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adri
#1 / 26
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 Struct & malloc problem : pointer expected Hi,
Hope someone can help solve this. I do not know what is wrong
with the try program below. The compiler indicated error on the
line with scanf statement -> pointer expected !
#include <stdio.h>
#include <stdlib.h>
typedef struct data
{
int x
double y;
} RECORD;
main()
{
int count;
RECORD *p;
p=(RECORD *) malloc(sizeof(RECORD)*200);
for (count=0; count< 200; count++)
{
print("Enter data : ");
scanf("%f",p[count]->y);
}
Quote: }
thx & rgds
adrian
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| Sun, 07 Dec 2003 13:52:39 GMT |
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Zoran Cutur
#2 / 26
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Struct & malloc problem : pointer expected
Quote: > Hi, > Hope someone can help solve this. I do not know what is wrong
> with the try program below. The compiler indicated error on the
> line with scanf statement -> pointer expected !
> #include <stdio.h>
> #include <stdlib.h>
> typedef struct data
> {
> int x
> double y;
> } RECORD;
since it simply hides that the type is a struct. It is
mostly easier to use the type if you know from it's name
what it is.
Quote: > main()
it's called
int main (void)
Quote: > { > int count; > RECORD *p;
struct data *p;
Quote: > p=(RECORD *) malloc(sizeof(RECORD)*200);
p = malloc(200 * sizeof *p);
No need to cast malloc as you have included stdlib.h.
Quote: > for (count=0; count< 200; count++)
> {
> print("Enter data : ");
> scanf("%f",p[count]->y);
that can only be done by passing a reference (a pointer)
to a variable to a function. So you should calculate
the address of p[count]->y so scand knows where in
memory to write data it reads from input.
Also %f specifies that a pointer to a float variable is
passed to scanf rather than a pointer to a double. For
doubles use %lf.
scanf("%lf", &p[count]->y);
Quote: > } > }
> thx & rgds
> adrian
--
"LISP is worth learning for the profound enlightenment experience
you will have when you finally get it; that experience will make you
a better programmer for the rest of your days." -- Eric S. Raymond
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| Sun, 07 Dec 2003 14:48:29 GMT |
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Morris Dove
#3 / 26
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Struct & malloc problem : pointer expected Quote:
> Hi, > Hope someone can help solve this. I do not know what is wrong
> with the try program below. The compiler indicated error on the
> line with scanf statement -> pointer expected !
> #include <stdio.h>
> #include <stdlib.h>
> typedef struct data
> {
> int x
> double y;
> } RECORD;
> main()
> {
> int count;
> RECORD *p;
> p=(RECORD *) malloc(sizeof(RECORD)*200);
> for (count=0; count< 200; count++)
> {
> print("Enter data : ");
> scanf("%f",p[count]->y);
> }
> }
Adrian...
Here's a minimal list of things that need fixing. I'll leave it up
to you to find out why (or ask more specific questions if you
can't):
[1] int main(void)
[2] p = malloc(sizeof(RECORD) * 200);
[3] if (p == NULL) /* Do error recovery here */
[4] scanf("%f",&p[count]->y);
In addition, your use of scanf() is fairly risky practice and makes
your program unreliable. Remember that failure of a compiler to
find errors doesn't mean that the code is good.
--
Morris Dovey
West Des Moines, Iowa USA
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| Sun, 07 Dec 2003 14:54:06 GMT |
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thomas haup
#4 / 26
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Struct & malloc problem : pointer expected Quote:
> Hi, > Hope someone can help solve this. I do not know what is wrong
> with the try program below. The compiler indicated error on the
> line with scanf statement -> pointer expected !
> #include <stdio.h>
> #include <stdlib.h>
> typedef struct data
> {
> int x
> double y;
> } RECORD;
> main()
> {
> int count;
> RECORD *p;
> p=(RECORD *) malloc(sizeof(RECORD)*200);
Fine so far (though I know there's a number of people out there going to beat You up for casting malloc()'s result - but never mind, the result remains correct anyway).
p now
- points to a block of memory large enough to hold 200 RECORDs
- always points to an object of type RECORD and size sizeof(RECORD)
- may be used like an array of 200 records
Since p[i] is equivalent to *(p+i) for any <type> *p (with <type> != void), p[i] will always yield a RECORD , not a RECORD * .
Quote: > for (count=0; count< 200; count++)
> {
> print("Enter data : ");
> scanf("%f",p[count]->y);
1) As stated above, p[count] is an object of type RECORD, which is a structure - not a pointer to a structure. To address its element y, You'll need to use p[count].y.
2) Scanf expects a pointer to the target object as its argument, thus You need to take the address of p[count].y using the prefix &-operator.
So the correct form for this line should be
scanf("%f",&p[count].y);
Quote: Cheers,
Thomas
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| Sun, 07 Dec 2003 13:09:04 GMT |
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Richard B
#5 / 26
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Struct & malloc problem : pointer expected Quote:
> Hope someone can help solve this. I do not know what is wrong > with the try program below. The compiler indicated error on the > line with scanf statement -> pointer expected ! > typedef struct data
> {
> int x
> double y;
> } RECORD;
> RECORD *p;
> p=(RECORD *) malloc(sizeof(RECORD)*200);
Miss the cast, change to sizeof *variable, like this:
p=malloc(200*sizeof *p);
It's more solid that way.
