Format
#include <stdlib.h>
void *calloc(size_t num, size_t size);
Language Level: ANSI
Threadsafe: Yes.
Description
The calloc() function reserves storage space for an array of num elements, each of length size bytes. The calloc() function then gives all the bits of each element an initial value of 0.
Return Value
The calloc() function returns a pointer to the reserved space. The storage space to which the return value points is suitably aligned for storage of any type of object. To get a pointer to a type, use a type cast on the return value. The return value is NULL if there is not enough storage, or if num or size is 0.
For more information about teraspace storage, see the ILE Concepts manual.
Example that uses calloc()
This example prompts for the number of array entries required, and then reserves enough space in storage for the entries. If calloc() is successful, the example prints out each entry; otherwise, it prints out an error.
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
long * array; /* start of the array
*/
long * index; /* index variable
*/
int i; /* index variable
*/
int num; /* number of entries of the array
*/
printf( "Enter the size of the array\n" );
scanf( "%i", &num);
/* allocate num entries */
if ( (index = array = (long *) calloc( num, sizeof( long ))) != NULL )
{
for ( i = 0; i < num; ++i ) /* put values in arr */
*index++ = i; /* using pointer no */
for ( i = 0; i < num; ++i ) /* print the array out */
printf( "array[%i ] = %i\n", i, array[i] );
}
else
{ /* out of storage */
perror( "Out of storage" );
abort();
}
}
/****************** Output should be similar to: **********************
Enter the size of the array
array[ 0 ] = 0
array[ 1 ] = 1
array[ 2 ] = 2
*/
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