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Top Forums Programming sizeof an array of structure without using 'sizeof' operator Post 302302542 by rvan on Tuesday 31st of March 2009 09:13:35 AM
Old 03-31-2009
fine. thank you.
#include <stdio.h>

#pragma pack(1)

main()
{

char begin;
struct ptr
{
int a;
char b;
int *p;
}abc;
char end;

printf("Using sizeof operator = %u\n", sizeof(abc));

printf("Begin address = %u \n",&begin);
printf("end address = %u \n",&end);
printf("first element in struct address = %u \n",&abc.a);
printf("sizeof the structure = %u \n",(&begin > &end) ? (&begin - &end - 1) : (&end - &begin - 1));

}
The sizeof the structure using 'sizeof' operator is 9 as we are using #pragma pack(1)

But if you use the (&begin - &end - 1) logic the size is 15 which is wrong. It is because the
address of end variable is say 1245036
and the address of the first element in the structure is 1245040
and the begin variable address is 1245052
.

The first element of the structure is in the address which is a multiple of 4 and within structure we have a 'char' which is not padded and the int * starts immediately after that. but once it comes out of the structure again the 'begin' variable is in multiple of 4. Does it mean #pragma pack(1) packs only the variable inside the structure to avoid structure padding? if it is so then why when we declare
#include <stdio.h>
#pragma pack(1)
struct array
{
int a;
char c;
int d;
int e;
int f;
int h;
};

int main()
{
char begin;
struct array a[100];
char end;
printf("Size of structure = %u ", (&begin > &end) ? (&begin - &end - 1)/100 : (&end - &begin - 1)/100);
}

Here both sizeof operator and the program logic gives the same output 21 which is correct. Am not clear why it is not behaving properly in the below mentioned code. Could someone explain it to me?

main()
{

char begin;
struct ptr
{
int a;
char b;
int *p;
}abc;
char end;
 

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array(3)						     Library Functions Manual							  array(3)

NAME
array - The array library interface SYNTAX
#include <array.h> DESCRIPTION
An allocated array variable keeps track of o a (nonzero) pointer to a dynamically allocated region of memory; o the number of bytes allocated (always positive); and o the number of bytes initialized (between 0 and the number of bytes allocated). There are two other possibilities for the state of an array variable: unallocated and failed. In both cases, there is no dynamically allo- cated region of memory. A new array variable is normally created as a static variable: #include "array.h" static array x; At this point it is unallocated. The array library provides various allocation and inspection functions. A new array variable can also be created dynamically. It must be initialized to all-0, meaning unallocated, before it is given to any of the array functions. It must be returned to the unallocated (or failed) state, for example with array_reset, before it is destroyed. These rules prevent all memory leaks. Expansion and inspection array x; t* p1 = array_allocate(&x,sizeof(t),pos); t* p2 = array_get(&x,sizeof(t),pos); t* p3 = array_start(&x); int64 len = array_length(&x,sizeof(t)); int64 bytes = array_bytes(&x); Truncation and deallocation array x; array_truncate(&x,sizeof(t),len); array_trunc(&x); array_reset(&x); array_fail(&x); Comparison array x; array y; if (array_equal(&x,&y)) /* arrays are equal... */ Concatenation array x; array y; array_cat(&x,&y); array_catb(&x,"fnord",5); array_cats(&x,"fnord"); array_cats0(&x,"fnord"); /* also append the */ array_cat0(&x); /* append */ array_cate(&x,"fnord",1,4); /* append "nor" */ ORIGINAL API DEFINITION
http://cr.yp.to/lib/array.html SEE ALSO
array_get(3), array_start(3), array_fail(3) array(3)
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