a simple struct code failing

[saurabh]

#include<stdio.h>
#include<stdlib.h>

Hi gurus,
I am trying to compile the following code ,And I am getting an error on the
line labeled /*This one*/.
The error says,
"expected expression before 'node'"
Can someone please explain me the reason?
struct node
{
int data;
struct node* next;

};

typedef struct node node;
node * make_n(node * nodeptr,int n)
{
int i=0;
node* newnode=NULL;
/*Please check the following line*/
nodeptr=(node*)malloc(sizeof node);/*This one*/
nodeptr->next=NULL;
nodeptr->data=1;
node * firstnode=nodeptr;
for(i=2;i<n+1;i++)
{
newnode=(node*)malloc(sizeof node);
newnode->data=(2*i);
newnode->next=NULL;
nodeptr->next=newnode;
nodeptr=newnode;

}
return firstnode;
}

[Bartc]

"saurabh" <sa*****@nowhere.nodomainwrote in message
news:ga**********@aioe.org...

#include<stdio.h>
#include<stdlib.h>

Hi gurus,
I am trying to compile the following code ,And I am getting an error on
the
line labeled /*This one*/.
The error says,
"expected expression before 'node'"
Can someone please explain me the reason?

nodeptr=(node*)malloc(sizeof node);/*This one*/

Change sizeof node to sizeof (node). And the same a few lines further down.
I think sizeof expects parentheses around type arguments.

--
Bartc

[John Bode]

On Sep 12, 12:12 pm, saurabh <[e-mail]:

#include<stdio.h>
#include<stdlib.h>

Hi gurus,
I am trying to compile the following code ,And I am getting an error on the
line labeled /*This one*/.
The error says,
"expected expression before 'node'"
Can someone please explain me the reason?
struct node
{
int data;
struct node* next;

};

typedef struct node node;

node * make_n(node * nodeptr,int n)
{
int i=0;
node* newnode=NULL;
/*Please check the following line*/
nodeptr=(node*)malloc(sizeof node);/*This one*/

When using the sizeof operator on a type name, the type name must be
enclosed in parentheses:

nodeptr = malloc(sizeof (node));

Personally, I prefer calling sizeof on the object being allocated
(note that the name of the object is *not* in parentheses):

nodeptr = malloc(sizeof *nodeptr);

It's a bit more straightforward to me.

[snip rest]

[Keith Thompson]

saurabh <sa*****@nowhere.nodomainwrites:
[...]

I am trying to compile the following code ,And I am getting an error on the
line labeled /*This one*/.
The error says,
"expected expression before 'node'"
Can someone please explain me the reason?
struct node
{
int data;
struct node* next;

};

typedef struct node node;

[...]

nodeptr=(node*)malloc(sizeof node);/*This one*/

[...]

The sizeof operator has two forms: "sizeof expr" and "sizeof (type)".
Since node is a type name, "sizeof node" needs to be "sizeof (node)".

But there's a better way to write that entire line:

nodeptr = malloc(sizeof *nodeptr);

Casting the result of malloc is unnecessary, and can mask errors in
some cases. And by using "sizeof *nodeptr", the line remains correct
even if nodeptr is later changed to some other type. (Don't worry
about dereferencing an uninitialized pointer; the argument of sizeof
isn't evaluated. (There are exceptions for variable-length arrays;
that needn't concern you here.))

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[viza]

On Fri, 12 Sep 2008 19:12:36 +0200, saurabh wrote:

Hi gurus,

Don't assume that anyone who posts here is an expert!

I am trying to compile the following code ,And I am getting an error on

node* newnode=NULL;
nodeptr=(node*)malloc(sizeof node);

You seem to have not declared nodeptr, but that is not the error you are
asking about:

sizeof works in two ways:

sizeof(type)
sizeof object

eg:
sizeof(int)

or:
int foo;
sizeof foo;

You can include extra parentheses in the second type, but you cannot omit
them in the first type.

HTH
viza

[Keith Thompson]

viza <to******@gm-il.com.obviouschange.invalidwrites:

On Fri, 12 Sep 2008 19:12:36 +0200, saurabh wrote:

[...]

