assign address to pointer array
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Peter Humphre
#1 / 10
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 assign address to pointer array I want to assign the address of a thing to an array of (addresses of
things), along the lines of:
#include <malloc.h>
int main(){
int *x;
int y=64;
x = (int*)malloc((size_t)3*sizeof(int));
x[1] = &y;
return 0;
Quote: }
but, to my understanding, x[1] is equivalent to *(x+1). So I need to
something like (x+1) = &y, but (x+1) isn't an l-value.
Obviously doing something stupid...
Thanks
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| Sun, 07 Dec 2003 11:57:14 GMT |
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Jack Klei
#2 / 10
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assign address to pointer array On Wed, 20 Jun 2001 03:57:14 GMT, "Peter Humphrey"
Quote: > I want to assign the address of a thing to an array of (addresses of > things), along the lines of: > #include <malloc.h>
<stdlib.h> for 12 years now.
Quote: > int main(){
> int *x;
> int y=64;
> x = (int*)malloc((size_t)3*sizeof(int));
Don't cast the return value of malloc() in C. The language does not
require it and it can hide errors. The cast to size_t is totally
unnecessary for several reasons. The first is that the prototype
tells the compiler that the parameter is a size_t so it will perform
the conversion automatically. The second is that the sizeof operator
always yields a size_t value anyway.
The best way to call malloc() (after including the proper <stdilb.h>)
is:
x = malloc(3 * sizeof *x);
Quote: > x[1] = &y; > return 0; > } > but, to my understanding, x[1] is equivalent to *(x+1). So I need to
> something like (x+1) = &y, but (x+1) isn't an l-value.
> Obviously doing something stupid...
> Thanks
have performed this assignment, *x[1] will be the value stored in y.
--
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++ ftp://snurse-l.org/pub/acllc-c++/faq
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| Sun, 07 Dec 2003 12:13:03 GMT |
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Peter Humphre
#3 / 10
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assign address to pointer array Jack
Thanks very much for reply. I'm mixing up c and c++ code. Sorry for
confusing the issue.
If I now compile the following code:
#include <stdlib.h>
int main(void){
int *x;
int y=64, z;
x = malloc(3*sizeof(int*));
x[1] = &y; /* line 6 */
z=*x[1]; /* line 7 */
return 0;
Quote: }
line 6 "int ' differs in levels of indirection from 'int *'"
line 7 "illegal indirection"
Thans again
Quote: > On Wed, 20 Jun 2001 03:57:14 GMT, "Peter Humphrey"
> > I want to assign the address of a thing to an array of (addresses of
> > things), along the lines of:
> > #include <malloc.h>
> There is no header named malloc.h in C. malloc() is prototyped in
> <stdlib.h> for 12 years now.
> > int main(){
> > int *x;
> > int y=64;
> > x = (int*)malloc((size_t)3*sizeof(int));
> Don't cast the return value of malloc() in C. The language does not
> require it and it can hide errors. The cast to size_t is totally
> unnecessary for several reasons. The first is that the prototype
> tells the compiler that the parameter is a size_t so it will perform
> the conversion automatically. The second is that the sizeof operator
> always yields a size_t value anyway.
> The best way to call malloc() (after including the proper <stdilb.h>)
> is:
> x = malloc(3 * sizeof *x);
> > x[1] = &y;
> > return 0;
> > }
> > but, to my understanding, x[1] is equivalent to *(x+1). So I need to
> > something like (x+1) = &y, but (x+1) isn't an l-value.
> > Obviously doing something stupid...
> > Thanks
> Why do you think that x[1] = &y; does not do what you want? Once you
> have performed this assignment, *x[1] will be the value stored in y.
> --
> 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++ ftp://snurse-l.org/pub/acllc-c++/faq
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| Sun, 07 Dec 2003 12:43:17 GMT |
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Jack Klei
#4 / 10
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assign address to pointer array On Wed, 20 Jun 2001 04:43:17 GMT, "Peter Humphrey"
Quote: > Jack
> Thanks very much for reply. I'm mixing up c and c++ code. Sorry for
> confusing the issue.
That's OK, but there is no malloc.h header in C++ either. In C++
malloc() is prototyped in <stdlib.h> or in newer ISO compilers
<cstdlib> could be included instead. In either case you do need the
cast in C++.
Quote: > If I now compile the following code:
Sorry, my mistake. Must be time to get some sleep. Let's try
again...
Quote: > #include <stdlib.h>
> int main(void){
> int *x;
This should be:
int **x;
...because x will point to an array of pointers to int. Therefore it
must be a pointer to pointer.
Quote: > int y=64, z;
> x = malloc(3*sizeof(int*));
This is still better written as:
x = malloc(3 * sizeof *x);
That way if you later decide you want to store pointers to double
instead of pointers to int, you only have to change the definition of
x from int **x to double **x, you don't have to hunt down and change
all the memory allocation calls.