Quote: > scanf("%f",p[count]->y);
p is a pointer to RECORD. p[count], therefore, is a RECORD. A RECORD is
not a pointer. Therefore, you can't apply -> to it. In fact, you
probably meant
scanf("%f",p[count].y);
instead. Note that scanf() is fraught with pitfalls, but can be useful
if you're careful indeed.
Richard
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| Sun, 07 Dec 2003 14:58:10 GMT |
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willem veenhove
#6 / 26
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Struct & malloc problem : pointer expected Quote:
> > scanf("%f",p[count]->y);
> p is a pointer to RECORD. p[count], therefore, is a RECORD.
> A RECORD is not a pointer. Therefore, you can't apply -> to it.
> In fact, you probably meant
> scanf("%f",p[count].y);
> instead.
*You* probably meant:
scanf("%f", &p[count].y);
because scanf does expect a pointer.
You should have a coffee before you start posting ;-)
willem
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| Sun, 07 Dec 2003 15:41:07 GMT |
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Zoran Cutur
#7 / 26
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Struct & malloc problem : pointer expected Once upon a while "Zoran Cutura"
...
Quote: > scanf("%lf", &p[count]->y);
So you have to write
scanf("%lf", &p[count].y);
--
"LISP is worth learning for the profound enlightenment experience
you will have when you finally get it; that experience will make you
a better programmer for the rest of your days." -- Eric S. Raymond
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| Sun, 07 Dec 2003 15:48:31 GMT |
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Des Walke
#8 / 26
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Struct & malloc problem : pointer expected
Quote:
> > typedef struct data
> > {
> > int x
> > double y;
> > } RECORD;
Just pointing out the missing ';' after attribute x in the declaration
of struct data. To OP, your compiler should have warned about this also,
so I guess this is just a typo in the mail message.
Des Walker
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| Sun, 07 Dec 2003 15:24:48 GMT |
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#9 / 26
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Struct & malloc problem : pointer expected
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| Wed, 18 Jun 1902 08:00:00 GMT |
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morgan mair fhe
#10 / 26
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Struct & malloc problem : pointer expected Quote: >The above typedef appears to be useless to many programmers
>since it simply hides that the type is a struct. It is
>mostly easier to use the type if you know from it's name
>what it is.
some of us actually learned how to use abstractions thank you
gosh no typedefs
no #defines
i think you should warn people about the evils of functions
you never know what a function might do
crippled coding standards to cope with crippled interfaces
or you could do your interfaces correctly
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| Sun, 07 Dec 2003 20:21:07 GMT |
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Gergo Bara
#11 / 26
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Struct & malloc problem : pointer expected
Quote: > >The above typedef appears to be useless to many programmers > >since it simply hides that the type is a struct. It is > >mostly easier to use the type if you know from it's name > >what it is. > some of us actually learned how to use abstractions thank you
Gergo
--
Give a woman an inch and she'll park a car in it.
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| Sun, 07 Dec 2003 20:28:22 GMT |
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#12 / 26
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Struct & malloc problem : pointer expected
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| Wed, 18 Jun 1902 08:00:00 GMT |
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Mark A. Odel
#13 / 26
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Struct & malloc problem : pointer expected
Quote: >> typedef struct data { >> int x >> double y; >> } RECORD; > The above typedef appears to be useless to many programmers
> since it simply hides that the type is a struct. It is
> mostly easier to use the type if you know from it's name
> what it is.
good thing. I write drivers almost exclusively. Many devices have registers
that are read-only/write-only at the same address. Before hardware is built,
I like to simulate my code and structures. Thus, for the simulated, in-memory
version of the device's register map is a struct. When I switch to the real
hardware I need only change the typedef from struct to union at a single
point.
E.g.
#ifdef REAL_HARDWARE
typdef union
#else /* SIMULATED HARDWARE */
typdef struct
#endif
{
volatile unsigned char command;
volatile unsigned char status;
Quote: } RegMap;
RegMap *pMap = (RegMap *) SOME_ADDR;
map->command = WRITE_SECTORS;
while (map->status & BSY);
So, for me, hiding struct-ness can be a good thing.
--
Warmest regards.
Optimize only if it runs too slowly or does not fit, spaces instead of tabs.
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| Sun, 07 Dec 2003 20:35:53 GMT |
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blis
#14 / 26
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Struct & malloc problem : pointer expected
Quote:
> > > scanf("%f",p[count]->y);
> > p is a pointer to RECORD. p[count], therefore, is a RECORD.
> > A RECORD is not a pointer. Therefore, you can't apply -> to it.
> > In fact, you probably meant
> > scanf("%f",p[count].y);
> > instead.
> *You* probably meant:
> scanf("%f", &p[count].y);
> because scanf does expect a pointer.
> You should have a coffee before you start posting ;-)
both of **You** probably meant:
scanf("%lf", &p[count].y);
since y is a double
Roger...
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| Sun, 07 Dec 2003 21:30:32 GMT |
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Richard B
#15 / 26
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Struct & malloc problem : pointer expected Quote:
> *You* probably meant: > scanf("%f", &p[count].y);
So I did.
Quote: > You should have a coffee before you start posting ;-)
Most definitely not. But I _should_ have my tea.
Richard
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| Sun, 07 Dec 2003 23:17:13 GMT |
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