> node* newnode=NULL;
nodeptr=(node*)malloc(sizeof node);

You seem to have not declared nodeptr, but that is not the error you are
asking about:

sizeof works in two ways:

sizeof(type)
sizeof object

eg:
sizeof(int)

or:
int foo;
sizeof foo;

[...]
Correction: it's
sizeof(type)
sizeof expression
(Actually in the second case the expression has to be in a particular
subset of expressions called a "unary-expression".)

Of course the expression can be the name of an object, but it doesn't
have to be; ``sizeof 42'' is equivalent to ``sizeof int''.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Bartc]

"viza" <to******@gm-il.com.obviouschange.invalidwrote in message
news:dY*******************@newsfe10.ams2...

On Fri, 12 Sep 2008 19:12:36 +0200, saurabh wrote:

>Hi gurus,

Don't assume that anyone who posts here is an expert!

You seem to have not declared nodeptr, but that is not the error you are
asking about:

You mean, the 'nodeptr' declared in this line:

>node * make_n(node * nodeptr,int n)

--
Bartc

[Ralf Damaschke]

Keith Thompson wrote:

Correction: it's
sizeof(type)
sizeof expression
(Actually in the second case the expression has to be in a particular
subset of expressions called a "unary-expression".)

Of course the expression can be the name of an object, but it doesn't
have to be; ``sizeof 42'' is equivalent to ``sizeof int''.

Aehem, they differ in that the first expression is legal, the second not.

-- Ralf

[viza]

On Fri, 12 Sep 2008 19:45:10 +0000, Bartc wrote:

"viza" <to******@gm-il.com.obviouschange.invalidwrote

>On Fri, 12 Sep 2008 19:12:36 +0200, saurabh wrote:

>>Hi gurus,

Don't assume that anyone who posts here is an expert!

>You seem to have not declared nodeptr, but that is not the error you
are asking about:

You mean, the 'nodeptr' declared in this line:

>>node * make_n(node * nodeptr,int n)

Sigh. All in all a worthwhile post there then...

[Keith Thompson]

Ralf Damaschke <rw****@gmx.dewrites:

Keith Thompson wrote:

>Correction: it's
sizeof(type)
sizeof expression
(Actually in the second case the expression has to be in a particular
subset of expressions called a "unary-expression".)

Of course the expression can be the name of an object, but it doesn't
have to be; ``sizeof 42'' is equivalent to ``sizeof int''.

Aehem, they differ in that the first expression is legal, the second not.

D'oh!

``sizeof 42'' is equivalent to ``sizeof(int)''.

But hey, I *could* have preceded it with ``#define int (int)''. 8-)}

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[CBFalconer]

Bartc wrote:

"saurabh" <sa*****@nowhere.nodomainwrote in message

>#include<stdio.h>
#include<stdlib.h>

I am trying to compile the following code ,And I am getting an
error on the line labeled /*This one*/. The error says,
"expected expression before 'node'"
Can someone please explain me the reason?

nodeptr=(node*)malloc(sizeof node);/*This one*/

Change sizeof node to sizeof (node). And the same a few lines further down.
I think sizeof expects parentheses around type arguments.

And after that fix your code. Basically you are confusing the use
of nodeptr and newnode. Also don't cast the returned value from
malloc.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.

[CBFalconer]

viza wrote:

saurabh wrote:

>I am trying to compile the following code ,And I am getting an

> node* newnode=NULL;
nodeptr=(node*)malloc(sizeof node);

You seem to have not declared nodeptr, but that is not the error
you are asking about:

Yes he has declared it. Look in the parameter list. Parameters
are basically initialized local variables.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.

[saurabh]

On Fri, 12 Sep 2008 19:49:43 -0400, CBFalconer wrote:

viza wrote:

>saurabh wrote:

>>I am trying to compile the following code ,And I am getting an

>> node* newnode=NULL;
nodeptr=(node*)malloc(sizeof node);

You seem to have not declared nodeptr, but that is not the error you
are asking about:

Yes he has declared it. Look in the parameter list. Parameters are
basically initialized local variables.

Thanks a lot guys,I understood this.