Quote: > x[1] = &y; /* line 6 */ > z=*x[1]; /* line 7 */ > return 0; > } > line 6 "int ' differs in levels of indirection from 'int *'"
> line 7 "illegal indirection"
> Thans again
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++ ftp://snurse-l.org/pub/acllc-c++/faq
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| Sun, 07 Dec 2003 12:53:33 GMT |
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#5 / 10
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assign address to pointer array
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| Wed, 18 Jun 1902 08:00:00 GMT |
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Peter Humphre
#6 / 10
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assign address to pointer array that's what I was missing. Thanks very much for your help....much
appreciated.
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| Sun, 07 Dec 2003 13:22:57 GMT |
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#7 / 10
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assign address to pointer array
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| Wed, 18 Jun 1902 08:00:00 GMT |
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Arthur H. Gol
#8 / 10
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assign address to pointer array Quote:
> Jack
> Thanks very much for reply. I'm mixing up c and c++ code. Sorry for
> confusing the issue.
> If I now compile the following code:
> #include <stdlib.h>
> int main(void){
> int *x;
ITYM int **x; It's supposed to be a pointer to a bunch of int *'s, not a
bunch of ints.
Quote: > int y=64, z;
> x = malloc(3*sizeof(int*));
You should always check the return value of malloc() [but you know that,
of course ;-)]
Quote: > x[1] = &y; /* line 6 */ > z=*x[1]; /* line 7 */ > return 0; > } > line 6 "int ' differs in levels of indirection from 'int *'"
> line 7 "illegal indirection"
i), so in order to store an int *, x must be of type int **, for
example.
HTH,
--ag
Quote:
> > On Wed, 20 Jun 2001 03:57:14 GMT, "Peter Humphrey"
> > > I want to assign the address of a thing to an array of (addresses of
> > > things), along the lines of:
> > > #include <malloc.h>
> > There is no header named malloc.h in C. malloc() is prototyped in
> > <stdlib.h> for 12 years now.
> > > int main(){
> > > int *x;
> > > int y=64;
> > > x = (int*)malloc((size_t)3*sizeof(int));
> > Don't cast the return value of malloc() in C. The language does not
> > require it and it can hide errors. The cast to size_t is totally
> > unnecessary for several reasons. The first is that the prototype
> > tells the compiler that the parameter is a size_t so it will perform
> > the conversion automatically. The second is that the sizeof operator
> > always yields a size_t value anyway.
> > The best way to call malloc() (after including the proper <stdilb.h>)
> > is:
> > x = malloc(3 * sizeof *x);
> > > x[1] = &y;
> > > return 0;
> > > }
> > > but, to my understanding, x[1] is equivalent to *(x+1). So I need to
> > > something like (x+1) = &y, but (x+1) isn't an l-value.
> > > Obviously doing something stupid...
> > > Thanks
> > Why do you think that x[1] = &y; does not do what you want? Once you
> > have performed this assignment, *x[1] will be the value stored in y.
Artie Gold, Austin, TX (finger the cs.utexas.edu account for more info)
--
I am looking for work. Contact me.
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| Sun, 07 Dec 2003 12:58:19 GMT |
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Matt Gessne
#9 / 10
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assign address to pointer array Quote:
> I want to assign the address of a thing to an array of (addresses of
> things), along the lines of:
> #include <malloc.h>
> int main(){
> int *x;
> int y=64;
> x = (int*)malloc((size_t)3*sizeof(int));
> x[1] = &y;
> return 0;
> }
> but, to my understanding, x[1] is equivalent to *(x+1). So I need to
> something like (x+1) = &y, but (x+1) isn't an l-value.
> Obviously doing something stupid...
> Thanks
If you're going to do that, you'd need
*(x+1)
And don't typecast malloc...
Leave it be void * as it's declared.
HTH
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| Sat, 13 Dec 2003 23:21:30 GMT |
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Martin Ambuh
#10 / 10
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assign address to pointer array Quote:
> I want to assign the address of a thing to an array of (addresses of
> things), along the lines of:
For your consideration:
#if 0
/* mha - The OP used the following, an inclusion of a non-standard
header. A correction follows. */
#include <malloc.h>
#endif
#include <stdlib.h> /* mha */
int main()
{
#if 0
/* mha - The OP used the following, a declaration of a
pointer-to-int, yet claimed he wanted "an array of (addresses
of things)." A correction follows. */
int *x;
#endif
int **x; /* mha */
int y = 64;
#if 0
/* mha - The OP used the following clutter. A replacement follows,
retaining the OP's magic number of 3, however */
x = (int *)malloc((size_t) 3 * sizeof(int));
#endif
x = malloc(3 * sizeof *x); /* mha */
/* mha - The OP did not check for success of malloc(). A
simple-minded check is added below. */
if (!x)
exit(EXIT_FAILURE); /* mha */
x[1] = &y;
/* mha - The OP did not free the allocated memory. Added below. */
free(x); /* mha */
return 0;
Quote: }
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| Sun, 14 Dec 2003 01:09:40 GMT |
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