Re: Noob Q: How to properly type cast in this case
- From: jt@xxxxxxxxxxx (Jens Thoms Toerring)
- Date: 29 May 2008 20:36:08 GMT
* Tong * <sun_tong_001@xxxxxxxxxxxxxxxxxxxxx> wrote:
I am trying to compile a big program, in which the following line gives me
the "lvalue required" error:
(Small_Int *)markedplace->data.dp = p;
The following is what I stripped out for the illustration purpose:
$ cat -n type_cast.c
1 #include <stdio.h>
2
3 typedef unsigned char Byte;
4 typedef signed char Small_Int;
5
6 typedef struct _list
7 {
8 struct _list *np; /* Pointer to next */
9 struct _list *lp; /* Pointer to last */
10 Byte type;
11 union {
12 void *dp; /* Pointer to data */
13
14 /* These fields used for a small amount of data */
15 struct {
16 Byte d1;
17 Byte d2;
18 Byte d3;
19 Byte d4;
20 } bytes;
21
22 }data;
23
24 } List;
25
26 List *markedplace;
27 Small_Int i;
28
29 main(int argc,char * argv[])
30 {
31 markedplace->data.dp = NULL;
32 markedplace->data.dp = &i;
33 (Small_Int *)markedplace->data.dp = &i;
34 }
$ gcc type_cast.c
type_cast.c: In function 'main':
type_cast.c:33: error: lvalue required as left operand of assignment
So we don't need to type cast for void pointers?
Every pointer to a object (but not functions) can be converted
to a void pointer and back again without problems (and even
without an explicit cast). The assignment
markedplace->data.dp = &i;
implicitely converts the value of the right side (which is a pointer
to a 'Small_Int') to a void pointer as if you had written explicitely
markedplace->data.dp = ( void * ) &i;
But you can't cast the left hand side. It stays a void pointer
whatever you do. If it would work otherwise why shouldn't also
e.g.
int x = 13;
( double ) x = 3.1419265359;
work in changing 'x', so it suddenly has a type of double?
Actually, the cast on the left hand side takes the value
stored in 'x' and converts that value to a double. And after
this operation the compiler suddenly sees
13.0 = 3.1419265359;
and complains rightly about "lvalue required as left operand
of assignment" since now you have something on the left hand
side that isn't something a value could be assigned to since
it's already a value.
Line 32 was actually how I
fixed it, but I want to know how to properly type cast if I have to.
By only casting values, not trying to "cast" objects. If you
do e.g.
int x;
double pi = 3.1419265359;
x = ( int ) pi;
you don't cast the variable 'pi', you only cast the value that
is stored in the 'pi'. So it's a valid cast since it doesn't try
to change anything about the type of 'pi'. And if you do e.g.
int i;
void *v = &i;
* ( int * ) v = 42;
then again you cast the value stored in 'v' (not 'v' itself) to
an int pointer and then write 42 to that address. As you see, a
cast can also appear on the left hand side, but only if a value
is acceptable at that place on the left hand side.
Regards, Jens
--
\ Jens Thoms Toerring ___ jt@xxxxxxxxxxx
\__________________________ http://toerring.de
.
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