03-06-2009
C dynamic pointer
Hi,
Can anyone tell me how i can declare and allocate dynamically an array of pointers to structured type?? Is declaration something like this:?
struct_name ** array;
Last edited by littleboyblu; 03-06-2009 at 12:00 PM..
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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)