

No.

*p = a;

stores the value of 'a' to the object 'p' is pointing to - i.e. to 'b'
- and

p = &a;

would re-point 'p' to 'a'. It stores the address of 'a' to 'p'.

--
Horst

[E. Robert Tisdale]

Garma wrote:

In this code,

int main(int argc, char* argv[]){
int a = 47;
int b;
int* p = &b; // p <-- address(b)
*p = a; // b <-- a
return 0;
}

Is there any problem of the code in C?
No.
In the statement "*p = a;",
is the value of 'a' just copied to the memory address of 'p'?
No.
*p is a reference to b -- *p is another name for b.

*p = a;

is the same as

b = a;
Or 'p' re-pointed to address of 'a'?

No. You would write

p = &a;

to "reseat" pointer p so that it points to a.

You must write

p = (int*)a;

to assign p the value of a (47);

[Jack Klein]

On Thu, 15 Jan 2004 16:23:17 -0800, "E. Robert Tisdale"
<E.**************@jpl.nasa.gov> wrote in comp.lang.c:

Garma wrote:

In this code,

int main(int argc, char* argv[]){
int a = 47;
int b;
int* p = &b; // p <-- address(b)
*p = a; // b <-- a
return 0;
}

Is there any problem of the code in C?
No.

In the statement "*p = a;",
is the value of 'a' just copied to the memory address of 'p'?

No.
*p is a reference to b -- *p is another name for b.

No, it is not a reference to b, it is a pointer that currently points
to b. There is no such thing as a reference in the C language.
*p = a;

is the same as

b = a;

No, it has the same effect as "b = a;". The execution might or might
not be the same, only the result is.

Or 'p' re-pointed to address of 'a'?

No. You would write

p = &a;

to "reseat" pointer p so that it points to a.

You must write

p = (int*)a;

to assign p the value of a (47);

Where does it say that such an assignment, with a cast, will result in
a pointer having the value of 47? The result of such an assignment is
implementation-defined, but might well involve a completely different
bit pattern. And of course any attempt to use the pointer results in
undefined behavior.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.learn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Barry Schwarz]

On Fri, 16 Jan 2004 00:03:16 GMT, "Garma" <Ga*****@nosysserv.com>
wrote:

In this code,

int main(){
int a = 47;
int b;
int* p;

p = &b;
*p = a;
return 0;
}

Is there any problem of the code in C?
In statement of "*p = a;", is just value of 'a' copied to the memory address
of 'p' or 'p' re-pointed to address of 'a'?

Neither. The value of a is copied to the area pointed to by p, which
in this case is b. The value stored in p is unchanged.

If you wanted p to point to a, then you would need p = &a;

If you wanted the value of a to be copied to the memory address of p,
that is to be placed in the memory that p occupies, you would need
p = (int*)a; but this is not guaranteed to work if 47 is not a valid
address for an int.
<<Remove the del for email>>

[E. Robert Tisdale]

Jack Klein wrote:

There is no such thing as a reference in the C language.

Tell that to Brian W. Kernighan and Dennis M. Ritchie,
"The C Programming Language", Chapter 5: Pointers and Arrays,
Section 1: Pointers and Addresses, page 90-1:

"Pointer references can also occur on the left side of assignments.
If px points to x, then

*px = 0

sets x to zero, and

*px += 1

increments it, as does

(*px)++

. . ."

[pete]

E. Robert Tisdale wrote:

Jack Klein wrote:

There is no such thing as a reference in the C language.
Tell that to Brian W. Kernighan and Dennis M. Ritchie,

It's too late now.
"The C Programming Language", Chapter 5: Pointers and Arrays,
Section 1: Pointers and Addresses, page 90-1: "Pointer references can also occur on the left side of assignments.

If it would have been in K&R2,
it would have been on page 94, but it's not.

--
pete

[Richard Bos]

Jack Klein <ja*******@spamcop.net> wrote:

On Thu, 15 Jan 2004 16:23:17 -0800, "E. Robert Tisdale"
<E.**************@jpl.nasa.gov> wrote in comp.lang.c:

You must write

p = (int*)a;

to assign p the value of a (47);
Where does it say that such an assignment, with a cast, will result in
a pointer having the value of 47? The result of such an assignment is
implementation-defined, but might well involve a completely different
bit pattern.

The bit pattern is immaterial; the same thing is true for

float f;
int a=47;

f=a;

In fact, the quoted line does assign to p, not exactly 47, but perhaps a
better phrasing is "that pointer value which has some relation to the
value 47". It's not quite the same thing as the int 47; but it should be
semantically related, if at all possible on the underlying architecture.
And of course any attempt to use the pointer results in
undefined behavior.

Of course.

Richard

[Richard Heathfield]

Jack Klein wrote:

On Thu, 15 Jan 2004 16:23:17 -0800, "E. Robert Tisdale"
<E.**************@jpl.nasa.gov> wrote in comp.lang.c:

*p is a reference to b -- *p is another name for b.

Tisdale is of course wrong about *p being a reference to b. If anything, it
is p, not *p, that refers to b. See below.
No, it is not a reference to b, it is a pointer that currently points
to b. There is no such thing as a reference in the C language.

Counter-example from 6.2.5 of C99:

"A pointer type may be derived from a function type, an object type, or an
incomplete type, called the referenced type. A pointer type describes an
object whose value provides a reference to an entity of the referenced
type."

<snip>

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Joe Wright]

Richard Heathfield wrote:

Jack Klein wrote:

On Thu, 15 Jan 2004 16:23:17 -0800, "E. Robert Tisdale"
<E.**************@jpl.nasa.gov> wrote in comp.lang.c:

*p is a reference to b -- *p is another name for b.

Tisdale is of course wrong about *p being a reference to b. If anything, it
is p, not *p, that refers to b. See below.

No, it is not a reference to b, it is a pointer that currently points
to b. There is no such thing as a reference in the C language.

Wrong. I think you mean "pass by reference" of variables as function
parameters. It is clear that pointers 'reference' other objects.
Counter-example from 6.2.5 of C99:

"A pointer type may be derived from a function type, an object type, or an
incomplete type, called the referenced type. A pointer type describes an
object whose value provides a reference to an entity of the referenced
type."

See I told you. Pointers are objects! Where did I put that old thread?
:-)
--
Joe Wright http://www.jw-wright.com
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---