help:question about sizeof?

[guoliang]

help:

i try the code:
"
char str[]={'h','e','l','l','o','\n','\0'};
char *pstr=str;
printf("sizeof=%d\n",sizeof(str));
printf("sizeof=%d\n",sizeof(pstr)); //i want to get sizeof(str) not
sizeof(void *)

"
how can i use pstr to get sizeof string which it point?

thx!

[Peter Nilsson]

"guoliang" <[e-mail]:

help:

i try the code:
"
char str[]={'h','e','l','l','o','\n','\0'};

char str[] = "Hello\n";

char *pstr=str;
printf("sizeof=%d\n",sizeof(str));

%d will tell printf to expect an int. You supply a size_t
which is unlikely to be a type that promotes to int.

printf("sizeof = %u\n", (int) sizeof(str));

printf("sizeof=%d\n",sizeof(pstr)); //i want to get
sizeof(str) not sizeof(void *)

I suspect you meant sizeof(char *), although void and character
pointers will have the same size.

"
how can i use pstr to get sizeof string which it point?

You can't.

http://c-faq.com/malloc/sizeof.html

--
Peter

[seni.yin]

fuction "strlen()"

[Martin Ambuhl]

guoliang wrote:

help:

i try the code:
"
char str[]={'h','e','l','l','o','\n','\0'};
char *pstr=str;
printf("sizeof=%d\n",sizeof(str));
printf("sizeof=%d\n",sizeof(pstr)); //i want to get sizeof(str) not
sizeof(void *)

"
how can i use pstr to get sizeof string which it point?

Well, the _string_ has a size of strlen(pstr)+1. You cannot use pstr to
get the size of the _array_ to which it points: such information is not
incorporated into the pointer or into the array. In addition to the
obvious, that the sizeof a pointer is just the sizeof a pointer,
remember that sizeof yields a compile-time constant. Since the sizes of
arrays pointed can change during execution, they can hardly be
compile-time constants.

[shaanxxx]

On Apr 13, 10:24 am, Martin Ambuhl <[e-mail]:

guoliang wrote:

help:

i try the code:
"
char str[]={'h','e','l','l','o','\n','\0'};
char *pstr=str;
printf("sizeof=%d\n",sizeof(str));
printf("sizeof=%d\n",sizeof(pstr)); //i want to get sizeof(str) not
sizeof(void *)

"
how can i use pstr to get sizeof string which it point?

Well, the _string_ has a size of strlen(pstr)+1. You cannot use pstr to
get the size of the _array_ to which it points: such information is not
incorporated into the pointer or into the array. In addition to the
obvious, that the sizeof a pointer is just the sizeof a pointer,
remember that sizeof yields a compile-time constant.

above statement hold true for c89. c99 has variable length array and
it needs run sizeof operator for variable length array.

6.5.3.4 The sizeof operator:
EXAMPLE 3 In this example, the size of a variable length array is
computed and returned from a
function:
#include <stddef.h>
size_t fsize3(int n)
{
char b[n+3]; // variable length array
return sizeof b; // execution time sizeof
}

Since the sizes of
arrays pointed can change during execution, they can hardly be
compile-time constants.

[shaanxxx]

On Apr 13, 10:24 am, Martin Ambuhl <[e-mail]:

guoliang wrote:

help:

i try the code:
"
char str[]={'h','e','l','l','o','\n','\0'};
char *pstr=str;
printf("sizeof=%d\n",sizeof(str));
printf("sizeof=%d\n",sizeof(pstr)); //i want to get sizeof(str) not
sizeof(void *)

"
how can i use pstr to get sizeof string which it point?

Well, the _string_ has a size of strlen(pstr)+1. You cannot use pstr to
get the size of the _array_ to which it points: such information is not
incorporated into the pointer or into the array. In addition to the
obvious, that the sizeof a pointer is just the sizeof a pointer,
remember that sizeof yields a compile-time constant.

above statement hold true for c89. c99 has variable length array and
it needs runtime sizeof operator for variable length array.

6.5.3.4 The sizeof operator:
EXAMPLE 3 In this example, the size of a variable length array is
computed and returned from a
function:
#include <stddef.h>
size_t fsize3(int n)
{
char b[n+3]; // variable length array
return sizeof b; // execution time sizeof

Since the sizes of
arrays pointed can change during execution, they can hardly be
compile-time constants.

[JimS]

On 12 Apr 2007 20:28:57 -0700, "Peter Nilsson" <ai***@acay.com.au>
wrote:

>%d will tell printf to expect an int. You supply a size_t
which is unlikely to be a type that promotes to int.

Agree.

printf("sizeof = %u\n", (int) sizeof(str));

Are you sure about that? %u is for unsigned int.

Jim

[Peter Nilsson]

On Apr 13, 4:28 pm, JimS <[e-mail]:

On 12 Apr 2007 20:28:57 -0700, "Peter Nilsson" <[e-mail]>

printf("sizeof = %u\n", (int) sizeof(str));

Are you sure about that? %u is for unsigned int.

True. The cast should have been (unsigned). Thanks.

--
Peter